Chemical Thermodynamics MCQ Questions & Answers in Physical Chemistry | Chemistry
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121.
Standard Gibb's free energy change for the isomerisation reaction $$cis{\text{ - }}2{\text{ - }}pentene \rightleftharpoons trans{\text{ - }}2{\text{ - }}pentene$$ is $$ - 3.67\,kJ/mol$$ at $$400\,K.$$ If more $$trans{\text{ - }}2{\text{ - }}pentene$$ is added to the reaction vessel, then
A
more $$cis{\text{ - }}2{\text{ - }}pentene$$ is formed
B
equilibrium remains unaffected
C
additional $$trans{\text{ - }}2{\text{ - }}pentene$$ is formed
D
equilibrium is shifted in forward direction
Answer :
more $$cis{\text{ - }}2{\text{ - }}pentene$$ is formed
According to Le-Chatelier’s principle, when we increase the concentration of $$trans{\text{ - }}2{\text{ - }}pentene,$$ then the reaction shifts in backward direction and hence,
the concentration of $$cis{\text{ - }}2{\text{ - }}pentene$$ increase to maintain the equilibrium constant $$K$$ constant at given temperature.
122.
The standard enthalpies of formation of $$C{o_2}\left( g \right),$$ $${H_2}O\left( l \right)$$ and glucose $$(s)$$ at $${25^ \circ }C$$ are $$ - 400\,kJ/mol,$$ $$ - 300kJ/mol$$ and $$ - 1300\,kJ/mol,$$ respectively. The standard enthalpy of combustion per gram of glucose at $${25^ \circ }C$$ is
A
$$+2900 kJ$$
B
$$-2900 kJ$$
C
$$-16.11 kJ$$
D
$$+16.11 kJ$$
Answer :
$$-16.11 kJ$$
The standard enthalpy of the combustion of glucose can be calculated by the eqn.
$$\eqalign{
& {C_6}{H_{12}}{O_6}\left( s \right) + 6{O_2}\left( g \right) \to 6C{O_2}\left( g \right) + 6{H_2}O\left( l \right) \cr
& \Delta {H_C} = 6 \times \Delta {H_f}\left( {C{O_2}} \right) + 6 \times \Delta {H_f}\left( {{H_2}O} \right) - \Delta {H_f}\left[ {{C_6}{H_{12}}{O_6}} \right] \cr
& \Delta {H^ \circ } = 6\left( { - 400} \right) + 6\left( { - 300} \right) - \left( { - 1300} \right) = - 2900kJ/mol \cr} $$
For one gram of glucose, enthalpy of combustion
$$\Delta {H^ \circ } = - \frac{{2900}}{{180}} = - 16.11\,kJ/gm$$
123.
Which of the following salts should cause maximum cooling when $$1$$ $$mole$$ of it is dissolved in the same volume of water ?
A
$$NaCl;\,\Delta {H^ \circ } = 5.35\,kJ\,mo{l^{ - 1}}$$
B
$$KN{O_3};\Delta {H^ \circ } = 53.5\,kJ\,mo{l^{ - 1}}$$
C
$$KOH;\,\Delta {H^ \circ } = - 56.0\,kJ\,mo{l^{ - 1}}$$
D
$$HBr\,;\,\Delta {H^ \circ } = - 83.3\,kJ\,mo{l^{ - 1}}$$
Dissolution of $$KN{O_3}$$ is endothermic, hence heat is absorbed and cooling is observed.
124.
The entropy change involved in the isothermal reversible expansion of 2 mole of an ideal gas from a volume of $$10\,d{m^3}$$ to a volume of $$100\,d{m^3}$$ at $${27^ \circ }C$$ is :
A
$$38.3\,J\,mo{l^{ - 1}}\,{K^{ - 1}}$$
B
$$35.8\,J\,mo{l^{ - 1}}\,{K^{ - 1}}$$
C
$$32.3\,J\,mo{l^{ - 1}}\,{K^{ - 1}}$$
D
$$42.3\,J\,mo{l^{ - 1}}\,{K^{ - 1}}$$
Answer :
$$38.3\,J\,mo{l^{ - 1}}\,{K^{ - 1}}$$
Entropy change for an isothermal reversible process is given by
$$\eqalign{
& \Delta S = nR\,{\text{l}}n\frac{{{V_2}}}{{{V_1}}} = 2 \times 8.314 \times 2.303\log \frac{{100}}{{10}} \cr
& = 38.3\,J\,mo{l^{ - 1}}{K^{ - 1}} \cr} $$
125.
