If $$\Delta G$$ standard is zero, this means :

the system is at equilibrium at standard conditions

the reaction is spontaneous at standard conditions

the reaction is non spontaneous at standard conditions

the reaction is both non spontaneous and at equilibrium

the reaction is both spontaneous and at equilibrium

MCQ Questions for Class 11 Medical Chemistry Thermodynamics Quiz 11

Hess's law is used to calculate:

0%
Enthalpy of reaction
0%
Entropy of reaction
0%
Work done in reaction
0%
All the above

Explanation

Hint: Hess’s law states that, if an overall reaction takes place in several steps, its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions, at the same temperature.

Correct Option: $A$

Explanation :

Hess’s law states that the change in enthalpy in a chemical reaction is the same whether the process is carried out in one step or in multiple steps. It is the outcome of the first law of thermodynamics. It is used for the calculation of standard reaction enthalpy.

Extra Information:

Hess’s law states that the standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction can be divided, while each occurs at the same temperature.

The enthalpy change for a reaction is independent of the number of ways a product can be obtained if the initial and final conditions are the same.

The negative enthalpy change for a reaction indicates an exothermic process, while positive enthalpy change corresponds to an endothermic process.

Hence from the above explanation, it is clear that it is used to calculate the Enthalpy of reaction

So, the Correct answer is Option A.

A sample of liquid in a thermally insulated container (a calorimeter) is stirred for 2 hr by a mechanical linkage to a motor in the surrounding. For this process:

0%
$w < 0; q = 0; \Delta U = 0$
0%
$w > 0; q < 0; \Delta U > 0$
0%
$w < 0; q > 0; \Delta U = 0$
0%
$w > 0; q = 0; \Delta U > 0$

The temperature in K at which

$\displaystyle \Delta G=0$ , for a given reaction with

$\displaystyle \Delta H=-20.5\ kJ{ mol }^{ -1 }$ and

$\displaystyle \Delta S=-50.0\ J{ K }^{ -1 }{ mol }^{ -1 }$ is:

0%
-410
0%
410
0%
2.44
0%
-2.44

Explanation

Given that,

$\displaystyle \Delta H = - 20.5\ kJ/mol = -20500\ J/mol$

$\Delta S=-50\ JK^{-1}mol^{-1}$

Now,

$\displaystyle \Delta G = \Delta H - T \Delta S$

$\displaystyle 0 = -20500 - T (-50.0)$

$\displaystyle 20500 = 50T$

$\displaystyle \Rightarrow T = 410\ K$

Calculate

${ \Delta H }/{ kJ }$ for the following reaction using the listed standard enthalpy of reaction data:

$2{ N }_{ 2 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \longrightarrow 2{ N }_{ 2 }{ O }_{ 5 }\left( s \right)$

${ N }_{ 2 }\left( g \right) +3{ O }_{ 2 }\left( g \right) +{ H }_{ 2 }\left( g \right) \longrightarrow 2HN{ O }_{ 3 }\left( aq \right) { \Delta H }/{ kJ }=-414.0$

${ N }_{ 2 }{ O }_{ 5 }\left( s \right) +{ H }_{ 2 }O\left( I \right) \longrightarrow 2HN{ O }_{ 3 }\left( aq \right) { \Delta H }/{ kJ }=-86.0$

$2{ H }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right) \longrightarrow 2{ H }_{ 2 }O\left( I \right) { \Delta H }/{ kJ }=-571.6$

0%
$-84.4$
0%
$-243.6$
0%
$-71.2$
0%
$-121.8$

Explanation

For the reaction,

$2{ N }_{ 2 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \longrightarrow 2{ N }_{ 2 }{ O }_{ 5 }\left( s \right)$

${ N }_{ 2 }\left( g \right) +3{ O }_{ 2 }\left( g \right) +{ H }_{ 2 }\left( g \right) \longrightarrow 2HN{ O }_{ 3 }\left( aq \right) { \Delta H }/{ kJ }=-414.0$

${ N }_{ 2 }{ O }_{ 5 }\left( s \right) +{ H }_{ 2 }O\left( I \right) \longrightarrow 2HN{ O }_{ 3 }\left( aq \right) { \Delta H }/{ kJ }=-86.0$

$2{ H }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right) \longrightarrow 2{ H }_{ 2 }O\left( I \right) { \Delta H }/{ kJ }=-571.6$

Applying Hess's law,

$\Delta { H }_{ r }=\left[ 2\left( -414 \right) +2\left( 86 \right) +571.6 \right]$

$\Delta { H }_{ r }=-84.4 kJ$

What can be used in combination with a calorimeter to compare the specific heats of two substances?

0%
Thermometer
0%
Conductivity tester
0%
Graduated cylinder
0%
Buret
0%
Salt bridge

The equilibrium constant

$\left( K \right)$ of a reaction may be written as:

0%
$K={ e }^{ { \Delta G }/{ RT } }$
0%
$K={ e }^{ { -\Delta { G }^{ } }/{ RT } }$
0%
$K={ e }^{ { \Delta H }/{ RT } }$
0%
$K={ e }^{ -{ \Delta { H }^{ } }/{ RT } }$

Explanation

Gibbs free energy

$\left( \Delta { G }^{ } \right)$ is related with equilibrium constant

$\left( K \right)$ as

$\Delta { G }^{ }=-RT\ln { K }$

$-\dfrac { \Delta { G }^{ } }{ RT } =\ln { K }$

$\therefore K={ e }^{ \dfrac { -\Delta { G }^{ } }{ RT } }$

$H_2(g) + \frac{ 1 }{ 2 }O_2(g)\longrightarrow 2H_2O(l); \Delta H = -86 kJ$

$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l)......... kJ(\pm?)$

0%
$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) + 172 kJ$ .
0%
$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) - 172 kJ$ .
0%
$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) +0 kJ$ .
0%
None of these

Explanation

Reaction:

$H_2(g) + \frac{ 1 }{ 2 }O_2(g)\longrightarrow 2H_2O(l); \Delta H = -86 kJ$ .................1
multiply it by 2 we will get,

$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l); \Delta H = -86*2 = -172 kJ$
which can be written as

$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) + 172 kJ$ .

