Equilibrium - Class 11 Engineering Chemistry - Extra Questions

An endothermic reaction which proceeds with decrease in volume will give maximum yield of the products at high $$P$$ and high $$T$$ .
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An endothermic reaction which proceeds with decrease in volume will give maximum yield of the products at high P and high T .

Since the reaction proceeds with decrease in volume, at high pressure, the equilibrium will shift in the forward direction which will reduce the effect of high temperature. The endothermic reaction absorbs heat. When temperature is high, the equilibrium will shift in the forward direction so as to reduce the effect of high temperature. Hence, under the conditions of high temperature and high pressure, the yield of the reaction will be maximum.


An exothermic reaction which proceeds with decrease in volume will give maximum yield at high $$P$$ and low $$T$$. 
If true enter 1, if false  enter 0.

An exothermic reaction which proceeds with decrease in volume will give maximum yield at high P and 
low T. Since the reaction proceeds with decrease in volume, at high pressure, the equilibrium will shift in the forward direction which will reduce the effect of high temperature. The exothermic reaction proceeds with evolution of heat. 
When temperature is low, the equilibrium will shift in the forward direction so as to reduce the effect of low temperature. Hence, under the conditions of low temperature and high pressure, the yield of the reaction will be maximum.



The solubility of a solute decrease with temperature when dissolution is exothermic.
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The solubility of a solute decrease with temperature when dissolution is exothermic.  
e.g., $$CaCl_2, Ca(OH)_2, (CH_3COO)_2Ca, NaOH, KOH$$ etc.

$$Ca(OH)_2(s) + aq \rightarrow Ca(OH)_2 (aq)  + Q  kcal$$.

Since heat is evolved, with increase in the temperature, the equilibrium shifts in the reverse direction so as to nullify the effect of increased temperature. Hence, the solubility decreases.


Addition of an inert gas at constant volume to the equilibrium mixture does not influence the equilibrium.
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Addition of an inert gas at constant volume to the equilibrium mixture does not influence the equilibrium. At constant volume, the number of moles of reactants and products is same or concentration remain same. Hence, addition of the inert gas will not alter the equilibrium.


For a reversible system at constant temperature, the value of $${K}_{c}$$ increases if the concentrations are changed at equilibrium.
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For a reversible system at constant temperature, the value of $${K}_{c}$$ remains same if the concentrations are changed at equilibrium. However, the equilibrium will shift in either forward direction or in the reverse direction to re-attain the equilibrium.


The degree of dissociation of $$PC{l}_{5}$$ of decreases with increase in pressure.
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The degree of dissociation of $$PC{l}_{5}$$ of decreases with increase in pressure.

$$PCl_5 \rightleftharpoons PCl_3 + Cl_2$$

$$\Delta n = (1+1)-1=-1$$

Thus, there is decrease in the number of moles for the reaction.
With increase in the pressure, the equilibrium will shift in the backward direction so as to nullify the effect of increased pressure. This suppresses the degree of dissociation of phosphorus penta chloride.


When a liquid and its vapours are at equilibrium and the pressure is suddenly decreased, cooling occurs.
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When a liquid and its vapors are at equilibrium and the pressure is suddenly decreased, cooling occurs.

$$liquid \leftrightharpoons vapor$$

When  the pressure is suddenly decreased, the equilibrium will shift to right so as to nullify the effect of decreased pressure. Thus, more liquid evaporates to form vapor. Evaporation is an endothermic process. Heat is absorbed in the process. This results in cooling.


Solubility of $$NaOH$$ increases with increase in temperature.
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Solubility of $$NaOH$$ decreases with increase in temperature.
The dissolution of NaOH in water is exothermic in nature. As the temperature increases, the equilibrium will shift in the backward direction so as to nullify the effect of increased temperature. This will decrease the solubility of NaOH.


The equilibrium constant does not depend on the initial concentration of reactants but depends on concentration of various species at equilibrium .
If true enter 1, if false  enter 0.

The equilibrium constant does not depend on the initial concentration of reactants but depends on concentration  of various species at equilibrium . The equilibrium constant is the ratio between the product of the molar concentrations of the products to that of the product of the molar concentrations of the reactants with each concentration term raised to a power equal to stoichiometric coefficient in the balanced chemical equation.  


Le Chatelier's principle finds extensive application in predicting the conditions for getting maximum yield of products in reactions of industrial importance.
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Le Chatelier's principle finds extensive application in predicting the conditions for getting maximum yield of products in reactions of industrial importance.
The application of Le Chatelier's principle in predicting the conditions for getting maximum yield of products in reactions of industrial importance is demonstrated as below.
Change
Shift of equilibrium to
Addition of reactant
Forward direction
Removal of product
Forward direction
Removal of reactant
Backward direction
Addition of product
Backward direction.

When, the equilibrium shifts in forward direction, the $$K_p$$ value increases. This helps in optimizing the conditions for obtaining maximum yield.
Nature of the reaction
Temperature
Shift of equilibrium to
Exothermic
Increased
backward direction
Exothermic
Decreased
forward direction
Endothermic
Increased
forward direction
Endothermic
Decreased
backward direction

Th unwanted side reactions can be minimized by favoring those conditions which shifts the equilibrium towards backward direction.


High pressure is favourable for those reversible reactions in which there is decrease in the number of molecules.
If true enter 1, if false  enter 0.

High pressure is favorable for those reversible reactions in which there is decrease in the number of molecules. As the pressure is high, there is a tendency to decrease the pressure. Hence, the equilibrium will shift in a direction in which there is a decrease in the number of moles of gaseous species. This will decrease the pressure of the system.


The equilibrium constants for the reaction, $${ N }_{ 2 }(g)+{ O }_{ 2 }(g)\leftrightharpoons 2NO(g)$$ at $${ 1727 }^{ o }C$$ and $${ 2227 }^{ o }C$$ are $$4.08\times { 10 }^{ -4 }$$ and $$3.6\times {10}^{-3}$$ respectively. Calculate the enthalpy change for the reaction. (Given $$R=1.987\ cal\ mol^{-1} K^{-1}$$)

we know the relation between enthalpy change and potential
 $$\Delta S = \dfrac{\Delta nEf }{\Delta T}$$

=$$\dfrac{3.1 \times 10^{-3}}{500}$$

$$6.01 \times 10^{-6}$$
this is answer


The value of equilibrium constant of a reaction depends upon the initial values of concentration of reactants.
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The value of equilibrium constant of a reaction is independent of the initial values of concentration of reactants. The equilibrium constant is the ratio between the product of molar concentrations of the products to that of the product of the molar concentrations of the reactants with each concentration term raised to a power equal to stoichiometric coefficient in the balanced chemical equation at equilibrium. 

