Thermodynamics — Page 7 | JEE Chemistry

Q61

The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is : (Specific heat of water liquid and water vapour are 4.2 kJ K

-

1 kg

-

1 and 2.0 kJ K

-

1 kg

-

1; heat of liquid fusion and vapourisation of water are 334 kJ

-

1 and 2491 kJ kg

-

1, respectively). (log 273 = 2.436, log 373 = 2.572, log 383 = 2.583)

A 7.90 kJ kg

-

1 K

-

1

B 2.64 kJ kg

-

1 K

-

1

C 8.49 kJ kg

-

1 K

-

1

D 9.26 kJ kg

-

1 K

-

1

Correct Answer

Option D

Solution

\Delta {S_1} = {{\Delta {H_{fusion}}} \over {273}} = {{334} \over {273}} = 1.22

\Delta {S_2} = 4.2\ell N\left( {{{363} \over {273}}} \right) = 1.31

\Delta {S_3} = {{\Delta {H_{vap}}} \over {373}} = {{2491} \over {373}} = 6.67

\Delta {S_4} = 2.0\ell n\left( {{{383} \over {373}}} \right) = 0.05

\Delta {S_{total}} = 9.26\,

kJ Kg $-$ 1 K $-$ 1

Q62

A process has

\Delta

H = 200 J mol–1 and

\Delta

S = 40 JK–1 mol–1. Out of the values given below, choose the minimum temperature above which the process will be spontaneous :

A 4 K

B 20 K

C 5 K

D 12 K

Correct Answer

Option C

Solution

\Delta

G =

\Delta

H $-$ T

\Delta

S T =

{{\Delta H} \over {\Delta S}} = {{200} \over {40}} = 5K

Q63

For the chemical reaction X

\rightleftharpoons

Y, the standard reaction Gibbs energy depends on temperature T (in K) as

\Delta

rGo (in kJ mol–1) = 120

- {3 \over 8}

T. The major component of the reaction mixture at T is :

A Y if T = 300 K

B Y if T = 280 K

C X if T = 350 K

D X if T = 315 K

Correct Answer

Option D

Solution

X $\rightleftharpoons$ Y Keq =

{{\left[ Y \right]} \over {\left[ X \right]}}

If Keq > 1 $\Rightarrow$ [Y] > [X] and Keq < 1 $\Rightarrow$ [Y] < [X] We know,

\Delta

Go = -RT ln(Keq) So when Keq > 1 then

\Delta

Go < 0 and Y is major. And when Keq < 1 then

\Delta

Go > 0 and X is major. Temperature at which Go = 0 is 120

- {3 \over 8}

T = 0 $\Rightarrow$ T = 320 K For T > 320 K

\Delta

Go < 0 and Y is major. For T < 320 K

\Delta

Go > 0 and X is major.

Q64

Two blocks of the same metal having same mass and at temperature T1 and T2, respectively, are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy,

\Delta

S, for this process is :

A 2Cp In

\left[ {{{{{\left( {{T_1} + {T_2}} \right)}^{{1 \over 2}}}} \over {{T_1}{T_2}}}} \right]

B 2Cp In

\left[ {{{\left( {{T_1} + {T_2}} \right)} \over {2{T_1}{T_2}}}} \right]

C Cp In

\left[ {{{{{\left( {{T_1} + {T_2}} \right)}^2}} \over {4{T_1}{T_2}}}} \right]

D 2Cp In

\left[ {{{\left( {{T_1} + {T_2}} \right)} \over {4{T_1}{T_2}}}} \right]

Correct Answer

Option C

Solution

When two blocks comes in contact with each other and attain thermal equilibrium then final temperature of the blocks, Tf =

{{{T_1} + {T_2}} \over 2}

\Delta

S =

\Delta

S1 +

\Delta

S2 = Cp

\ln \left( {{{{T_f}} \over {{T_1}}}} \right)

+ Cp

\ln \left( {{{{T_f}} \over {{T_2}}}} \right)