Based on the following thermochemical equations

${H}_{2}(g)+C(s)\longrightarrow CO(g)+{H}_{2}(g)$ ;

$\Delta H=133kJ {mol}^{-1}$

$CO(g)+\cfrac{1}{2}{O}_{2}(g)\longrightarrow {CO}_{2}(g)$ ;

$\Delta H=-282kJ{mol}^{-1}$

${H}_{2}(g)+\cfrac{1}{2}{O}_{2}(g)\longrightarrow {H}_{2}O(g)$ ;

$\Delta H=-242kJ{mol}^{-1}$

$C(s)+{O}_{2}(g)\longrightarrow {CO}_{2}(g)$ ;

$\Delta H=xkJ{mol}^{-1}$
The value of

$x$ will be:

0%
$393.0$
0%
$655.0$
0%
$-393.0$
0%
$-655.0$

Explanation

On adding Eqs

$(i), (ii), (iii)$ we get

$C(s)+{O}_{2}(g)\longrightarrow {CO}_{2}(g)$ ;

$\Delta H=(131-282-242)kJ$

$=-393.0kJ$

I :

$\displaystyle { S }_{ 8 }\left( s \right) +8{ O }_{ 2 }\left( g \right) \rightarrow 8{ SO }_{ 2 }\left( g \right) \Delta H=-2374.6kJ$
II :

$\displaystyle { S }_{ 8 }\left( s \right) +12{ O }_{ 2 }\left( g \right) \rightarrow 8{ SO }_{ 3 }\left( g \right) \Delta H=-3165.8.6kJ$
From the information given above, determine the heat of reaction for the combustion of sulfur dioxide.

$\displaystyle 2{ SO }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right) \rightarrow 2{ SO }_{ 3 }\left( g \right)$

0%
-5540.4 kJ
0%
-1385.1 kJ
0%
-791.2 kJ
0%
-197.8 kJ
0%
-899.2 kJ

Explanation

$\displaystyle \displaystyle { S }_{ 8 }\left( s \right) +8{ O }_{ 2 }\left( g \right) \rightarrow 8{ SO }_{ 2 }\left( g \right) \Delta H=-2374.6kJ$

Reverse above reaction,

$\displaystyle \displaystyle 8{ SO }_{ 2 }\left( g \right) \rightarrow { S }_{ 8 }\left( s \right) +8{ O }_{ 2 }\left( g \right) \Delta H=2374.6kJ$ ...... (1)

$\displaystyle \displaystyle { S }_{ 8 }\left( s \right) +12{ O }_{ 2 }\left( g \right) \rightarrow 8{ SO }_{ 3 }\left( g \right) \Delta H=-3165.8.6kJ$ ......(2)

Add (1) and (2),

$\displaystyle \displaystyle 8{ SO }_{ 2 }+ 4{ O }_{ 2 }\left( g \right) \rightarrow 8{ SO }_{ 3 }\left( g \right) \Delta H=2374.6-3165.8.6kJ = -791.26 kJ$
Divide above equation by 4,

$\displaystyle \displaystyle 2{ SO }_{ 2 }+ { O }_{ 2 }\left( g \right) \rightarrow 4{ SO }_{ 3 }\left( g \right) \Delta H=\dfrac { -791.26 kJ}{4} = -197.8 kJ$

Bond	Average Bond Energy (kJ/mol)
$C \equiv O$	1075
$C=O$	728
$C-Cl$	326
$Cl-Cl$	243

From the above given data, calculate the heat of reaction for the following reaction :

$CO+{ Cl }_{ 2 }\rightarrow CO{ Cl }_{ 2 }$

0%
$+62$ kJ
0%
$-62$ kJ
0%
$-409$ kJ
0%
$+706$ kJ

Explanation

$CO+Cl_2 \rightarrow COCl_2$

Heat of reaction = Sum of bond Energy of products - sum of bond energy of reactant.

Heat of reaction =

$[728 + 2\times 326 ] - [1075 + 243] = 1380 - 1318 = +62\ kJ$

Hence, the correct option is A.

Name the apparatus used to measure the heat absorbed or released by a reaction.

0%
Centrifuge
0%
Barometer
0%
Balance
0%
Calorimeter
0%
Battery

Statement 1: The

$\Delta H_{reaction}$ of a particular reaction can be arrived at by the summation of the

$\Delta H_{reaction}$ values of two or more reactions that, added together, will give the

$\Delta H_{reaction}$ of the particular reaction.
Statement 2: Hess's Law conforms to the First Law of Thermodynamics, which states that the total energy of the universe is a constant.

0%

Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1.
0%

Both Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1.
0%

Statement 1 is correct but Statement 2 is not correct.
0%

Statement 1 is not correct but Statement 2 is correct.
0%

Both the Statement 1 and Statement 2 are not correct.