For complete combustion of ethanol, $${C_2}{H_5}OH\left( l \right) + 3{O_2}\left( g \right) \to 2C{O_2}\left( g \right) + 3{H_2}O\left( l \right),$$ the amount of heat produced as measured in bomb calorimeter, is $$1364.47\,kJ\,mo{l^{ - 1}}$$ at $${25^ \circ }C.$$ Assuming ideality the enthalpy of combustion, $${\Delta _c}H,$$ for the reaction will be :
$$\left( {R = 8.314\,kJ\,mo{l^{ - 1}}} \right)$$
A
$$ - 1366.95\,kJ\,mo{l^{ - 1}}$$
B
$$ - 1361.95\,kJ\,mo{l^{ - 1}}$$
C
$$ - 1460.95\,kJ\,mo{l^{ - 1}}$$
D
$$ - 1350.50\,kJ\,mo{l^{ - 1}}$$
Answer :
$$ - 1366.95\,kJ\,mo{l^{ - 1}}$$
$$\eqalign{
& {C_2}{H_5}OH\left( l \right) + 3{O_2}\left( g \right) \to 2C{O_2}\left( g \right) + 3{H_2}O\left( l \right) \cr
& {\text{Bomb calorimeter gives }}\Delta U{\text{ of the reaction}} \cr
& {\text{Given,}}\,\,\Delta U = - 1364.47\,kJ\,mo{l^{ - 1}} \cr
& \Delta {n_g} = - 1 \cr
& \Delta H = \Delta U + \Delta {n_g}RT \cr
& = - 1364.47 - \frac{{1 \times 8.314 \times 298}}{{1000}} \cr
& = - 1366.93\,kJ\,mo{l^{ - 1}} \cr} $$
126.
The factor of $$\Delta G$$ values is important in metallurgy. The $$\Delta G$$ values for the following reactions at $${800^ \circ }C$$ are given as
$${S_2}\left( s \right) + 2{O_2}\left( g \right) \to 2S{O_2}\left( g \right),$$ $$\Delta G = - 544\,kJ$$
$$2Zn\left( s \right) + {S_2}\left( s \right) \to 2ZnS\left( s \right),$$ $$\Delta G = - 293\,kJ$$
$$2Zn\left( s \right) + {O_2}\left( g \right) \to 2ZnO\left( s \right),$$ $$\Delta G = - 480\,kJ$$
The $$\Delta G$$ for the reaction, $$2ZnS\left( s \right) + 3{O_2}\left( g \right) \to $$ $$2ZnO\left( s \right) + 2S{O_2}\left( g \right)$$ will be
$$\Delta G = - RT\,\,{\text{ln}}\,\,{K_p}\,{\text{or}}\,{K_p} = {e^{ - \,\frac{{\Delta G}}{{RT}}}}$$
129.
From given following equations and $$\Delta {H^ \circ }$$ values, determine the enthalpy of reaction at $$298\,K$$ for the reaction :
$${C_2}{H_4}\left( g \right) + 6{F_2}\left( g \right) \to $$ $$2\,\,C{F_4}\left( g \right) + 4HF\left( g \right)$$
$${H_2}\left( g \right) + {F_2}\left( g \right) \to 2HF\left( g \right);$$ $$\Delta H_1^ \circ = - 537\,kJ$$
$$C\left( s \right) + 2{F_2}\left( g \right) \to C{F_4}\left( g \right);$$ $$\Delta H_2^ \circ = - 680\,kJ$$
$$2C\left( s \right) + 2{H_2}\left( g \right) \to {C_2}{H_4}\left( g \right);$$ $$\Delta H_3^ \circ = 52\,kJ$$
130.
In thermodynamics, a process is called reversible when
A
surroundings and system change into each other.
B
there is no boundary between system and surroundings.
C
the surroundings are always in equilibrium with the system.
D
the system changes into the surroundings spontaneously.
Answer :
the surroundings are always in equilibrium with the system.
In a reversible process, the driving and the opposite forces are nearly equal, hence the system and the surroundings always remain in equilibrium with each other.