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 .

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.  

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 $${K}_{c}$$ 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.

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

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 . 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.
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 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.

we know the relation between enthalpy change and potential

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. 

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. 

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

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, 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. 

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 $$H2O$$ ).

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.

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.

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".

A.  $$\Delta H$$ is 0 so connect with 1 , 3

	$$H_2$$	$$+H_2$$	$$\rightleftharpoons$$	$$2HI$$
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. $$=0.5-0.39$$
$$=0.11\ atm$$

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.

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

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

$$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$$  
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. 

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

(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
(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. 

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. 

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

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 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. 

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.

Increase the rate of backward reaction

No effect

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 $$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(6.35)^2}{2.115 \times 0.355} = 53.76$$

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

$$K_c =\dfrac {[CH_3OH]}{[CO][H_2]^2}, K_p=\dfrac {p_{CH_3OH}}{P_{CO} \times p_{H_2}^2}$$
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 $$Q_c=K_c$$ .

Increase the rate of forward reaction

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

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

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

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

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.

(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.

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.

$${ { N }_{ 2 } }(g)+{ { 3H }_{ 2 } }(g)\rightleftharpoons { \quad 2NH }_{ 3 }(g);\quad \Delta H=-ve$$ 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.

$$\cfrac { { K }_{ 10+{ T }^{ o }C } }{ { K }_{ { T }^{ o }C } } =2$$ 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.

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

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.

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.

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

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

(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.

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

(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.

(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.

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.

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.

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

The given reaction is :-

$$K$$ 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.

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

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

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

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