Explanation

Statement 1: The

$\displaystyle \Delta H_{reaction}$ of a particular reaction can be arrived at by the summation of the

$\displaystyle \Delta H_{reaction}$ values of two or more reactions that, added together which give the

$\displaystyle \Delta H_{reaction}$ of the particular reaction.
Statement 2: Hess's Law conforms to the First Law of Thermodynamics, which states that the total energy of the universe is a constant.
Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1.
Consider following example

$\displaystyle C(s) + \dfrac {1}{2} O_2 (g) \rightarrow CO (g) \: \: \: \Delta H_{1, reaction}$

$\displaystyle CO(g) + \dfrac {1}{2} O_2 (g) \rightarrow CO_2 (g) \: \: \: \Delta H_{2, reaction}$

$\displaystyle C(s) + O_2 (g) \rightarrow CO_2 (g) \\ \: \: \: \Delta H_{3, reaction} = \Delta H_{1, reaction} + \Delta H_{2, reaction}$

Calculate the enthalpy ,

$\Delta H$ , for the given reaction using the the given bond energies ?

$C_{2}H_{4} + Cl_{2}\rightarrow ClH_{2}C-CH_{2}Cl$

Bond energies	kJ/ mol
$C-C$	$347$
$C = C$	$612$
$C-Cl$	$341$
$C-H$	$414$
$Cl-Cl$	$243$

0%
$\Delta H = -800 kJ$
0%
$\Delta H = -680 kJ$
0%
$\Delta H = -174 kJ$
0%
$\Delta H = +174 kJ$
0%
$\Delta H = +200 kJ$

Explanation

$\displaystyle \Delta H^o _{rxn} = \Sigma \Delta H^o _{\text {reactant bonds}} - \Sigma \Delta H^o _{\text {product bonds}} \\ \Delta H^o _{rxn}= [4\Delta H^o _{C-H}+ \Delta H^o _{C=C}+Cl-Cl] - [\Delta H^o _{C-C}+4\Delta H^o _{C-H}+2\Delta H^o _{C-Cl}]$

$\displaystyle \Delta H^o _{rxn} = [ \Delta H^o _{C=C}+Cl-Cl] - [\Delta H^o _{C-C}+2\Delta H^o _{C-Cl}]$

$\displaystyle \Delta H^o _{rxn} = [612 + 243] - [347 + 2(341)]$

$\displaystyle \Delta H^o _{rxn} = 855-1029$

$\displaystyle \Delta H^o _{rxn} = -174kJ$

The spontaneity of a reaction is indicated by:

0%
enthalpy change
0%
entropy change
0%
gibbs free energy change
0%
activation energy
0%
specific heat capacity

At the standard temperature and pressure (STP) of

$0$

$^oC$ and

$1$ atmosphere, the Gibbs free energy change is ______.

0%
$-247 kJ/mol.$
0%
$-257 kJ/mol.$
0%
$-237 kJ/mol.$
0%
$-227 kJ/mol.$

Explanation

At the standard temperature and pressure (STP) of

$0$

$^oC$ and

$1$ atmosphere, the Gibbs free energy change is

$-237 kJ/mol$ . The negative value means that the reaction should be spontaneous in the forward direction.

$H_2(g)$ +

$\dfrac{1}{2}$

$O_2(g) \rightleftharpoons H_2O(l)$

$G^0 =$

$-237 \ kJ/mol$

Hence, the correct option is

$\text{C}$

The reverse of a spontaneous reaction is ......... .

0%
always spontaneous
0%
always non spontaneous
0%
sometimes spontaneous
0%
sometimes non spontaneous
0%
There is no way of telling

The important considerations in deciding if a reaction will be spontaneous are :

0%
stability & state of reactants
0%
energy gained & heat evolved
0%
exothermic energy & randomness of the products
0%
endothermic energy & randomness of the products
0%
endothermic energy & structure of the products

What is the value of

$\Delta H$ for the reaction

$X + 2Y \rightarrow 2Z$ ?

$W+X\rightarrow 2Y$ ;

$\Delta H=- 400$ kcal/mol

$2W + 3X \rightarrow 2Z + 2Y$ ;

$\Delta H=- 150$ kcal/mol

0%
$- 550$ kcal/mol
0%
$+50$ kcal/mol
0%
$- 50$ kcal/mol
0%
$+ 650$ kcal/mol
0%
$+250$ kcal/mol

Explanation

Multiply equation

$\displaystyle W+X\rightarrow 2Y$ with

$2$ .

$\displaystyle 2W+2X\rightarrow 4Y$ ;

$\displaystyle \Delta H= -400 \times 2 = -800$ kcal .....(1)

$\displaystyle 2W + 3X \rightarrow 2Z + 2Y$

$\displaystyle \Delta H= -150$ kcal .....(2)

Subtracting equation (1) from equation (2) we get,

$\displaystyle X + 2Y \rightarrow 2Z$

$\displaystyle \Delta H= -150 -(-800)=+650$ kcal

$\Delta G$ standard is zero, this means :

0%
the reaction is spontaneous at standard conditions
0%
the reaction is non spontaneous at standard conditions
0%
the system is at equilibrium at standard conditions
0%
the reaction is both non spontaneous and at equilibrium
0%
the reaction is both spontaneous and at equilibrium

Explanation

$\Delta G = 0$ , it means the reaction is at equilibrium at standard condictions.
As

$\Delta G = negative$ , means spontaneous.

$\Delta G = positive$ , means non-spontaneous.

Which of the following drives spontaneous reactions?

0%
Low enthalpy values and high entropy values.
0%
Low enthalpy values and low entropy values.
0%
High enthalpy values and low entropy values.
0%
High enthalpy values and high entropy values.
0%
High temperatures and low pressures.

Explanation

Low enthalpy and high entropy drives spontaneous reactions.
These leads to negative value of Gibbs free energy change.

