Chemical Thermodynamics MCQ Questions & Answers in Physical Chemistry | Chemistry
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41.
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 temperatures of involved substances
D
the physical states of reactants and products
Answer :
the nature of intermediate reaction steps
Enthalpy change for a reaction does not depend upon the nature of intermediate reaction steps.
42.
Heat of neutralization of a strong acid $$HA$$ and a weaker acid $$HB$$ with $$KOH$$ are $$ - 13.7$$ and $$ - 12.7\,k\,cal\,mo{l^{ - 1}}.$$ When $$1\,mole$$ of $$KOH$$ was added to a mixture containing $$1\,mole$$ each of $$HA$$ and $$HB,$$ the heat change was $$ - 13.5\,kcal.$$ In what ratio is the base distributed between $$HA$$ and $$HB.$$
A
3 : 1
B
1 : 3
C
4 : 1
D
1 : 4
Answer :
4 : 1
Let $$x$$ mole of $$KOH$$ be neutralized by the strong acid $$HA.$$ Then, moles neutralized by $$HB = 1 - x$$
Hence, $$ - 13.7 \times x + \left( { - 12.7} \right) \times \left( {1 - x} \right) = - 13.5$$
$$ \Rightarrow \,x = 0.8;\,\,\frac{x}{{1 - x}} = \frac{{0.8}}{{0.2}} = \frac{4}{1} = 4:1$$
43.
If the heat change at constant volume for decomposition of silver oxide is $$80.25\,kJ,$$ what will be the heat change at constant pressure ?
44.
The pressure-volume work for an ideal gas can be calculated by using the expression $$W = - \int\limits_{{V_t}}^{{V_f}} {{P_{ex}}\,dV.} $$ The work can also be calculated from the $$PV - $$ plot by using the area under the curve within the specified limits. When an ideal gas is compressed (i) reversibly or (ii) irreversibly from volume $${V_i}$$ to $${V_f}.$$ Choose the correct option.
A
$${W_{{\text{reversible}}}} = {W_{{\text{irreversible}}}}$$
B
$${W_{{\text{reversible}}}} < {W_{{\text{irreversible}}}}$$
C
$${W_{{\text{reversible}}}} > {W_{{\text{irreversible}}}}$$
D
$${W_{{\text{reversible}}}} = {W_{{\text{irreversible}}}} + {P_{ex}}\,.\Delta V$$
In both the curves, work done on the gas is represented by the shaded area and the area under the curve is always more in irreversible compression.
Hence, $${W_{{\text{irreversible}}}} > {W_{{\text{reversible}}}}$$
45.
Dissolution of ammonium chloride in water is an endothermic reaction, yet it is a spontaneous process. This is due to the fact that
A
$$\Delta H\,\,{\text{is}}\,\, + ve,\Delta S\,\,{\text{is}}\,\, - ve$$
B
$$\Delta H\,\,{\text{is}}\,\, - ve,\Delta S\,\,{\text{is}}\,\, + ve$$
C
$$\Delta H\,\,{\text{is}}\,\, + ve,\Delta S\,\,{\text{is}}\,\, + ve\,\,$$ $${\text{and}}\,\,\Delta H < T\Delta S$$
D
$$\Delta H\,\,{\text{is}}\,\, + ve\,\,{\text{and}}\,\,\Delta H > T\Delta S$$
$$\Delta G$$ will be negative under these conditions.
46.
For a reaction : $$X \to Y + Z$$
Absolute entropies are $$X = 120\,J\,{K^{ - 1}}\,mo{l^{ - 1}},$$ $$Y = 213.8\,J\,{K^{ - 1}}\,mo{l^{ - 1}}$$ and $$Z = 197.9\,J\,{K^{ - 1}}\,mo{l^{ - 1}}.$$
What will be the entropy change at $$298\,K$$ and $$1\,atm?$$
47.
The value for $$\Delta U$$ for the reversible isothermal evaporation of $$90\,g$$ water at $$100{\,^ \circ }C$$ will be $$\left( {\Delta {H_{{\text{evap}}}}} \right.$$ of water $$ = 40.8\,kJ\,mo{l^{ - 1}},$$ $$\left. {R = 8.314\,J\,{K^{ - 1}}\,mo{l^{ - 1}}} \right)$$
48.
At what temperature liquid water will be in equilibrium with water vapour ?
$$\Delta {H_{{\text{vap}}}} = 40.73\,kJ\,mo{l^{ - 1}},$$ $$\Delta {S_{{\text{vap}}}} = 0.109\,kJ\,{K^{ - 1}}\,mo{l^{ - 1}}$$
A
282.4$$\,K$$
B
373.6$$\,K$$
C
100$$\,K$$
D
400$$\,K$$
Answer :
373.6$$\,K$$
$${\text{At equilibrium}}\,\,\Delta G = 0\,\,$$ $${\text{for}}\,\,\Delta G = \Delta H - T\Delta S$$
$${\text{or}}\,\,T = \frac{{\Delta H}}{{\Delta S}} = \frac{{40.73}}{{0.109}} = 373.6\,K$$
49.
For which of the process, $$\Delta S$$ is negative ?
A
$${H_{2\left( g \right)}} \to 2{H_{\left( g \right)}}$$
B
$${N_{2\left( {g,1\,atm} \right)}} \to {N_{2\left( {g,8\,atm} \right)}}$$
C
$$2S{O_{3\left( g \right)}} \to 2S{O_{2\left( g \right)}} + {O_{2\left( g \right)}}$$
D
$${C_{{\text{(diamond)}}}} \to {C_{{\text{(graphite)}}}}$$