For the general chemical reaction $$aA + bB \rightleftharpoons cC +dD$$, the equilibrium constant expression is 

$$K= \frac {K_f}{K_b} = \frac {[C]^c [D]^d}{[A]^a [B]^b }$$

Here,  $$K_f$$ and $$K_b$$ are the velocity constant of forward and backward reactions respectively. Thus, the value of the equilibrium constant is independent of the initial concentrations.


The reaction : $$CaC{ O }_{ 3 }\left( g \right) \rightleftharpoons CaO\left( s \right) +C{ O }_{ 2 }\left( g \right)$$, would proceed to completion if the reaction vessel is connected to bottle containing $$KOH$$ solution or carried out in open vessel.
If true enter 1, if false enter 0.

The reaction $$CaC{ O }_{ 3 }\left( g \right) \rightleftharpoons CaO\left( s \right) +C{ O }_{ 2 }\left( g \right)$$ would proceed to completion if the reaction vessel is connected to bottle containing $$KOH$$ solution or carried out in open vessel. This is an equilibrium reaction. 

If one of the products is continuously removed from the reaction mixture as soon as it is formed, more and more of the reactants will participate in the reaction till the limiting reactant is completely consumed. In the above reaction, the $$CO_2$$ gas is either removed by carrying out the reaction in the open vessel so that entire $$CO_2$$ escapes in the atmosphere or the $$CO_2$$ is dissolved in KOH solution, thus it is removed from the equilibrium mixture. Due to this, the reaction goes to completion.


The dissociation of $$CaC{O}_{3}$$ is suppressed at high pressure.
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The dissociation of $$CaC{O}_{3}$$ is suppressed at high pressure.

$$CaCO_3(s) \rightleftharpoons CaO (s) + CO_2 (g)$$

$$\Delta n = ({0+1})-0=1$$

Thus, the reaction proceeds with increase in the number of moles of gaseous species. At high pressure, the equilibrium will shift in the reverse direction so that the effect of increased pressure is minimized. This suppresses the dissociation.


In an equilibrium reaction, the concentration of reactants decreases initially and then becomes constant.
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In an equilibrium reaction, the concentration of reactants decreases initially and then becomes constant. Initially more and more reactants are converted to products. The rate of forward reaction is much higher than the rate of the backward reaction. Due to this, the concentration of reactants decreases with progress of reaction. This continues till an equilibrium is established at which the rates of forward and backward reactions are equal. Whatever amount of reactants are consumed in the forward reaction, an equivalent amount is formed in the reverse reaction. The concentration of the reactants becomes constant.


In a reversible reaction, some amount of heat energy is liberated in the forward reaction. Name the reaction. What change in temperature favours the forward reactions?

In a reversible reaction, evolution of heat energy indicates that it is an exothermic reaction and $$\Delta H=-ve$$. In exothermic reactions, heat is released with the products. When separated molecules join together, enough energy is released to overcompensate for the energy required to break reactant bonds. 
For example $$A+B\rightleftharpoons C+D+Energy$$ 
Hence, the low temperature favours the formation of products. If the temperature is increased the equilibrium will shift to the left (using Le Chatelier's principle). If temperature is decreased, the reaction will proceed forward to produce more heat (which is lacking). The effect of temperature on equilibrium will also change the value of the equilibrium constant.


What is the effect of pressure on the equilibrium of the nitrogen and oxygen to give nitric oxide?

The reaction has been investigated in the temperature range 223-298 K at apressure of 200 Torr of $$N_2$$ carrier gas. The influence of water vapor has been studied at 298 K. The yield of $$HNO_3$$ was found to increase in the presence of water vapour (by 90% at about 3 Torr of $$H2O$$).
 So accurate pressure increases the formation of Nitric acid or presssure affects the formation of Nitric acid.


State the Le Chatelier's principle?

Explanation

According to Le Chateleir's principle,

If a system's equilibrium is disrupted by a change in one or more of the deciding factors (such as temperature, pressure, or concentration), the system will adjust to a new equilibrium by counteracting the effect of the change as much as possible.

For example: $$ H_2 + I_2 \rightleftharpoons 2 HI$$
If we remove $$H_2$$ from the reaction then to maintain the equilibrium reaction shift backwards that is towards $$H_2$$ to maintain the concentration of $$H_2$$.

The principle of Le Chatelier is an observation concerning reaction chemical equilibria.



State Le Chatelier's principle.

Explanation:

Part 1: Statement:

Le Chatelier’s principle states that “when some changes in external physical conditions like pressure, temperature or concentration of products and reactants is applied to an equilibrium system, the equilibrium will shift in such a way that it can compensate the changes made in the reaction.

Part 2: Reasons:

  • The agents that can change the position of chemical equilibrium in a reaction are temperature, pressure, and concentration of either reactants or products or both.

Part 3:

Effects of concentration change:

  • If the concentration of the reactants is reduced, then the equilibrium position changes to the right and vice versa.
  • If the concentration of products is reduced, then the equilibrium shifts to the left and vice versa.

Effects of temperature change:

  • If the temperature is increased, then the reaction will be favored where the temperature is reduced i.e., the endothermic reaction is favored. Similarly, if the temperature is decreased, then the reaction will favor where the temperature is increased i.e., an exothermic reaction is favored

Effects of pressure change:

  • Likely to that of temperature, if the pressure is increased, then the reaction will favor where the pressure is reduced and if the pressure is decreased, then the reaction will favor where the pressure is increased.


State Le-chatlier's principle.

According to Le-Chatlier's principle, "if a system at equilibrium is subjected to a disturbance or stress, then the equilibrium shifts in the direction that tends to nullify the effect of the disturbance or stress".


Which measurable property becomes constant in water $$\rightleftharpoons $$ water vapour equilibrium at constant temperature?

Vapour pressure
Vapour pressure becomes constant in water-vapour equilibrium.