$\displaystyle \Delta G = \Delta H - T \Delta S$

From the heats of reaction of these individual reactions:

$A + B \rightarrow 2C$

$\triangle H = -500 kJ$

$D + 2B \rightarrow E$

$\triangle H = -700 kJ$

$2D + 2A \rightarrow F$

$\triangle H = +50 kJ$
Find the heat of reaction for

$F + 6B$

$\rightarrow 2E + 4C$

0%
+450 kJ
0%
-1100 kJ
0%
+2350 kJ
0%
-350 kJ
0%
-2450 kJ

Explanation

$\displaystyle 2D + 2A \rightarrow F$

$\displaystyle \triangle H = +50 kJ$
Reverse above equation

$\displaystyle F \rightarrow 2D + 2A$

$\displaystyle \triangle H = -50 kJ$ ......(1)

$\displaystyle D + 2B \rightarrow E$

$\displaystyle \triangle H = -700 kJ$
Multiply above equation with (2)

$\displaystyle 2D + 4B \rightarrow 2E$

$\displaystyle \triangle H = -1400 kJ$ ......(2)

$\displaystyle A + B \rightarrow 2C$

$\displaystyle \triangle H = -500 kJ$
Multiply above equation with (2)

$\displaystyle 2A + 2B \rightarrow 4C$

$\displaystyle \triangle H = -1000 kJ$ ......(3)
Add equations (1), (2) and (3)

$\displaystyle F + 6B \rightarrow 2E + 4C$

$\displaystyle \triangle H = -50 kJ -1400 kJ -1000 kJ = - 2450 kJ$

For a reaction to be spontaneous, the sign on delta G should be :

0%
positive
0%
there should be no sign
0%
negative
0%
spontaneity is not related to Gibbs Free Energy
0%
positive or Negative

Explanation

$\triangle G = negative$ for spontaneous reaction.

$\triangle G = Positive$ for non-spontaneous reaction.

$\triangle G = 0$ for equilibrium.

Which of the following is most likely to produce a spontaneous reaction?

0%
Negative Enthalpy
0%
Positive Enthalpy
0%
Negative Entropy
0%
Positive Entropy
0%
Negative Enthalpy and positive Entropy

Explanation

For the spontaneous reaction, the negative enthalpy and positive entropy is most likely.
As for spontaneous reaction the value of

$\Delta G$ must be negative.
Hence, as per formula,

$\Delta G= \Delta H - T \Delta S$ ,
So, if

$\Delta H$ and

$\Delta S$ are negative and positive respectively then the value of the

$\Delta G$ must be negative whatever the temperature.

If the gibbs free energy is negative than reaction will be?

0%
always positive
0%
sometimes negative
0%
non-spontaneous
0%
spontaneous
0%
None of the above

Explanation

(D) : Always negative.

$\Delta G = 0$ , it means the reaction is equilibrium at standard conditions.
A negative value of

$\Delta G$ , means spontaneous.
A positive value of

$\Delta G$ , means non-spontaneous.

Determine

$\Delta G^{o}$ for the following reaction.

$CO(g)+\cfrac{1}{2}O_2(g)\rightarrow CO_2(g)$ ;

$\Delta H^{o} = -282.84 kJ$

[Given:

$S^{o} _{CO_2}=213.8$ JK

$^{-1}$ mol

$^{-1}$ ,

$S^{o}\ _{ CO(g)} = 197.9$ J K

$^{-1}$ mol

$^{-1}$

$S^{o} \ _{O_2}=205.8$ J K

$^{-1}$ mol

$^{-1}$ ]

0%
$-157.33$ kJ
0%
$+201.033$ kJ
0%
$-256.91$ kJ
0%
$+257.033$ kJ

Explanation

The reaction is:

$\displaystyle CO(g)+\frac{1}{2}O_2(g)\rightarrow CO_2(g)$

The standard entropy change for the reaction is:

$\displaystyle \Delta S^o = S^0_{CO_2} - [S^o_{CO} + 0.5 S^o_{O_2}]$

$\displaystyle \Delta S^0 = 213.8 - [197.9 + 0.5 \times 205.8] = -87 \: J/K/mol$

The change in standard Gibbs free energy for the reaction is:

$\displaystyle \Delta G^o = \Delta H^0 -T\Delta S^o$

$\displaystyle \Delta G^o = −282840 - 298 \times (-87) = -256914 \: J/mol = -256.914 \: kJ/mol$

An ideal mono-atomic gas of given mass is heated at constant pressure. In this process, the fraction of supplied energy used for the increase of the internal energy of the gas is

0%
$\dfrac{3}{8}$
0%
$\dfrac{3}{5}$
0%
$\dfrac{3}{4}$
0%
$\dfrac{2}{5}$

Explanation

Let the change in temperature of the gas be

$\Delta T$ .

Thus, the amount of heat supplied under constant pressure is

$Q = C_P\Delta T$

The change in internal energy is

$\Delta U = C_V \Delta T$

Thus, the fraction of energy used for internal energy is

$\dfrac{\Delta U}{Q}= \displaystyle \frac{C_V}{C_P}$

For a mono-atomic ideal gas,

$C_V = 3R/2$ and

$C_P = 5R/2$

Thus,

$\displaystyle \frac{\Delta U}{Q} = \frac{C_V}{C_P} = \frac{3R/2}{5R/2}=\frac{3}{5}$

How much energy is needed to convert

$100\ g$ of ice at

$263\ K$ to liquid water at a temperature of

$283\ K$ ?

