Chemical Thermodynamics MCQ Questions & Answers in Physical Chemistry | Chemistry

For a cyclic process, net change in internal energy is zero. Change in internal energy does not depend on the path by which final stage is reached.

Internal energy and molar enthalpy are state functions. Work (reversible or irreversible) is a path function.

$$\eqalign{ & {\text{For the reaction}} \cr & 2\,ZnS \to 2\,Zn + {S_2};\,\Delta {G_1}^ \circ = 293\,kJ...\left( {\text{i}} \right) \cr & 2\,Zn + {O_2} \to 2\,ZnO;\,\Delta {G_2}^ \circ = - 480\,kJ...\left( {{\text{ii}}} \right) \cr & {S_2} + 2\,{O_2} \to 2\,S{O_2};\Delta {G_3}^ \circ = - 544\,kJ...\left( {{\text{iii}}} \right) \cr & \Delta {G^ \circ }\,{\text{for the reaction}} \cr & 2\,ZnS + 3\,{O_2} \to 2\,ZnO + 2\,S{O_2} \cr & {\text{can be obtained by adding eqn}}{\text{.}}\,\left( {\text{i}} \right){\text{,}}\left( {{\text{ii}}} \right){\text{and}}\left( {{\text{iii}}} \right) \cr & \Rightarrow \Delta {G^ \circ } = 293 - 480 - 544 = - 731\,kJ \cr} $$

To calculate average enthalpy of $$C – H$$  bond in methane following informations are needed
(i) dissociation energy of $${H_2}$$  i.e.
$$\frac{1}{2}{H_2}\left( g \right) \to H\left( g \right);\,\Delta H = x\left( {{\text{suppose}}} \right)$$
(ii) Sublimation energy of $$C\left( {graphite} \right)$$   to $$C\left( g \right)$$
$$\eqalign{ & C\left( {graphite} \right) \to C\left( g \right);\Delta H = y\left( {{\text{suppose}}} \right) \cr & {\text{Given}} \cr & C\left( {graphite} \right) + 2{H_2}\left( g \right) \to C{H_4}\left( g \right); \cr & \Delta H = 75\,kJ\,mo{l^{ - 1}} \cr} $$

$$\eqalign{ & \frac{1}{2}{N_2} + \frac{3}{2}{H_2} \to N{H_3} \cr & \frac{1}{2}{x_1} + \frac{3}{2}{x_2} - 3{x_3} \cr} $$

$$\eqalign{ & {P_1}{V_1} = {P_2}{V_2} \cr & \therefore \,\,V = \frac{{5\,bar \times 2d{m^3}}}{{1\,bar}} \cr & = 10\,d{m^3} \cr & {\text{Work,}}\,\,W = - {P_{ext}}\left( {\left( {{V_{final}} - {V_{initial}}} \right.} \right) \cr & = - 1\,bar\left( {10 - 2} \right) \cr & = - 1 \times {10^5}Pa \times 8 \times {10^{ - 3}}{m^3} \cr & = - 800\,J \cr} $$

For reversible isothermal expansion,
$$\eqalign{ & w = - nRT\ln \frac{{{V_2}}}{{{V_1}}} \cr & \therefore \,\left| w \right| = nRT\ln \frac{{{V_2}}}{{{V_1}}} \cr & \left| w \right| = nRT\left( {\ln \,{V_2} - \ln {V_1}} \right) \cr & \left| w \right| = nRT\ln \,{V_2} - nRT{V_1} \cr & y = mx + c \cr} $$
So, slope of curve 2 is more than curve 1 and intercept of curve 2 is more negative than curve 1.

$$\eqalign{ & 3{H_2}O\left( l \right) \to 3{H_2}O\left( g \right); \cr & \Delta n = 3,\,\Delta E = \Delta H - \Delta nRT \cr & = 30 - 3 \times \frac{2}{{1000}} \times 500 \cr & = 27\,kcal \cr} $$

$$\left( {\text{A}} \right):\Delta {n_g} = 2 - 2 = 0\,;$$     hence $$\Delta H = \Delta U$$
$$\left( {\text{B}} \right):\Delta {n_g} = 2 - 1 = 1\,;$$     hence $$\Delta H = \Delta U + RT$$
$$\left( {\text{C}} \right):\Delta {n_g} = 2 - 4 = - 2\,;$$     hence $$\Delta H = \Delta U - 2RT$$
$$\left( {\text{D}} \right):\Delta {n_g} = 5 - 2 = 3;$$     hence $$\Delta H = \Delta U + 3RT$$

$$\eqalign{ & {H_2}\left( g \right) + C{l_2}\left( g \right) \to 2HCl\left( g \right) \cr & {\Delta _r}S = \sum {{S_m}^ \circ } \left( P \right) - \sum {{S_m}^ \circ \left( R \right)} \cr} $$
$${\Delta _r}{S^ \circ } = 2 \times {S_m}^ \circ \left( {HCl} \right) - $$     $$\left[ {{S_m}^ \circ \left( {C{l_2}} \right) + {S_m}^ \circ \left( {{H_2}} \right)} \right]$$
          $$ = \left( {2 \times 186.7} \right) - \left( {223 + 130.6} \right)$$
          $$ = 373.4 - 353.6$$
          $$ = + 19.8\,J{K^{ - 1}}\,mo{l^{ - 1}}$$