Variation of equilibrium constant $$K$$ with temperature $$T$$ is given by van't Hoff equation:
$$\log { K } =\log { A } -\cfrac { \Delta { H }^{ o } }{ 2.303RT } $$
A graph between $$\log { K } $$ and $${T}^{-1}$$ was a straight line as shown in the figure and having $$\theta =\tan ^{ -1 }{ (-0.5) } $$ and $$OP=10$$. Calculate:
(a) $$\Delta { H }^{ o }$$ (Standard heat of reaction) when $$T=298K$$
(b) $$A$$ (pre-exponential factor)
(c) Equilibrium constant $$K$$ at $$298K$$
(d) $$K$$ at $$798K$$, if $$\Delta { H }^{ o }$$ is independent of temperature
657997_2c70abb70e474ff9a5291a27f83787ea.png

$$\mathbf{Explanation:}$$

$$\boldsymbol{a)\ log_{10}K=log_{10}A-\frac{\Delta H^{o}}{2.303RT}}$$

It is an equation of a straight line of the $$y=c+mx$$

Slope  $$m\ tan \theta=$$ $$\cfrac{\Delta H^{o}}{2.303R}$$

$$0.5=\cfrac{\Delta H^{0}}{2.303}\times 8.314$$

$$\mathbf{\Delta H^{o}=9.574Jmol^{-1}}$$


$$b)\ Intercept, D^{'}=log_{10}A=10A=10^{10}$$

$$\mathbf{D^{'}=10^{10}}$$


$$c)\ logK=10-\cfrac{9.574}{2.303\times 8.314\times 298}$$

$$\mathbf{K=9.96\times 10^{9}}$$


$$d)\ log\cfrac{K_{2}}{K_{1}}=\cfrac{\Delta H}{20303}\times R(\cfrac{1}{T_{1}}-\cfrac{1}{T_{2}})$$

$$\cfrac{log(K_{2})}{9.96\times 10^{9}}=1.05\times 10^{-3}$$

$$\mathbf{log(K_{2})=9.98\times 10^{6}}$$

$$\mathbf{K_{2}=e^{9.98\times 10^{6}}}$$


Match the reaction equilibrium from List I with the factors which can affect it from List II.

Using of le chatelier principle 
Le Chatlier's principle (also known as "Chatelier's principle" or "The Equilibrium Law") states that when a system experiences a disturbance (such as concentration, temperature, or pressure changes), it will respond to restore a new equilibrium state.
A.  $$\Delta H $$ is 0 so connect with 1 , 3

B.  $$\Delta H $$ is positive  1,4

C. $$\Delta H $$ is negative 1,2 

D. $$\Delta H $$ is positive 1 ,4 


Explain the effect of change in concentrations on the following equillibria:
a) $$PCl_{ 5(g) }\rightleftharpoons PCl_{ 3(g) }{ Cl }_{ 2(g) }$$
b) $$2SO_{ 2(g) }+{ O }_{ 2(g) }\rightleftharpoons 2SO_{ 3(g) }$$


1344272_1131909_ans_15a602ec09344d9b93c8f61a5f1ceb18.jpg


The $$K_p$$ for the reaction; $$H_2+I_2\rightleftharpoons 2HI$$, at $$460^{o}C$$ is $$49$$. If the initial pressure of $$H_2$$ and $$I_2$$ is $$0.5\ atm$$ respectively, determine the partial pressure of each gas at equilibrium.

$$H_2$$$$+H_2$$$$\rightleftharpoons$$$$2HI$$
Initial$$0.5$$$$0.5$$$$0$$
At eqm.$$0.5-x$$$$0.5-x$$$$2x$$
$$K_p=\dfrac{(2x)^2}{(0.5-x)^2}=49$$ or $$\dfrac{2x}{0.5-x}=7$$
or $$2x=3.5-7x$$ or $$9x=3.5$$
or $$x=\dfrac{3.5}{9}=0.39$$
$$\therefore$$ Pressure of $$H_2$$ and $$I_2$$ at eqm. $$=0.5-0.39$$
$$=0.11\ atm$$


Why does a gas fizz when a soda water bottle is opened?

The amount of the gas dissolved is very high due to high pressure. On opening the bottle, the pressure tends to decrease to atmospheric pressure. So the solubility decrease i.e. the dissolved gas escapes out.


(a) What would be the effect of an increase in pressure on following equilibria:
(i) $$CO(g) + 2H_2(g)\rightleftharpoons CH_3OH(g)$$
(ii) $$2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)$$
(iii)$$CO_2(g) + C(s)\rightleftharpoons 2CO(g)$$
(iv)$$COCI_2(g)\leftrightarrow CO(g)+CI_2(g)$$
(b)In above cases if temperature is increased then what should be the direction o f reaction


1359809_1574944_ans_36b9617fab8846fc85bee1effd13ea48.jpg


In a gaseous reaction, $$A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)}$$. Predict the effect of addition of inert gas if addition is made at (a) constant volume, (b) constant pressure.


1760677_1743579_ans_cadc59c378734a1cad07ae0b40c777b8.jpg


How do the following help in bringing about a chemical change?
Pressure 


Some chemical reactions take place when reactants are subjected to high pressure.e.g: Nitrogen and hydrogen when subjected to high pressure produce ammonia gas.

1696882_1803970_ans_e232247f546d43d3b017d92b4a00799b.jpg


What is the effect of increasing pressure on the equilibrium?
$$N_2 + 3H_2 \overset{\rightharpoonup}{\leftharpoondown} \ 2NH_3 \ ?$$

Equilibrium will shift in the forward direction forming more of ammonia.


Name the factors that affect equilibrium position of a reversible reaction.

(i) Temperature (ii) Pressure, (iii) Concentration, (iv) Catalyst.


Under what conditions the synthesis of $$SO_3$$ is more ? Explain. 

$$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$$ 
The conditions for more synthesis of $$SO_3$$ are given as below: 
(i) Increase in concentration of $$SO_2$$ or $$O_2$$: 
If the concentration of $$SO_2$$ or $$O_2$$ will increase, then according to Le-Chatelier's principle, the equilibrium will shift to forward direction and more $$SO_3$$ will be formed. 
(ii) Decrease in temperature: A decrease in temperature favours the exothermic reaction so the equilibrium will shift to forward direction and more amount of $$SO_3$$ will be formed. 
(iii) Increase in Pressure: An increase in pressure will favour the reactions that decrease the number of gaseous molecules. The equilibrium will shift to the right and the yield of $$SO_3$$ will be increased. 
(iv) Removal of $$SO_3$$ from reaction vessel. 


What will be the effect on equilibrium when $$\Delta \ n_g$$ is negative and pressure is decreased? 

When $$\Delta \ n_g$$ is negative and pressure is reduced, the equilibrium will be shifted to left and the yield of product will decrease. 