${C}_{ice}=0.49\ cal/({g}^{o}C)$

${C}_{water}=1.00\ cal/({g}^{o}C)$

$\Delta {H}_{fus}= 79.8\ cal/{g}$

$\Delta {H}_{vap}=540\ cal/g$

0%
$9470\ cal$
0%
$3288\ cal$
0%
$2288\ cal$
0%
$727.3\ cal$

Explanation

ice

$\rightarrow$ liquid water

${C}_{ice}=0.49\ cal/({g}^{o}C)$

${C}_{water}=1.00\ cal/({g}^{o}C)$

$\Delta {H}_{fus}= 79.8\ cal/{g}$

$\Delta {H}_{vap}=540\ cal/g$

ice

$\rightarrow$ melt (fusicm)

$\rightarrow$ water

$(263\,k)$ At

$0^{\circ}c(237\,k)$

$283\,k$

ENERGY

$\displaystyle = \int_{263}^{273}C_{ice}dt+\Delta H_{fus}+\int_{273}^{283}C_{water}dt$

$= 0.49(273-263)+79.8+1(283-273)$

$= 94.7\,cal\,g$

For 100 g ice energy

$= 94.7\times 100\,cal$

$= 9470\,cal$

1165071_526799_ans_c885e6ef080346219202af0d05f59a6c.jpg

$H_2(g)+Cl_2(g)=2HCl(g)$ ;

$\Delta H(298K)=-22.06$ kcal. For this reaction,

$\Delta U$ is equal to:

0%
$-22.06+2\times 10^{-3}\times 298\times 2$ kcal
0%
$-22.06+2\times 298$ kcal
0%
$-22.06-2\times 298\times 4$ kcal
0%
$-22.06$ kcal

Explanation

For a chemical reaction-

$\Delta H = \Delta E +\Delta n_{g} RT$

$\Delta E = \Delta E- \Delta n_{g} RT$

$\Delta E = -22.06Kcal$

As the

$\Delta n_{g} = 0$

$Mg(s) \rightarrow \rightarrow \rightarrow Mg^{2+}$ ; Energy = A KJ/mol.

$Cl_{2}(g) \rightarrow \rightarrow \rightarrow 2Cl^{-}$ ; Energy = B KJ/mol.
What is the

$\triangle H_{f}$ of

$MgCl_{2}$ if lattice enthalpy involved is C KJ/mol ?

0%
$A+B+C$ KJ/mol.
0%
$A-B+C$ KJ/mol.
0%
$A+B-C$ KJ/mol.
0%
$-(A+B+C)$ KJ/mol.

Explanation

$Mg(s)\rightarrow Mg(g)\rightarrow Mg^{ 2+ }(g)$ Energy = AkJ/mol

$Cl_{ 2 }\rightarrow 2Cl(g)\rightarrow 2Cl^{ 2- }(g)$ Energy = BkJ/mol

Adding two equations:

$Mg+{ Cl }_{ 2 }\rightarrow Mg{ Cl }_{ 2 }$

$U$ =CkJ/mol

$\Delta { H }_{ f }=\Delta { H }_{ sub }+IE+0.5\Delta { H }_{ diss }+{ E }_{ ea }+{ U }_{ lattice\quad enthalpy }$

$\Delta { H }_{ f }$ = A+B+C KJ/mol

What is the necessary condition for the sponataneity of a process?

0%
$\Delta S>0$
0%
$\Delta E<0$
0%
$\Delta H<0$
0%
$\Delta G<0$

Explanation

For spontaneity of any process

$\Delta G<0$ .

$\Delta S$ need not be necessarily greater than 0.

The equilibrium constant of a reaction is

$0.008$ at

$298$ K. The standard free energy change of the reaction at the same temperature is :

0%
$+11.96 \ kJ$
0%
$-11.96\ kJ$
0%
$-5.43 \ kJ$
0%
$-8.46 \ kJ$

Explanation

The equilibrium constant of a reaction is

$0.008$ at

$298$ K.
The standard free energy change of the reaction

$\Delta G^o = -RTln K = - 8.314 \times 298 \times ln \ 0.008 = 11962 \: J = 11.96 \: kJ$

Note:

$R$ is the ideal gas constant (8.314 J/mol/K) and

$T$ is the absolute temperature (

$298$ K). Hence, the standard free energy change of the reaction at the same temperature is

$+11.96\ kJ$ .

Which of the following conditions are true about spontaneous process?

0%
$\Delta (G_{system})_{T,P}<0$
0%
$\Delta S_{system} + \Delta S_{Surroundings} >0$
0%
$\Delta (G_{system})_{T,P}=0$
0%
$\Delta S_{system} + \Delta S_{Surroundings} <0$

Explanation

For spontaneous process,

$\Delta (G_{system})_{T,P}<0$ and

$\Delta S_{system} + \Delta S_{Surroundings} >0$

The process is spontaneous at the given temperature, if:

0%
$\Delta H$ is $+ve$ and $\Delta S$ is $-ve$
0%
$\Delta H$ is $-ve$ and $\Delta S$ is $+ve$
0%
$\Delta H$ is $+ve$ and $\Delta S$ is $+ve$
0%
$\Delta H$ is $+ve$ and $\Delta S$ is equal to zero

Explanation

Any process will be spontaneous if

$\Delta S=+ve$ and

$\Delta H=-ve$ and this condition make

$\Delta G=-ve$ . If

$\Delta G=\Delta H+T\Delta S$ is negative then the reaction is spontaneous.

Hence, option B is correct.

