Practice Start Test

Thermodynamics

JEE Chemistry · 103 questions · Page 3 of 11 · Click an option or "Show Solution" to reveal answer

Q21
(ΔH−ΔU\Delta H - \Delta U) for the formation of carbon monoxide (CO) from its elements at 298 K is : (R = 8.314 J K–1 mol–1)
A –1238.78 J mol–1
B 1238.78 J mol–1
C –2477.57 J mol–1
D 2477.57 J mol–1
Correct Answer
Option B
Solution

For the reaction,

C(s)+12O2(g)→CO{C_{\left( s \right)}} + {1 \over 2}{O_{2\left( g \right)}} \to CO
ΔH=ΔU+ΔnRT\Delta H = \Delta U + \Delta nRT

or

     ΔH−ΔU=ΔnRT\,\,\,\,\,\Delta H - \Delta U = \Delta nRT
Δn=1−12=12;  \Delta n = 1 - {1 \over 2} = {1 \over 2};\,\,
ΔH−ΔU=12×8.314×298\Delta H - \Delta U = {1 \over 2} \times 8.314 \times 298
=1238.78 Jmol−1= 1238.78\,Jmo{l^{ - 1}}
Q22
For a particular reversible reaction at temperature T, ∆H and ∆S were found to be both +ve. If Te is the temperature at equilibrium, the reaction would be spontaneous when :
A Te > T
B T > Te
C Te is 5 times T
D T = Te
Correct Answer
Option B
Solution

At equilibrium

ΔG=0\Delta G = 0

Hence,

ΔG=ΔH−TeΔS=0\Delta G = \Delta H - {T_e}\Delta S = 0

∴\therefore

      ΔH=TeΔS   \,\,\,\,\,\,\Delta H = {T_e}\Delta S\,\,\,

or

    Te=ΔHΔS  \,\,\,\,{T_e} = {{\Delta H} \over {\Delta S}}\,\,

For a spontaneous reaction

ΔG\Delta G

must be negative which is possible only if

ΔH<TΔS\Delta H < T\Delta S

or

   T>ΔHΔS;Te<T\,\,\,T > {{\Delta H} \over {\Delta S}};{T_e} < T
Q23
A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0oC. As it does so, it absorbs 208J of heat. The values of q and w for the process will be : (R = 8.314 J/mol K) ( l n 7.5 = 2.01)
A q = – 208 J, w = – 208 J
B q = – 208 J, w = + 208 J
C q = + 208 J, w = + 208 J
D q = + 208 J, w = – 208 J
Correct Answer
Option D
Solution

By Ist law of thermodynamics, q =

Δ\Delta

E – W At const T, ∆E = 0 q = – W Heat absorbed = 208 J ∴\therefore q = +208 J W = – 208 J

Q24
For the reactions 2C + O2 →\to 2CO2; ΔH\Delta H = -393 J 2Zn + O2 →\to 2ZnO; ΔH\Delta H = -412 J
A carbon can oxidise Zn
B oxidation of carbon is not feasible
C oxidation of Zn is not feasible
D Zn can oxidise carbon
Correct Answer
Option D
Solution
ΔH\Delta H

negative shows that the reaction is spontaneous. Higher negative value for

ZnZn

shows that the reaction is more feasible.

Q25
For the reaction, A(g) + B(g) → \to C(g) + D(g), Δ\Delta Ho and Δ\Delta So are, respectively, − 29.8 kJ mol−1 and −0.100 kJ K−1 mol−1 at 298 K. The equilibrium constant for the reaction at 298 K is :
A 1.0 × \times 10−-10
B 1.0 × \times 1010
C 10
D 1
Correct Answer
Option D
Solution

We have

Δ\Delta

G

∘^\circ

=

Δ\Delta

H

∘^\circ

−- T

Δ\Delta

S

∘^\circ
Δ\Delta

G

∘^\circ

= −- 29.8 −- 298 (−- 0.1) = 0

Δ\Delta

G

∘^\circ

= −- 2.302 RT log Keq log Keq = 0 ⇒\Rightarrow Keq = 1

Q26
A heat engine absorbs heat Q1 at temperature T1 and heat Q2 at temperature T2. Work done by the engine is J (Q1 + Q2). This data :
A violates 1st law of thermodynamics
B violates 1st law of thermodynamics if Q1 is +ve
C violates 1st law of thermodynamics if Q1 is -ve
D does not violates 1st law of thermodynamics
Correct Answer
Option A
Solution

According to first law of thermodynamics energy can neither be created nor destroyed although it can be converted from one form to another.

NOTE : Carnot cycle is based upon this principle but during the conversion of heat into work some mechanical energy is always converted to other form of energy hence this data violates

11

st. law of thermodynamics.

Q27
The internal energy change when a system goes from state A to B is 40 kJ/mole. If the system goes from A to B by a reversible path and returns to state A by an irreversible path what would be the net change in internal energy?
A > 40 kJ
B < 40 kJ
C Zero
D 40 kJ
Correct Answer
Option C
Solution

For a cyclic process the net change in the internal energy is zero because the change in internal energy does not depend on the path.

Q28
The enthalpy change for a reaction does not depend upon :
A use of different reactants for the same product
B the nature of intermediate reaction steps
C the differences in initial or final temperature of involved substances
D the physical states of reactants and products
Correct Answer
Option B
Solution

Enthalpy change for a reaction does not depend upon the nature of intermediate reaction steps.

Q29
If at 298 K the bond energies of C - H, C - C, C = C and H - H bonds are respectively 414, 347, 615 and 435 kJ/mol, the value of enthalpy change for the reaction H2C = CH2(g) + H2(g) →\to H3C - CH3(g) at 298 K will be :
A - 250 kJ
B + 125 kJ
C - 125 kJ
D + 250 kJ
Correct Answer
Option C
Solution
CH2=CH2(g)+H2(g)→CH3−CH3C{H_2} = C{H_2}\left( g \right) + {H_2}\left( g \right) \to C{H_3} - C{H_3}

Enthalpy change == Bond energy of reactants −- Bond energy of products.

ΔH=1(C=C)+4(C−H)+1(H−H)−\Delta H = 1\left( {C = C} \right) + 4\left( {C - H} \right) + 1\left( {H - H} \right) -
                           1(C−C)−6(C−H)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\left( {C - C} \right) - 6\left( {C - H} \right)
=1(C=C)+1(H−H)−1(C−C)−2(C−H)= 1\left( {C = C} \right) + 1\left( {H - H} \right) - 1\left( {C - C} \right) - 2\left( {C - H} \right)
=615+435−347−2×414= 615 + 435 - 347 - 2 \times 414
=1050−1175= 1050 - 1175
=−125 kJ.= - 125\,kJ.
Q30
In an irreversible process taking place at constant T and P and in which only pressure-volume work is being done, the change in Gibbs free energy (dG) and change in entropy (dS), satisfy the criteria :
A (dS)V, E > 0, (dG)T, P < 0
B (dS)V, E = 0, (dG)T, P = 0
C (dS)V, E = 0, (dG)T, P > 0
D (dS)V, E < 0, (dG)T, P < 0
Correct Answer
Option A
Solution

For spontaneous reaction,

dS>0dS > 0

and

dGdG

should be negative i.e.

<0.< 0.
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