Which of the following reactions will get affected by increasing the pressure ? Also, mention whether change will cause the reaction to go into forward or backward direction.
(i) $$COCl_2(g) \rightarrow CO(g) + Cl_2(g)$$ 
(ii) $$CH_4(g) + 2H_2S(g) \rightarrow CS_2(g) + 4H_2(g)$$ 
(iii) $$CO_2(g) + C(s) \rightarrow 2CO(g)$$ 
(iv) $$2H_2(g) + CO(g) \rightarrow CH_3OH(g)$$ 
(v) $$CaCO_3(s) \rightarrow  CaO(s) + CO_2(g)$$ 
(vi) $$4NH_3(g) + SO_2(g) \rightarrow 4NO(g) + 6H_2O(g)$$

(i) $$COCl_2(g) \rightarrow CO(g) + Cl_2(g)$$ 
If pressure increases, the reaction will go into backward direction. 
(ii) $$CH_4(g) + 2H_2S(g) \rightarrow CS_2(g) + 4H_2(g)$$
If pressure increases, the reaction will go into backward direction. 
(iii) $$CO_2(g) + C(s) \rightarrow 2CO(g)$$
If the pressure increases, the reaction will go into backward direction. 
(iv) $$2H_2(g) + CO(g) \rightarrow CH_3OH(g)$$ 
If pressure increases, the reaction will go in to forward direction. 
(v) $$CaCO_3(s) \rightarrow  CaO(s) + CO_2(g)$$ 
If pressure increases, the reaction will go in to backward direction. 
(vi) $$4NH_3(g) + SO_2(g) \rightarrow 4NO(g) + 6H_2O(g)$$
If pressure increases, the reaction will go into backward direction. 


For the system, $$N_2 + O_2 \rightarrow  2NO $$, -44 kcal, at equilibrium, what will be the effect of the following: 
(i) Increasing the temperature
(ii) Decreasing the pressure
(iii) Increasing the concentration of NO 
(iv) Presence of catalyst

For the system, 
$$N_2 + O_2 \rightarrow  2NO$$, -44 kcal
At equilibrium
(i) Increasing the temperature : An increase in temperature favours the endothermic reaction because it takes up energy so it will favour the forward reaction and the yield of NO will increase. 
(ii) Decreasing the pressure : Since the number of molecules of reactants and products are same, so decrease in pressure will not affect the equilibrium. 
(iii) Increasing the concentration of NO : On increasing the concentration of NO, it will favour backward reaction and the yield of NO will increase. 
(iv) Presence of catalyst : It increases the rate of reaction by lowering the activation energy. 


Describe the factors which affect chemical equilibrium? 

There are following factors which affect chemical equilibrium : 
(1) Concentration: If at equilibrium, the concentration of reactants is increased, then rate of forward reaction increases and more amount of product is obtained. If there is increase in concentration of products, then rate of backward reaction increases, more amount of reactant is obtained. 
(2) Pressure : The effect of pressure on equilibrium takes place when the reactants are in gaseous state and in those reactions in which number of molecules of reactants and products are different. 
Example: $$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$$ 
                    1 mol    3 mol              2 mol 
In this reaction, the forward reaction is accompanied by a decrease in the total number of moles of reactants. If the pressure of the system is increased, then the equilibrium will shift in that direction where pressure decreases i. e., decrease in number of moles taken i. e., in the favour of formation of ammonia. 
If pressure decreases, the equilibrium will shift in the direction where pressure increases i.e., increase in number of moles occurs i. e., in backward direction. So, a decrease in pressure will favour dissociation of 
$$NH_3$$ to form $$N_2$$ and $$H_2$$. 
When there is no change in number of molecules of reactants and products in a reaction, then there is no effect of pressure on it. 
(3) Temperature: High temperature favors endothermic reaction and low temperature favors exothermic reaction.
(4) Catalyst : A catalyst increases the rate of the chemical reaction by making available a new low energy pathway for the conversion of reactants to products. It increases the rate of forward and reverse reactions that pass through the same transition state and does not affect equilibrium. Catalyst lowers the activation energy for the forward and backward reactions by exactly the same amount. Catalyst does not affect the equilibrium composition of a reaction mixture. It does not appear in the balanced chemical equation or in the equilibrium constant expression. 


Write the examples of the reaction in which
(a) product increases on increasing the pressure. 
(b) product increases on increasing the temperature. 

(a) $$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$$
 
(b) $$N_2O_4(g) \rightleftharpoons 2NO_2(g)$$ 


The equilibrium for the synthesis of ammonia is
$$N_2 + 3H_2 \rightleftharpoons NH_3 + x kJ$$,
What is the effect of pressure, temperature and concentration on this equilibrium? 

Effect of pressure : 
High pressure favours the formation of ammonia. An increase in pressure will favour the reaction that decreases the number of gas molecules. The equilibrium will shift to the right and the yield of ammonia will be increased.
 
Effect of temperature: 
Low temperature favours the formation of ammonia, a decrease in temperature favours the exothermic reaction because it releases energy, it will favour the forward reaction and the yield of ammonia will be increased. 

Effect of concentration: 
High concentration of reactants favours the formation of ammonia. According to Le-Chatelier's principle, if the concentration of $$N_2$$ or $$H_2$$ is increased, then equilibrium will shift to forward direction and the yield of ammonia will be increased. 


What will be the effect of temperature and pressure on the following equilibrium?
$$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$$

Effect of temperature: 
Low temperature favours the forward reaction. A decrease in temperature favours the exothermic reaction because it releases energy. So, it will favour the forward reaction. The yield of ammonia will increase. 

Effect of pressure: 
An increase in pressure will favours the reaction that decreases the number of gaseous molecules. There are fewer molecules of product then reactants so it will favour the forward reaction. The yield of ammonia will increase. 


What is the effect of reducing the volume on the system described below?

On reducing the volume, the pressure will increase. By Le Chatelier's principle, equilibrium will shift to the side accompanied by decrease of pressure i.e. decrease in the number of gaseous moles i.e. backward direction.


$$H_{2}(g)+I_{2}(g)\rightleftharpoons 2HI(g)+$$ Heat
The following circumstances influence the reaction
Increase the temperature.

Increase the rate of backward reaction


$$H_{2}(g)+I_{2}(g)\rightleftharpoons 2HI(g)+$$ Heat
The following circumstances influence the reaction
Increase the pressure

No effect


At normal temperature and pressure, 5.29mL hydrogen reacts with 0.040 gm iodine at $$444^0C$$ and gives 6.35 mL hydrogen iodide. Calculate equilibrium constant for the synthesis of HI at this temperature. 