Calculate enthalpy for formation of ethylene from the following data:

(I)

$C(graphite) + O_2 (g) \rightarrow CO_2 (g); \ \ \ \Delta H = -393.5 kJ$
(II)

$H_2(g) + \dfrac{1}{2} O_2 (g) \rightarrow H_2O(l); \ \ \ \ \Delta H = - 286.2 kJ$
(III)

$C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2 O(l); \ \ \ \ \Delta H = - 1410.8 kJ$

0%
54.1 kJ
0%
44.8 kJ
0%
51.4 kJ
0%
48.4 kJ

Explanation

(I)

$C(graphite) + O_2 (g) \rightarrow CO_2 (g)$ ---- 1

(II)

$H_2(g) + \dfrac{1}{2} O_2 (g) \rightarrow H_2O(l)$ ----- 2

(III)

$C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2 O(l)$ ---- 3

Desired reaction:

$2C(grapite)+2H_2(g)\rightarrow C_2H_4(g)$

The desired above reaction can be obtained by multiplying eqn. 1 and 2 by 2 and adding them followed by subtracting eqn. 3 from the sum.

Enthalpy of formation can be calculated as:

$\Delta H_f=(2\times Reaction\ I+2\times Reaction\ II-Reaction\ III)$

$\Delta H_f=(-2\times393.5-2\times286.2+1410.8)=51.4\ kJ$

$1000\ K$ , from the data :

$N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}(g)$ ;

$\triangle H = -123.77\ kJ\ mol^{-1}$

Substance	$N_{2}$	$H_{2}$	$NH_{3}$
$P/R$	$3.5$	$3.5$	$4$

Calculate the heat of formation of

$NH_{3}$ at

$300\ K$ .

0%
$-44.42\ kJ\ mol^{-1}$
0%
$-88.85\ kJ\ mol^{-1}$
0%
$+ 44.42\ kJ\ mol^{-1}$
0%
$+ 88.85\ kJ\ mol^{-1}$

Explanation

Fro, Kirchhoff's equation

$\triangle H_{2}(1000\ K) = \triangle H_{1}(300\ K) + \triangle C_{p}(1000 - 300)$

Here,

$\triangle H_{2} (1000\ K) = -123.77\ kJ\ mol^{-1}$

$\triangle H_{1} (300\ K) = ?$

$\triangle C_{p} = 2C_{p}(NH_{3}) - [C_{p}(N_{2}) + 3C_{p}(H_{2})]$

$= -6R$

$= -6\times 8.314 \times 10^{-3}kJ$

$\therefore -123.77 = \triangle H_{1} (300\ K) + 6\times 8.314\times 10^{-3} \times 700$

$\triangle H_{1} (300\ K) = -88.85\ kJ$

For two moles of

$NH_{3}$

$\therefore \triangle H_{f}(NH_{3}) = \dfrac {\triangle H_{1}(300 K)}{2}$

$= -\dfrac {88.85}{2}$

$= -44.42\ kJ\ mol^{-1}$

For

$A\rightarrow B,\, \Delta H=4\,\text{kcal mol}^{-1}$ ,

$\Delta S=10 \,\text{cal mol}^{-1}K^{-1}$ . Reaction is spontaneous when temperature is:

0%
$400$ K
0%
$300$ K
0%
$500$ K
0%
None of these

Explanation

For a spontaneous reaction

$\Delta H-T\Delta S<0$

Reaction will be spontaneous if

$T>\dfrac{\Delta H}{\Delta S}$

Reaction will be spontaneous if

$T>\dfrac{4000}{10}$

$T>400\ K$

For spontaneity of cell, which is correct?

0%
$\Delta G=0, \ \Delta H=0$
0%
$\Delta G=-ve, \ \Delta H=0$
0%
$\Delta G=+ve,\ \Delta H=0$
0%
$\Delta G=-ve$

Explanation

For any spontaneous process, Gibbs free energy should be negative.

$\Delta G <0$ and

$E_{cell}$ should be positive for this case.

Given that the bond energies of

$:N\equiv N$ is

$946$ kJ

$mol^{-1}$

$H-H$ is

$435$ kJ

$mol^{-1}$ ,

$N-N$ is

$159$ kJ

$mol^{-1}$ , and

$N-H$ is

$389$ kJ

$mol^{-1}$ , the heat of formation of hydrazine in the gas phase in kJ

$mol^{-1}$ is:

0%
$833$
0%
$101$
0%
$334$
0%
$1264$

Explanation

$N_2+2H_2\rightarrow N_2H_4$

$\Delta H_f=B.E(reactants)-B.E(products)$

$\Delta H_f=946+2\times435-(4\times389+159)=101\ kJ/mol$

Hence option

$B$ is correct.

A heat engine operating between 227 deg C and 27 deg C absorbs 1 kcal of heat from the 227 deg C reservoir per cycle. Calculate
(1) the amount of heat discharged into the low temperature reservoir.
(2) the amount of work done per cycle.
(3) the efficiency of cycle.

0%
0.4 kcal, 0.6 kcal, 40%
0%
0.6 kcal, 0.4 kcal, 40%
0%
0.4 kcal, 0.6 kcal, 60%
0%
0.7 kcal, 0.4 kcal, 40%

Explanation

Calculation of work done:

$\displaystyle T' = 227+273=500$ K

$\displaystyle T=27+273=300$ K

$\displaystyle q'=1$ 1 kcal

$\displaystyle \dfrac {w}{q'} = \dfrac {T'-T}{T'}$

$\displaystyle w=q(\dfrac {T'-T}{T'})$
Putting the various values in the above reaction

$\displaystyle w=1(\dfrac {500-300}{500})=1 \times \dfrac {200}{500}=0.4$ kcal.

Calculation of heat discharged q

$\displaystyle q'-q=w$

$\displaystyle 1-q=0.4$

$\displaystyle q=0.6$ kcal

Calculation of efficiency

$\displaystyle \eta = \dfrac {w}{q'}=0.4$ kcal or 40%

For a system in equilibrium,

$\Delta G = 0$ , under conditions of constant ________.

0%
temperature and pressure
0%
temperature and volume
0%
pressure and volume
0%
energy and volume

Explanation

For a system in equilibrium,

$\Delta G=0$ , under conditions of constant temperature and pressure.