= Cp

\ln \left( {{{{T_1} + {T_2}} \over {2{T_1}}}} \right)

+ Cp

\ln \left( {{{{T_1} + {T_2}} \over {2{T_2}}}} \right)

= Cp

\ln \left( {{{{{\left( {{T_1} + {T_2}} \right)}^2}} \over {4{T_1}{T_2}}}} \right)

Q65

For the equilibrium, 2H2O

\rightleftharpoons

H3O+ + OH

-

, the value of

\Delta

Go at 298 K is approximately :

A

-

80 kJ mol–1

B 100 kJ mol

-

1

C

-

100 kJ mol

-

1

D 80 kJ mol–1

Correct Answer

Option D

Solution

2H2O $\rightleftharpoons$ H3O+ + OH $-$ Here Keq = Kw(H2O) At 298 K, Kw(H2O) = 10-14 $\therefore$ Keq = 10-14

\Delta

Go = -RTln(Keq) = -RTln(Kw) = -2.303RT log (10-14) = -2.303 $\times$ 8.314 $\times$ 298 $\times$ $\times$ (-14) = 80 kJ/mol

Q66

The reaction, MgO(s) + C(s)

\to

Mg(s) + CO(g), for which

\Delta

rHo + 491.1 kJ mol–1 and

\Delta

rSo = 198.0 JK–1 mol–1, is not feasible at 298 K. Temperature above which reaciton will be feasible is :

A 2480.3 K

B 2040.5 K

C 2380.5 K

D 1890.0 K

Correct Answer

Option A

Solution

We know,

\Delta

Go =

\Delta

Ho - T

\Delta

So For a reaction to be spontaneous

\Delta

Go must be negative i.e., T

\Delta

So >

\Delta

Ho $\Rightarrow$ T >

{{\Delta H^\circ } \over {\Delta S^\circ }}

$\Rightarrow$ T >

{{491.1 \times 1000} \over {198}}

$\Rightarrow$ T > 2480.3 K

Q67

The standard reaction Gibbs energy for a chemical reaction at an absolute temperature T is given by

\Delta

rGo = A – BT Where A and B are non-zero constants. Which of the following is TRUE about this reaction?

A Exothermic if B < 0

B Endothermic if A > 0

C Exothermic if A > 0 and B < 0

D Endothermic if A < 0 and B > 0

Correct Answer

Option B

Solution

\Delta

Go =

\Delta

Ho - T

\Delta

So Given that A and B are non-zero constants, i.e., A =

\Delta

Ho, B =

\Delta

So If

\Delta

Ho is positive means reaction is endothermic.

Q68

For a reaction, 4M(s) + nO2(g)

\to

2M2On(s) the free energy change is plotted as a function of temperature. The temperature below which the oxide is stable could be inferred from the plot as the point at which :

A the free energy change shows a change from negative to positive value

B the slope changes from positive to negative

C the slope changes from negative to positive

D the slope changes from positive to zero

Correct Answer

Option A

Solution

\Delta

G =

\Delta

H – T

\Delta

S

\Delta

G = –ve (stable oxide)

\Delta

G = +ve (unstable oxide)

Q69

Five moles of an ideal gas at 1 bar and 298 K is expanded into vacuum to double the volume. The work done is :

A Zero

B -RT

\ln {{{V_2}} \over {{V_1}}}

C CV (T2 – T1)

D – RT (V2 – V1)

Correct Answer

Option A

Solution

As the expansion is done in vacuum that is in absence of pext so pext = 0 $\therefore$ W = - pext

\Delta

V = 0

Q70

The process that is NOT endothermic in nature is :

A Ar(g) + e-

\to

Ar-(g)

B H(g) + e-

\to

H-(g)

C Na(g)

\to

Na+(g) + e-

D O-(g) + e-

\to

O2-(g)

Correct Answer

Option B

Solution

H(g) + e- $\to$ H-(g) is exothermic rest of all endothermic process.