Chemical Thermodynamics MCQ Questions & Answers in Physical Chemistry | Chemistry

The energy involved in the conversion of
$$\eqalign{ & \frac{1}{2}C{l_2}\left( g \right)\,\,{\text{to}}\,\,C{l^{ - 1}}\left( {aq} \right)\,{\text{is}}\,{\text{given}}\,{\text{by}} \cr & \Delta H = \frac{1}{2}{\Delta _{diss}}H_{C{l_2}}^{\left( - \right)} + {\Delta _{eg}}H_{Cl}^{\left( - \right)} + {\Delta _{hyl}}H_{Cl}^{\left( - \right)} \cr} $$
Substituting various values from given data, we get
$$\eqalign{ & \Delta H = \left( {\frac{1}{2} \times 240} \right) + \left( { - 349} \right) + \left( { - 381} \right)kJ\,mo{l^{ - 1}} \cr & = \left( {120 - 349 - 381} \right)\,kJ\,mo{l^{ - 1}} \cr & = - 610\,kJ\,mo{l^{ - 1}} \cr} $$
i.e., the correct answer is (B)

$$\eqalign{ & {C_6}{H_6}\left( {\text{l}} \right) + \frac{{15}}{2}{O_2}\left( g \right) \to 6C{O_2}\left( g \right) + 3{H_2}O\left( {\text{l}} \right) \cr & \Delta {n_g} = 6 - \frac{{15}}{2} = - \frac{3}{2} \cr & \Delta H = \Delta U + \Delta {n_g}RT \cr & = - 3263.9 + \left( { - \frac{3}{2}} \right) \times 8.314 \times 298 \times {10^{ - 3}} \cr & = - 3263.9 + \left( { - 3.71} \right) \cr & = - 3267.6\,kJ\,mo{l^{ - 1}} \cr} $$

There is no phase change in formation. A new product is formed during the reaction.

$$\left( {\text{A}} \right):C{H_{4\left( g \right)}} + 2{O_{2\left( g \right)}} \to C{O_{2\left( g \right)}} + 2{H_2}O$$        shows combustion reaction
$$\left( {\text{B}} \right):{H_{2\left( g \right)}} \to 2{H_{\left( g \right)}}$$    shows bond dissociation
$$\left( {\text{C}} \right):NaC{l_{\left( s \right)}} \to Na_{\left( g \right)}^ + + Cl_{\left( g \right)}^ - $$      shows dissociation of $$NaCl$$
$$\left( {\text{D}} \right):NaC{l_{\left( s \right)}} \to Na_{\left( {aq} \right)}^ + + Cl_{\left( {aq} \right)}^ - $$       shows dissolution of $$NaCl$$

$$\eqalign{ & {\text{Volume of graphite}} = \frac{{{\text{Mass}}}}{{{\text{Density}}}} = \frac{{12}}{{2.25}} \cr & {\text{Volume of diamond}} = \frac{{12}}{{3.31}} \cr & {\text{Change in volume,}} \cr & \Delta V = \left( {\frac{{12}}{{3.31}} - \frac{{12}}{{225}}} \right) \times {10^{ - 3}}\,L \cr & \,\,\,\,\,\,\,\,\,\, = - 1.91 \times {10^{ - 3}}\,L \cr & \Delta {G^ \circ } = {\text{work done}} = - p\Delta V \cr & p = - \frac{{\Delta {G^ \circ }}}{{\Delta V}} \cr & = \frac{{1895\,J\,mo{l^{ - 1}}}}{{1.91 \times {{10}^{ - 3}} \times 101.3}} \cr & = 9794\,atm\left[ {\because \,\,1\,atm = {{10}^5} \times 1.013\,Pa} \right] \cr & = 9.92 \times {10^8}\,Pa \cr} $$

Given conditions are boiling conditions for water due to which
$$\eqalign{ & \Delta {S_{total}} = 0 \cr & \Delta {S_{system}} + \Delta {S_{surroundings}} = 0 \cr & \Delta {S_{system}} = - \Delta {S_{surroundings}} \cr & {\text{For}}\,{\text{process,}}\,\,\Delta {S_{system}} > 0 \cr & \Delta {S_{surroundings}} < 0 \cr} $$

$$\eqalign{ & C{H_4}\left( g \right) + 2\,{O_2}\left( g \right) \to C{O_2}\left( g \right) + 2{H_2}O\left( l \right) \cr & \Delta H = - 890\,kJ\,\,\,....\left( {\text{i}} \right) \cr & 2\,{H_2}O\left( l \right) \to 2\,{H_2}O\left( g \right); \cr & \Delta H = 2 \times 40.5 = 81\,kJ\,\,\,...\left( {{\text{ii}}} \right) \cr & {\text{From}}\,\,\left( {\text{i}} \right) + \left( {{\text{ii}}} \right),{\text{we get}} \cr & C{H_4}\left( g \right) + 2\,{O_2}\left( g \right) \to C{O_2}\left( g \right) + 2{H_2}O\left( g \right) \cr & \Delta H = - 890 + 81 = - 809\,kJ \cr} $$

For a spontaneous process, total change in entropy ( system + surrounding ) should be positive.

For non spontaneous reaction
$$\eqalign{ & \Delta G = + ve \cr & \Delta G = \Delta H - T\Delta S\,\,{\text{and}} \cr & \Delta S = 121\,J{K^{ - 1}} \cr & {\text{For}}\,\,\Delta G = + ve \cr} $$
$$\Delta H$$  has to be positive. Hence the reaction is endothermic.
The minimum value of $$\Delta H$$  can be obtained by putting $$\Delta G = 0$$
$$\eqalign{ & \Delta H = T\Delta S = 298 \times 121\,J \cr & = 36.06\,kJ \cr} $$

A spontaneous process is an irreversible process and may only be reversed by some external energy.