Chemical Thermodynamics MCQ Questions & Answers in Physical Chemistry | Chemistry

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 Ist law of thermodynamics.

The sign of $$\Delta H$$  is reversed.

The heat required to raise the temperature of body by $$1K$$ is called thermal capacity or heat capacity.

The Gibbs free energy of a reaction, $${\Delta _r}G$$  is related to the composition of the reaction mixture and the standard reaction Gibbs free energy $${\Delta _r}{G^ \circ }$$  as
$${\Delta _r}G = {\Delta _r}{G^ \circ } + RT\,\ln \,Q$$
where, $$Q = $$  reaction quotient
At equilibrium $$Q = K$$  and $${\Delta _r}G = 0.$$
Therefore, the above reaction becomes
$$\eqalign{ & 0 = {\Delta _r}{G^ \circ } + RT\,\ln \,K \cr & {\Delta _r}{G^ \circ } = - RT\,\ln \,K \cr & {\text{or}}\,\,{\Delta _r}{G^ \circ } = - 2.303\,RT\,\log \,K \cr & K = {\text{equilibrium constant}} \cr} $$

$$\eqalign{ & {\text{Given, for reaction}} \cr & \left( {\text{i}} \right)\,\,{H_2}O\left( \ell \right) \to {H^ + }\left( {aq.} \right) + O{H^ - }\left( {aq.} \right)\,;\Delta {H_r} = 57.32\,kJ \cr & \left( {{\text{ii}}} \right)\,{H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to {H_2}O\left( \ell \right)\,;\,\Delta {H_r} = - 286.20\,kJ \cr & {\text{For reaction (i)}} \cr & \Delta {H_r} = \Delta {H^ \circ }_f\left( {{H^ + }.aq} \right) + \Delta {H^ \circ }_f\left( {O{H^ - }.aq} \right) - \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right) \cr & 57.32 = 0 + \Delta {H^ \circ }_f\left( {O{H^ - },aq} \right) - \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right)\,\,\,\,\,...\left( {{\text{iii}}} \right) \cr & {\text{For reaction (ii)}} \cr & \Delta {H_r} = \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right) - \Delta {H^ \circ }_f\left( {{H_2},g} \right) - \frac{1}{2}\Delta {H^ \circ }_f\left( {{O_2},g} \right) \cr & - 286.20 = \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right) \cr & {\text{On replacing this value in equ}}{\text{. (iii) we have}} \cr & 57.32 = \Delta {H^ \circ }_f\left( {O{H^ - },aq} \right) - \left( { - 286.20} \right) \cr & \Delta {H^ \circ }_f = - 286.20 + 57.32 \cr & = - 228.88\,kJ \cr} $$

In closed insulated container a liquid is stirred with a paddle to increase the temperature, therefore it behaves as adiabatic process, so for it $$q = 0.$$
Hence, from first law of thermodynamics
$$\eqalign{ & \Delta E = q + W \cr & {\text{if}},\,\,q = 0 \cr} $$
$$\therefore \,\,\Delta E = W$$   but not equal to zero.

At equilibrium Gibbs free energy change $$\left( {\Delta G} \right)$$  is equal to zero.
$$\Delta G = \Delta H - T\Delta S$$
$$0 = 30 \times {10^3}\left( {J\,mo{l^{ - 1}}} \right)$$     $$ - T \times 105\left( {J\,{K^{ - 1}}mo{l^{ - 1}}} \right)$$
$$\eqalign{ & \therefore \,\,T = \frac{{30 \times {{10}^3}}}{{105}}K \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 285.71\,K \cr} $$

Entropy change at constant temperature
$$\eqalign{ & = 2.303nR\,\,{\text{log}}\frac{{{V_1}}}{{{V_2}}} = 2.303 \times 2 \times 2\,\,{\text{log}}\frac{{20}}{2} \cr & = 9.2\,cal\,{K^{ - 1}}mo{l^{ - 1}} \cr} $$

$$C{H_2} = C{H_2}\left( g \right) + {H_2}\left( g \right) \to C{H_3} - C{H_3}$$
Enthalpy change = Bond energy of reactants - Bond energy of products.
$$\eqalign{ & \Delta H = 1\left( {C = C} \right) + 4\left( {C - H} \right) + \left( {H - H} \right) - 1\left( {C - C} \right) - 6\left( {C - H} \right) \cr & = 1\,\left( {C = C} \right) + 1\left( {H - H} \right) - 1\left( {C - C} \right) - 2\left( {C - H} \right) \cr & = 615 + 435 - 347 - 2 \times 414 \cr & = 1050 - 1175 \cr & = - 125\,kJ \cr} $$

There is no change in volume from $$A \to B.$$