No of moles of Iodine  = $$\frac{Mass}{Molecular \ mass}  = \frac{0.0401 }{254}  =0.000157$$
Volume of iodine = $$ 0.0001578 \times 224 \times 1000$$
$$H_2 + I_2 \rightleftharpoons 2HI$$
Initial volume 5.29      3.53       0
Volume at equilibrium      (5.29 - x)     (3.53 - x)   [2x=(6.35)]
or
(5.29 - 3.175)(3.53 - 3.175)6.35 $$\Big[ x = \frac{6.35}{2} = 3.175\Big]$$
= 2.115 = 0.355
Equilibrium constant $$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(6.35)^2}{2.115 \times 0.355} = 53.76$$


A process is given below. What happens to the process if it subjected to a change given in the brackets?
$$N_2(g)+O_2(g) \rightleftharpoons 2NO(g)-180.7\ kJ$$ ( Pressure is increased and temperature is decreased).

Pressure has no effect. Decrease of temperature will shift the equilibrium in the backward direction.


The following reaction has attained equilibrium
$$CO(g) +2H_2(g) \rightleftharpoons CH_3OH(g), \Delta H^o=-92.0\ kJ\ mol^{-1}$$
What will happen if Volume of the reaction vessel is suddenly reduced to half?

$$K_c =\dfrac {[CH_3OH]}{[CO][H_2]^2}, K_p=\dfrac {p_{CH_3OH}}{P_{CO} \times p_{H_2}^2}$$
When volume of the vessel is reduced to half, the concentration of each reactant or product becomes double.
Thus 
$$Q_c =\dfrac {2[CH_3OH]}{2[CO] \times \left\{ 2[H_2]\right\}^2}=\dfrac 14 K_c$$
As $$Q_c < K_c$$, equilibrium will shift in the forward direction, producing more of $$CH_3OH$$ to make $$Q_c=K_c$$.


$$H_{2}(g)+I_{2}(g)\rightleftharpoons 2HI(g)+$$ Heat
The following circumstances influence the reaction
Increase the concentration of $$H_{2}$$.

Increase the rate of forward reaction


The following reaction has attained equilibrium
$$CO(g) +2H_2(g) \rightleftharpoons CH_3OH(g), \Delta H^o=-92.0\ kJ\ mol^{-1}$$
What will happen if the partial pressure of hydrogen is suddenly doubled?

Qp=pCH3OHpCO×(2pH2)2=14KpQp=pCH3OHpCO×(2pH2)2=14Kp.


Hydrogen gas is obtained from natural gas by partial oxidation with steam according to the following endothermic reaction 
$$CH_4(g)+H_2O(g) \rightleftharpoons CO(g)+3H_2(g)$$
How will the value of $$K_p$$ and composition of the equilibrium mixture be affected by increasing the pressure?

By Le Chatelier's principle, equilibrium will shift in the backward direction.


Hydrogen gas is obtained from natural gas by partial oxidation with steam according to the following endothermic reaction 
$$CH_4(g)+H_2O(g) \rightleftharpoons CO(g)+3H_2(g)$$
How will the value of $$K_p$$ and composition of the equilibrium mixture be affected by increasing the temperature?

By Le Chatlier's principle, equilibrium will shift in the backward direction.


The following system is in equilibrium 
$$SO_2Cl_2+Heat \rightleftharpoons SO_2Cl_2$$, What will happen to the temperature of the system if some $$Cl_2$$ is added into at constant volume? Give reason.

Temperature of the system will increase because on adding $$Cl_2$$, equilibrium will shift in the backward direction producing more heat.


List I and List II contain four entries each. Entries of List I are to be matched with entries of List II. One or more than one entries of List I may match with the same entry of List II.

Concept used: Le Chatelier's principle
Decreasing the volume will cause backward shift if change in gaseous moles is positive (i.e., $$\Delta n_{g}>0$$).
$$K_{p}= K_{c}(RT)^{\Delta n}$$  
So $$Kp < Kc$$ only if $$\Delta n_{g}<0$$.
Increasing the pressure will favour backward reaction if $$\Delta n_{g}>0$$.
Decrease of temperature favours forward reaction for exothermic reactions.


List I and List II contains four entries each. Entries of List I are to be matched some entries of List II. One or more than one entries of List I may have the match with the same entry of List II.

(A) Since there is no change in the number of moles on going from reactants to products, the increase in the pressure will not affect the equilibrium.
(B) The number of moles of product is higher than the number of moles of reactants. Hence, when pressure is increased, the equilibrium will shift in the backward direction so that there will be decrease in the number of moles.
(C) The reaction proceeds with decrease in the number of moles.
The reaction is endothermic. Hence, when temperature and pressure are increased,
 the equilibrium will shift in the forward direction.
(D) Two changes are simultaneously carried out which have opposite effects.
When the pressure is decreased, it will shift the reaction in the backward direction (as the reaction proceeds with decrease in the number of moles).
When nitrogen (one of the reactants) is added, the reaction will shift in the forward direction.
Since, the relative magnitude of these effects is not known, the direction in which the equilibrium will shift cannot be predicted.


Column I and Column II contains four entries each.Entries of Column I are to be matched some entries of Column II.One or more than one entries of Column I may have the matching with the same entries of Column II.

For
(A) both are gaseous so $$K_c = C_p/C_r;  C_p = K_c*C_r$$ so linear relationship.
(B)Reactant is solid so its conc. remain same and $$K_c = C_p$$ it has no relation with reactants conc., also products conc. will increase with time as well.
(C) both are gaseous and $$K_c = C_p^2/C_r;  C_p^2 = K_c*C_r$$ so parabolic relationship.
(D) Reactant is solid so its conc. remain same and $$K_c = C_p$$ it has no relation with reactants conc., also products conc. will increase with time as well.


List I and List II contains four entries each. Entries of List I are to be matched some entries of List II. One or more than one entries of List I may have the match with the same entry of List II.

$${ { N }_{ 2 } }(g)+{ { 3H }_{ 2 } }(g)\rightleftharpoons { \quad 2NH }_{ 3 }(g);\quad \Delta H=-ve$$ is an exothermic reaction so K  increases with decrease in temperature.
$${ { N }_{ 2 } }(g)+{ { O }_{ 2 } }(g)\rightleftharpoons 2NO(g);\quad \Delta H=+ve$$ as here energy is absorbed so K 
 increases with increase in temperature. also no. of molecules are same in both side so P has no effect.