$\Delta G= \Delta G^o + RTlnK$

When,

$\Delta G=0$ ,

$\Delta G^o =- RTlnK$

Given,

$C(s)+{O}_{2}(g)\rightarrow {CO}_{2}(g); \Delta H=-395\ kJ$

$S(s)+{O}_{2}(g)\rightarrow {SO}_{2}(g);\Delta H=-295\ kJ$

$C{S}_{2}(l)+3{O}_{2}(g)\rightarrow {CO}_{2}(g)+2{SO}_{2}(g);\Delta H=-1110\ kJ$

The heat of formation of

$C{S}_{2}(l)$ is:

0%
$250\ kJ$
0%
$62.5\ kJ$
0%
$31.25\ kJ$
0%
$125\ kJ$

Explanation

Since, the formation reaction of

$CS_2$ is,

$C(s) + 2S(s) \rightarrow CS_2(l)$

Since, the formation reaction of

$CS_2$ can also be derived using the given reaction as,

$1^{st} \text{reaction} +2\times 2^{nd} \text{ reaction} - 3^{rd} \text{ reaction}$

Hence, the enthalpy of the reaction,

$[1^{st} \text{reaction} +2\times 2^{nd} \text{ reaction} - 3^{rd} \text{ reaction}]$

$=\Delta H_{1^{st} \text{reaction}} +2\times \Delta H_{2^{nd} \text{ reaction}} -\Delta H_{3^{rd} \text{ reaction}}$

$=\text{heat of formation of CS}_2(l)$

Hence,

$\Delta H_{f(CS_2)} = -395 + 2\times (-295) - (-1110) = 125kJ$

Hence, the answer is option

$D$ .

What will be the heat formation of methane; if the heat of combustion of carbon is '

$-x$ '

$kJ$ , heat of formation of water is '

$-y$ '

$kJ$ and heat of combustion of methane is '

$-z$ '

$kJ$ ?

0%
$(-x-y-z)kJ$
0%
$(-z-x+2y)kJ$
0%
$(-x-2y-z)kJ$
0%
$(-x-2y+z)kJ$

Explanation

Formation of methane

$\Rightarrow \quad C+2{ H }_{ 2 }O\rightarrow { CH }_{ 4 }+{ O }_{ 2 }$

$\therefore$ Heat of formation of methane

$={ \Delta }_{ c }\left( C \right) -{ \Delta }_{ c }\left( { CH }_{ 4 } \right) -2\times { \Delta }_{ f }\left( { H }_{ 2 }O \right)$

$=-x-\left( -z \right) -2y$

$=\left( -x+z-2y \right) kJ$

$=(-x-2y+z)kJ$

Which one of the following is not applicable for a thermochemical equation?

0%
It tell about physical state of reactants and products
0%
It tells whether the reaction is spontaneous
0%
It tells whether the reaction is exothermic or endothermic
0%
It tells about the allotropic form (if any) of the reactants

Super cooled water is liquid water that has been cooled below its normal freezing point. This state is thermodynamically :

0%
unstable and tends to freeze into ice spontaneously
0%
stable and tends to freeze Into ice spontaneously
0%
stable and tends to fuse into liquid spontaneously
0%
unstable and tends to fuse into liquid spontaneously

Explanation

Super cooled water is liquid water that has been cooled below its normal freezing point. This state is thermodynamically unstable and tends to freeze into ice spontaneously. Below normal freezing point, water cannot exist in liquid state. It has to exist in solid state. Due to this, the conversion liquid water

$\displaystyle \rightarrow$ solid ice is spontaneous.

${H}_{2}+\cfrac{1}{2}{O}_{2}\rightarrow {H}_{2}O;\Delta H=-68.09kcal$

$K+{H}_{2}O+water \rightarrow KOH(aq.)+\cfrac{1}{2}{H}_{2};\Delta H=-48.0kcal$

$KOH+water\rightarrow KOH(aq); \Delta H=-14.0kcal$
the heat of formation of

$KOH$ is:

0%
$-68.39+48-14.0$
0%
$-68.39-48.0+14.0$
0%
$+68.39-48.0+14.0$
0%
$+68.39+48.0-14.0$

Explanation

(i)

${ H }_{ 2 }+\dfrac { 1 }{ 2 } { O }_{ 2 }\rightarrow { H }_{ 2 }O\quad \quad \quad ;\quad { \Delta H }_{ f }=-68.09Kcal$

(ii)

$K+{ H }_{ 2 }O+water\rightarrow KOH\left( aq \right) +\dfrac { 1 }{ 2 } { H }_{ 2 }\quad ;\quad { \Delta H }_{ f }=-48Kcal$

(iii)

$KOH+water\rightarrow KOH\left( aq \right) \quad \quad ;\quad { \Delta H }_{ f }=-14.0Kcal$

(i) + (ii) - (iii) gives heat of formation of

$KOH$

$\therefore$ Heat of formation of

$KOH=\left( -68.09-48.0+14.0 \right) Kcal$

From the following reactions at

$298\ K$ .

(A)

$CaC_{2}(s) + 2H_{2}O(l) \rightarrow Ca(OH)_{2}(s) + C_{2}H_{2} (g);\ \Delta H^{\circ}= - 127.9\ kJ\ mol^{-1}$

(B)

$Ca(s) + \dfrac {1}{2} O_{2}(g) \rightarrow CaO(s) ;\ \Delta H=- 635.1kJ\ mol^{-1}$

(C)

$CaO(s) + H_{2}O(l) \rightarrow Ca(OH)_{2}(s);\ \Delta H=- 65.2\ kJ\ mol^{-1}$

(D)

$C(s) + O_{2}(s) \rightarrow CO_{2}(s) ;\ \Delta H=- 393.5\ kJ\ mol^{-1}$

(E)

$C_{2}H_{2}(g) + \dfrac {5}{2}O_{2}(g) \rightarrow 2CO_{2}(g) + H_{2}O(l);\ \Delta H= - 1299.58\ kJ\ mol^{-1}$

Calculate the heat of formation of

$CaC_{2}(s)$ at

$298\ K$ .