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List I and List II contains four entries each. Entries of List I are to be matched some entries of List II. One or more than one entries of List I may have the match with the same entry of List II.

$$\cfrac { { K }_{ 10+{ T }^{ o }C } }{ { K }_{ { T }^{ o }C } } =2$$ means as we increase temp. value of K increases so reaction goes forward means a endothermic reaction.
$$\cfrac { { K }_{ 10+{ T }^{ o }C } }{ { K }_{ { T }^{ o }C } } =1/2$$ means as we increase temp. value of K decreases so reaction goes backward means a exothermic reaction.
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Statement: The equilibrium mole obtained during the experiment for the given reaction in 1 L vessel is shown below the reactants and products.
                               $$2A\left(g\right)     +      B\left(g\right)  \rightleftharpoons     3C\left(g\right)    +    3D$$
At equilibrium:      8 mol    16 mol        4 mol     4 mol

If the reaction in the forward direction is to be made, the initial concentration of reactants and products should be such that the reaction quotient should be less than four.

If the given statement is true enter 1 if false enter 0.

The value of the equilibrium constant K is as calculated below.

$$K = \dfrac {[C]^3[D]^3}{[A]^2[B]} =\dfrac {4^3 \times 4^3}{8^2 \times 16 } = 4$$.

If the reaction in the forward direction is to be made, the initial concentration of reactants and products should be such that reaction quotient should be less than four. This will ensure that more reactants and lesser products are present in the reaction mixture which will derive the equilibrium in the forward direction.

Hence, the given statement is $$true$$ and the correct entry is $$1$$


A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
How do rates of evaporation and condensation change initially?

In the closed vessel at constant temperature, the rate of evaporation will remain constant. Initially, the rate of condensation will be lower as lesser number of molecules per unit volume are present in the vapour phase.
Due to this, the number of collisions per unit volume with the liquid face decreases.


A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
What is the initial effect of the change on vapour pressure?

As liquid is in closed container so if volume is increases it means vapor above the liquid converted into liquid again therefore the vapor density as well as vapor pressure will decreases.


A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
What happens when equilibrium is restored finally and what will be the final vapour pressure?

At equilibrium, the rate of evaporation is equal to the rate of condensation. The final vapour pressure is equal to the original value.


Match the column. The codes for the lists have choices $$(A),\,(B),\,(C)\; and\;(D)$$ out of which only ONE is correct.

(A) For the reaction $$\;N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$$ increase in pressure accelerate the forward reaction.

(B) For the reaction $$\;2NO(g)\rightleftharpoons N_2(g)+O_2(g)$$ at constant volume introduction of inert gas, has no effect.

(C) For the reaction $$\;CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)$$ $$K_p=P_{total}$$
Le-chateliers principle is applicable and for $$P_{total} = P_{CO2}$$, which is equal to Kp for the given reaction.

(D) For the reaction $$\;PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$$ increase in volume favors forward reaction.


At 473 K, equilibrium constant $$\displaystyle { K }_{ c }$$ for decomposition of phosphorus pentachloride, $$\displaystyle { PCl }_{ 5 }$$ is $$\displaystyle 8.3\times { 10 }^{ -3 }$$. If decomposition is depicted as,
$$\displaystyle { PCl }_{ 5 }\left( g \right) \rightleftharpoons { PCl }_{ 3 }\left( g \right) +{ Cl }_{ 2 }\left( g \right) $$ $$\displaystyle { \Delta  }_{ r }{ H }^{ \ominus  }=124.0kJ{ mol }^{ -1 }$$
(a) Write an expression for $$\displaystyle { K }_{ c }$$ for the reaction.
(b) What is the value of $$\displaystyle { K }_{ c }$$ for the reverse reaction at the same temperature?
(c) What would be the effect on $$\displaystyle { K }_{ c }$$ if (i) more $$\displaystyle { PCl }_{ 5 }$$ is added (ii) pressure is increased (iii) the temperature is increased ?

(a) The equilibrium constant expression is
$$\displaystyle K_c = \frac {[PCl_3][Cl_2]}{[PCl_5]} $$
(b) The equilibrium constant expression for the reverse reaction is
$$\displaystyle K_c(reverse) = \frac {1}{8.3 \times 10^{-3}} = 120.48 $$
(c) (i) When more $$\displaystyle PCl_5 $$ is added, the direction of equilibrium is affected but the value of equilibrium constant is unaffected as $$\displaystyle  K_c$$ is constant at a given temperature.
(ii) When pressure is increased, the value of the equilibrium constant remains unaffected.
(iii) When temperature is increased, the value of equilibrium constant is increased. The reaction is endothermic. When temperature is increased, $$\displaystyle  K_f$$ is increased. Hence, $$\displaystyle K_c= \frac {K_f}{K_r} $$ is also increased.


Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

$$\displaystyle { CH }_{ 4 }\left( g \right) +{ H }_{ 2 }O\left( g \right) \rightleftharpoons CO\left( g \right) +3{ H }_{ 2 }\left( g \right) $$

(a) Write as expression for $$\displaystyle { K }_{ p }$$ for the above reaction.

(b) How will the values of $$\displaystyle { K }_{ p }$$ and composition of equilibrium mixture be affected by
(i) increasing the pressure,
(ii) Increasing the temperature and
(iii) Using a catalyst?

(a) The equilibrium constant expression is $$\displaystyle K_p =\frac {P_{CO} \times P^3_{H_2}}{P_{CH_4}P_{H_2O}} $$. 

(b) (i) As the pressure increases, the equilibrium will shift to left so that less number of moles are produced.

This is according to Le-Chatelier's principle (Refer image)

(ii) For exothermic reaction, with increase in temperature, the equilibrium will shift in the backward direction.

(iii) When a catalyst is used, the equilibrium is not disturbed. The equilibrium is quickly attained.

398192_423272_ans_51187c72b5554c0a95fedbc88bd5a9c3.jpg


Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction.
(i) $$\displaystyle { COCl }_{ 2 }\left( g \right) \rightleftharpoons CO\left( g \right) +{ Cl }_{ 2 }\left( g \right) $$
(ii) $$\displaystyle { CH }_{ 4 }\left( g \right) +2{ S }_{ 2 }\left( g \right) \rightleftharpoons { CS }_{ 2 }\left( g \right) +2{ H }_{ 2 }S\left( g \right) $$
(iii) $$\displaystyle { CO }_{ 2 }\left( g \right) +C\left( s \right) \rightleftharpoons 2CO\left( g \right) $$
(iv) $$\displaystyle 2{ H }_{ 2 }\left( g \right) +CO\left( g \right) \rightleftharpoons { CH }_{ 3 }OH\left( g \right) $$
(v) $$\displaystyle { CaCO }_{ 3 }\left( s \right) \rightleftharpoons CaO\left( s \right) +{ CO }_{ 2 }\left( g \right) $$
(vi) $$\displaystyle 4{ NH }_{ 3 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \rightleftharpoons 4NO\left( g \right) +6{ H }_{ 2 }O\left( g \right) $$

(i) The change in the number of moles of the gaseous species $$\displaystyle \Delta n = 2-1=1 $$.
When pressure is increased, the equilibrium reaction is affected. The direction of reaction will shift into backward direction.
(ii)The change in the number of moles of the gaseous species $$\displaystyle \Delta n = 3-3=0 $$. 
When the pressure is increased, the equilibrium reaction is not affected.
(iii) The change in the number of moles of the gaseous species $$\displaystyle \Delta n = 2-1=1 $$.
When pressure is increased, the equilibrium reaction is affected. The direction of reaction will shift into backward direction.
(iv) The change in the number of moles of the gaseous species $$\displaystyle \Delta n = 1-3=-2 $$.
When pressure is increased, the equilibrium reaction is affected. The direction of reaction will shift into forward direction.
(v) The change in the number of moles of the gaseous species $$\displaystyle \Delta n = 1-0=1 $$.
When pressure is increased, the equilibrium reaction is affected. The direction of reaction will shift into backward direction.
(vi) The change in the number of moles of the gaseous species $$\displaystyle \Delta n =  10-9=1$$.
When pressure is increased, the equilibrium reaction is affected. The direction of reaction will shift into backward direction.


Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?
(a) $$\displaystyle { PCl }_{ 5 }\left( g \right) \rightleftharpoons { PCl }_{ 3 }\left( g \right) { Cl }_{ 2 }\left( g \right) $$
(b) $$\displaystyle CaO\left( s \right) +{ CO }_{ 2 }\left( g \right) \rightleftharpoons { CaCO }_{ 3 }\left( s \right) $$
(c) $$\displaystyle 3Fe\left( s \right) +4{ H }_{ 2 }O\left( g \right) \rightleftharpoons { Fe }_{ 3 }{ O }_{ 4 }\left( s \right) +4{ H }_{ 2 }\left( g \right) $$

(a) The change in the number of moles of the gaseous species $$\displaystyle \Delta n = 1+1-1=1 >0 $$. Hence, the number moles of reaction product increase by decrease in pressure due to increase in volume.
(b) The change in the number of moles of the gaseous species  $$\displaystyle \Delta n = 0-1=-1= < 1 $$.  Hence, the number moles of reaction product decrease by decrease in pressure due to increase in volume.
(c) The change in the number of moles of the gaseous species  $$\displaystyle \Delta n = 4-4=0 $$.  Hence, the number moles of reaction product remains same due to decrease in pressure due to increase in volume.


State Le-chatelier's principle and apply it to the following equilibrium.
$$N_{2_{(g)}} + 3H_{2_{(g)}} \rightleftharpoons 2NH_{3_{(g)}}; \triangle H = -92.0\ KJ$$

Le Chatelier's principle:
If an external stress is applied to a reacting system at equilibrium, the system will adjust itself in such a way that the effect of the stress is reduced.
$$\displaystyle N_{2_{(g)}} + 3H_{2_{(g)}} \rightleftharpoons 2NH_{3_{(g)}}; \triangle H = -92.0\ KJ$$
The forward reaction is endothermic ( as the enthalpy change is negative) and the reverse reaction is exothermic. If the temperature of system is increased, the equilibrium will shift from left to right. So more and more of N$$\displaystyle _2$$  and H$$\displaystyle _2$$ will be converted to NH$$\displaystyle _3$$. The yield of NH$$\displaystyle _3$$ will increase at high temperature. The decrease in temperature will favor the reverse reaction.
In the forward reaction, the number of moles decreases and in the reverse reaction, the number of moles increases. With increase in the pressure (by reducing the volume at constant temperature), the forward reaction is favored as the number of moles will decrease. This will increase the yield of NH$$\displaystyle _3$$. When pressure is decreased, the reverse reaction is favored.


State Le Chatelier's principle and apply it to the following equilibrium.
$$2SO_{2_{(g)}} + O_{2_{(g)}} \rightleftharpoons 2SO_{3_{(g)}}; \triangle H = -189 k.J$$

Le Chatelier's principle:
If an external stress is applied to a reacting system at equilibrium, the system will adjust itself in such a way that the effect of the stress is reduced.
$$\displaystyle 2SO_{2_{(g)}} + O_{2_{(g)}} \rightleftharpoons 2SO_{3_{(g)}}; \triangle H = -189 k.J$$
The forward reaction is endothermic ( as the enthalpy change is negative) and the reverse reaction is exothermic. If the temperature of system is increased, the equilibrium will shift from left to right. So more and more of SO$$\displaystyle _2$$ will be converted to SO$$\displaystyle _3$$. The yield of SO$$\displaystyle _3$$ will increase at high temperature.
The decrease in temperature will favor the reverse reaction.
In the forward reaction, the number of moles decreases and in the reverse reaction, the number of moles increases. With increase in the pressure (by reducing the volume at constant temperature), the forward reaction is favored as the number of moles will decrease. This will increase the yield of SO$$\displaystyle _3$$. When pressure is decreased, the reverse reaction is favored.


When $$3.06\ g$$ of $${ NH }_{ 4 }HS(s)$$ is introduced into a $$two\ litre$$ evacuated flask at $${27}^{o}C$$, $$30$$% of the solid decomposes into gaseous ammonia and hydrogen sulphide.
(i) Calculate $${K}_{c}$$ and $${K}_{p}$$ for the reaction at $${27}^{o}C$$.
(ii) What would happen to the equilibrium when more $${ NH }_{ 4 }HS(s)$$ is introduced into the flask?

$${ NH }_{ 4 }HS(s)\rightleftharpoons { NH }_{ 3 }(g)+{ H }_{ 2 }S(g)$$
Moles of $${ NH }_{ 4 }HS=\cfrac { 3.06 }{ 51 } =0.06$$

Degree of dissociation $$=0.3$$

(i) At equilibrium
$$\left[ { NH }_{ 3 }(g) \right] =\cfrac { 0.3\times 0.06 }{ 2 } ;\left[ { H }_{ 2 }S(g) \right] =\cfrac { 0.3\times 0.06 }{ 2 } $$

$${ K }_{ c }=\left[ { NH }_{ 3 }(g) \right] \left[ { H }_{ 2 }S(g) \right] =8.1\times { 10 }^{ -5 }$$

Now applying,
$${ K }_{ p }={ K }_{ c }{ \left( RT \right)  }^{ \Delta n }=8.1\times { 10 }^{ -5 }\times { \left( 0.082\times 300 \right)  }^{ 2 }=0.049$$

(ii) If more amount of $$NH_4HS$$ is introduced in the flask at equilibrium then, the reaction will favour forward reaction in order to decrease the concentration of $$NH_4HS$$. This is in accordance to Le-chatelier's principle.