0%
$-59.82\ kJ\ mol^{-1}$
0%
$+59.82\ kJ\ mol^{-1}$
0%
$-190.22\ kJ\ mol^{-1}$
0%
$+190.22\ kJ\ mol^{-1}$

Standard heats of formation for

$C{Cl}_{4},{H}_{2}O,{CO}_{2}$ and

$HCl$ at

$298K$ are

$-25.5,-57.8,-94.1$ and

$-22.1kJ/mol$ respectively.
For the reaction, what will be

$\Delta H$ ?

$C{Cl}_{4}+2{H}_{2}O\rightarrow {CO}_{2}+4HCl$

0%
$36.4\ kJ$
0%
$20.7\ kJ$
0%
$-20.7\ kJ$
0%
$-41.4\ kJ$

Explanation

For the given reaction,

$CCl_4 +2H_2O \rightarrow CO_2 + 4HCl$

$\Delta H = \Delta H_{f(CO_2)} +$

$4\times$

$\Delta H_{f(HCl)}$ -

$\Delta H_{f(CCl_4)}$ -

$2\times$

$\Delta H_{f(H_2O)}$
=>

$\Delta H = -94.1 + 4\times (-22.1) - (-25.5) - 2\times (-57.8) = -41.4 KJ$
Hence, answer is option

$D$ .

$S+{O}_{2}\rightarrow {SO}_{2}; \Delta H=-298.2kJ$

${ SO }_{ 2 }+\cfrac { 1 }{ 2 }{O}_{2} \rightarrow { SO_3 }; \Delta H=-98.7kJ$

${SO}_{3}+{H}_{2}O\rightarrow {H}_{2}{SO}_{4};\Delta H=-130.2kJ$

${H}_{2}+\cfrac { 1 }{ 2 } { O }_{ 2 }\rightarrow {H}_{2}O;\Delta H=-227.3kJ$
The heat of formation of

${H}_{2}{SO}_{4}$ will be:

0%
$-754.4kJ$
0%
$+320.5kJ$
0%
$-650.3kJ$
0%
$-433.7kJ$

Explanation

Since, on adding all the 4 reaction given in question we will get a final reaction,

$S + 2O_2 + H_2\rightarrow H_2SO_4$
Which is basically the formation reaction of

$H_2SO_4$
Hence, the heat of formation of

$H_2SO_4$ = sum of heat of reactions of all four reaction, i.e,

$\Delta H_{f(H_2SO_4)} = \Delta H_1 + \Delta H_2 + \Delta H_3 +\Delta H_4 = -298.2 -98.7 - 130.2 -227.3 = -754.4KJ.$
hence, answer is option

$A$ .

$\Delta {H}_{f(x)},\Delta {H}_{f(y)},\Delta {H}_{f(R)}$ and

$\Delta {H}_{f(S)}$ denote the enthalpies of formation of

$x,y,R$ and

$S$ respectively. The enthalpy of the reaction

$x+y\rightarrow R+S$ is:

0%
$\Delta {H}_{f(x)}+\Delta {H}_{f(y)}$
0%
$\Delta {H}_{f(R)}+\Delta {H}_{f(S)}$
0%
$\Delta {H}_{f(x)}+\Delta {H}_{f(y)}-\Delta {H}_{f(R)}-\Delta {H}_{f(S)}$
0%
$\Delta {H}_{f(R)}+\Delta {H}_{f(S)}-\Delta {H}_{f(x)}-\Delta {H}_{f(y)}$

Explanation

For any reaction the enthalpy change of the reaction is given by,

$\Delta H = \text{sum of heat of formation of products in stoichiometric ratio - sum of heat of formation of reactants in stoichiometric ratio}$
i.e, for any reaction

$aA + bB \rightarrow cC + dD$

$\Delta H = c\Delta H_{f(C)} + d\Delta H_{f(D)} - a\Delta H_{f(A)} - b\Delta H_{f(B)}$
Hence, for the given reaction,

$x+y\rightarrow R+S$ ,

$\Delta H = \Delta H_{f(R)} + \Delta H_{f(S)} - \Delta H_{f(x)} - \Delta H_{f(y)}$
Hence, answer is option

$D$ .

What are the most favourable conditions for the reaction;

${ SO }_{ 2 }(g)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)\leftrightharpoons { SO }_{ 3 }(g);\Delta { H }^{ o }=-ve$ to occur?

0%
Low temp and high press
0%
Low temp and low press
0%
High temp and low press
0%
High temp and high press

Explanation

Negative value of enthalpy change suggests exothermic reaction. Such reactions are favored at low temperature as heat released during the reaction raised the temperature and nullifies the effect of low temperature.

$\displaystyle \Delta n = \text { number of moles of gaseous products } - \text { number of moles of gaseous reactants }$

$\displaystyle \Delta n = 1 - (1+ \dfrac {1 }{ 2 } )$

$\displaystyle \Delta n =-\dfrac {1 }{ 2 }$
Since

$\displaystyle \Delta n < 0$ , the reaction is favored at high pressure. Increasing the pressure will shift the reaction towards forward direction (one with less number of moles of gaseous species). This will nullify the effect of increase in pressure.

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 11 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 11 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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