For the reaction, $$S{ O }_{ 2 }(g)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)\rightleftharpoons { SO }_{ 3 }(g)$$, $${K}_{p}$$ is $$32\ atm^{ -1/2 }$$ at $$800K$$ and $$\Delta H$$ for the reaction is $$-187.9kJ$$ $${mol}^{-1}$$. Calculate its value at $$900K$$, if it is assumed that $$\Delta H$$ remains constant over this range of temperature.

The given reaction is :-
$${ SO }_{ 2 }+\dfrac { 1 }{ 2 } { O }_{ 2 }\rightleftharpoons { SO }_{ 3 }\left( g \right) $$
                    $${ K }_{ P }=32{ atm }^{ -1/2 }at\quad 800K$$
$$\triangle H=-187.9KJ/mol$$
By Vant' Hoff equation :-
$$log\left( \dfrac { { \left( { K }_{ P } \right)  }_{ 2 } }{ { \left( { K }_{ P } \right)  }_{ 1 } }  \right) =\dfrac { \triangle H }{ 2.303R } \left( \dfrac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } }  \right) $$
$$\Rightarrow log\left( { { K }_{ P } }_{ 2 } \right) -log\left( { { K }_{ P } }_{ 1 } \right) =\dfrac { -187.9\times 1000 }{ 2.303\times 8.314 } \left( \dfrac { 900-800 }{ 900\times 800 }  \right) $$
$$\Rightarrow log{ { K }_{ P } }_{ 2 }=\dfrac { -187.9\times 1000\times 100 }{ 2.303\times 8.314\times 900\times 800 } +log\left( 32 \right) $$
$$\Rightarrow log{ { K }_{ P } }_{ 2 }=1.36+1.50$$
$$\Rightarrow { K }_{ { P }_{ 2 } }=Antilog\left( 2.86 \right) =724.44$$


Write the effect of temperature pressure and concentration on the following reaction $$2SO_{2}+O_{2} \rightleftharpoons 2So_{3};\Delta H=-198\ kJ/mol$$


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In a gaseous reaction; $$A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$$, the increase in temperature causes the change in the concentrations of $$A, B, C$$ and $$D$$. The concentrations of $$C$$ and $$D$$ also change on addition of some amount of $$A$$. Does the value of $$K$$ change in either of the two situations?

$$K$$ change in the first case only but it remains the same in the second case as $$K$$ depends only on the temperature not on the concentration of the reactants or products.


The reaction:

$$Sb_2S_3(s)+3H_2(g)\rightleftharpoons 2Sb(s)+3H_2S(g)$$

It was studied by analysing the equilibrium mixture for the amount of $$H_2S$$ produced. A vessel whose volume was $$2.5$$ litre was filled with $$0.01$$ mole of $$Sb_2S_3$$ and $$0.01$$ mole of $$H_2S$$. After the mixture came to equilibrium in the closed vessel at $$440^o$$C, the gaseous mixture was removed and the $$H_2S$$ was dissolved in water. Sufficient $$Pb^{2+}$$ ions were added to react completely with the $$H_2S$$ to precipitate $$PbS$$. If $$1.029$$g of $$PbS$$ was obtained, what is the value of $$K_c$$ at $$440^o$$C?


1639538_1740141_ans_352068f4c4bc420585a3e85cdfb33354.jpg


A tenfold increase in pressure on the reaction, $$N_2(g)+3H_2\rightleftharpoons 2NH_3(g)$$ at equilibrium results in .......... in $$K_p$$.


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What is the equilibrium constant for this dissolving process?


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Derive the best conditions for dissociation of $$NH_{3}.$$ Given $$2NH_{3} \rightleftharpoons N_{2} + 3H_{2}; \triangle H = +91.94\ kJ$$.

Low pressure favours reaction showing an increase in moles i.e., forward reaction.

High temperature favours reaction showing an endothermic nature i.e., forward reaction.

Removal of $$N_2$$ or $$H_{2}$$ also favours forward reaction.


The chemical equation of one of the different stages of manufacturing sulphuric acid by the contact process is given below. Find out the influence of the following factors in the reaction given below.
$$2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3}(g)+heat$$

Increase the amount of oxygen. 

On increasing the amount of oxygen, rate of forward reaction increases.


Following data is given for the reaction: $$CaCO_{3}(s)\rightarrow CaO(s)+CO_{2}(g)$$
$$\Delta _{f}H^{\circleddash}[CaO(s)]=-635.1kJ$$ $$mol^{-1}$$
$$\Delta _{f}H^{\circleddash}[CO_{2}(g)]=-393.5kJ$$ $$mol^{-1}$$
$$\Delta _{f}H^{\circleddash}[CaCO_{3}(s)]=-1206.9kJ$$ $$mol^{-1}$$
Predict the effect of temperature on the equilibrium constant of the above reaction.

$$\Delta _{r}H^{\circleddash}=\Delta _{f}H^{\circleddash}[CaO(s)]+\Delta _{f}H^{\circleddash}[CO_{2}(g)]-\Delta _{f}H^{\circleddash}[CaCO_{3}(s)]$$
$$\therefore \Delta _{f}H^{\circleddash}=178.3kJ$$ $$mol^{-1}$$.
The reaction is endothermic. Hence, according to Le Chatelier’s principle, reaction will proceed in forward direction on increasing temperature.


The chemical equation of one of the different stages of manufacturing sulphuric acid by contact process is given below. Find out the influence of the following factors in the reaction given below.
$$2SO_{2}(g)+O_{2}(g)\rightleftharpoons 2SO_{3}(g)+heat$$

Pressure is increased

When pressure of the reaction is increased, rate of forward reaction increases.


For an exothermic reaction, what happens to the equilibrium constant if temperature is increased?

$$K=k_f /k_b$$. In exothermic reaction, with increase of temperature $$k_b$$ increases much more than $$k_f$$. Hence $$K$$ decrease.


Class 11 Engineering Chemistry Extra Questions