Chemical Thermodynamics MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & {H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to {H_2}O\left( l \right); \cr & \Delta H = - 285.8\,kJ\,\,\,....\left( {\text{i}} \right) \cr & {H^ + }\left( {aq} \right) + O{H^ - }\left( {aq} \right) \to {H_2}O\left( l \right); \cr & \Delta H = - 57.3\,kJ\,\,\,....\left( {{\text{ii}}} \right) \cr & \frac{1}{2}{H_2}\left( g \right) + aq \to {H^ + }\left( {aq} \right) + {e^ - };\Delta H = 0 \cr & {\text{(by convention)}}\,\,\,...\left( {{\text{iii}}} \right) \cr & \left( {\text{i}} \right){\text{ - }}\left( {{\text{ii}}} \right){\text{ - }}\left( {{\text{iii}}} \right)\,\,{\text{gives,}} \cr} $$
$$\frac{1}{2}{H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) + {e^ - } + aq \to $$       $$O{H^ - }\left( {aq} \right)$$
$$\Delta H = - 285.8 + 57.3 = - 228.5\,kJ$$

$$\eqalign{ & \Delta {H^ \circ } = \Delta {U^ \circ } + \Delta nRT \cr & \Delta {H^ \circ } = - 772.7 + \frac{{5 \times 8.314 \times 298}}{{1000}} \cr & = - 760.3\,kJ\,mo{l^{ - 1}} \cr} $$

$$\eqalign{ & {\text{From }}{1^{{\text{st}}}}\,{\text{law of thermodynamics }} \cr & {q_{sys}} = \Delta U - w = 0 - \left[ { - {P_{ex{t^ \bullet }}}\Delta V} \right] \cr & \,\,\,\,\,\,\,\,\,\, = 3.0\,atm \times \left( {2.0L - 1.0L} \right) \cr & \,\,\,\,\,\,\,\,\,\, = 3.0L - atm \cr & \therefore \,\,\Delta {S_{surr.}} = \frac{{{{\left( {{q_{rev.}}} \right)}_{surr.}}}}{T} = - \frac{{{q_{sys.}}}}{T} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \frac{{3.0 \times 101.3J}}{{300K}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 1.013\,J/K \cr} $$

$$\eqalign{ & {\text{Isobutane + oxygen}} \to C{O_2} + {H_2}O \cr & \Delta H = - 2870\,kJ\,mo{l^{ - 1}}\,......\left( {\text{i}} \right) \cr & n{\text{ - butane + oxygen}} \to {\text{C}}{{\text{O}}_2} + {H_2}O \cr & \Delta H = - 2878\,kJ\,mo{l^{ - 1}}\,\,\,....\left( {{\text{ii}}} \right) \cr & \left( {{\text{ii}}} \right){\text{ - }}\left( {\text{i}} \right);n{\text{ - butane}} - {\text{Isobutane,}} \cr & \Delta H = \left( { - 2878 + 2870} \right) \cr & = - 8\,kJ\,mo{l^{ - 1}} \cr} $$

$$\eqalign{ & \Delta H = \Delta U + \Delta nRT \cr & \Delta n = {n_p} - {n_R} \cr & {\text{Now,}}\,\,\Delta n = 2 - \frac{5}{2} = - \frac{1}{2} \cr & \therefore \,\,\Delta H = \Delta U - \frac{1}{2}RT \cr & {\text{Thus,}}\,\,\Delta U = \Delta H + \frac{1}{2}RT \cr & \therefore \,\,\Delta U > \Delta H \cr} $$

$$\eqalign{ & \mathop {HCl}\limits_{{\text{Strong acid}}} \to \mathop {{H^ + }}\limits_{{\text{(Complete ionisation)}}} + C{l^ - }\,\,\,\,\,...\left( {\text{i}} \right) \cr & N{H_4}OH \rightleftharpoons \mathop {NH_4^ + }\limits_{{\text{Weak base}}} + O{H^ - } \cr & \Delta H = x\,kJ\,mo{l^{ - 1}}\,\,\,...\left( {{\text{ii}}} \right) \cr & {H^ + } + O{H^ - } \to {H_2}O \cr & \Delta H = - 55.90\,kJ\,mo{l^{ - 1}}\,\,\,...\left( {{\text{iii}}} \right) \cr} $$
( from neutralisation of strong acid and strong base )
$$\eqalign{ & {\text{From equation (i), (ii) and (iii)}} \cr & N{H_4}OH + HCl \to NH_4^ + + C{l^ - } + {H_2}O \cr & \Delta H = - 51.46\,kJ\,mo{l^{ - 1}} \cr & \therefore \,\,x + \left( { - 55.90} \right) = - 51.46 \cr & x = - 51.46 + 55.90 \cr & \,\,\,\, = 4.44\,kJ\,mo{l^{ - 1}} \cr & \therefore \,{\text{Enthalpy of ionisation of}}\,\,N{H_4}OH\,\,{\text{is}} \cr & \Rightarrow 4.44\,kJ\,mo{l^{ - 1}} \cr} $$

For free expansion ( i.e. in vacuum ), $${P_{{\text{ext}}}} = 0.$$
Thus, $$W = - {P_{{\text{ex}}}}\,\Delta V = 0$$
For adiabatic change, $$q = 0$$
$$\therefore \,\,\Delta U = q + W = 0$$     which is true for isothermal process where $$T$$  is constant i.e., $$\Delta T = 0.$$

$$\eqalign{ & S + \frac{3}{2}{O_2} \to S{O_3} + 2x\,kcal\,\,\,\,...\left( {\text{i}} \right) \cr & S{O_2} + \frac{1}{2}{O_2} \to S{O_3} + y\,kcal\,\,\,\,...\left( {{\text{ii}}} \right) \cr & {\text{Now, subtract eq}}{\text{.}}({\text{ii}}){\text{ from }}({\text{i}}),{\text{we get,}} \cr & S + {O_2} \to S{O_2} + \left( {2x - y} \right)kcal \cr} $$
∴ Heat of formation of $${\text{ }}S{O_2}$$  is equal to $$\left( {2x - y} \right)kcal.$$

Among the given reactions only in the case of
$$C\left( {{\text{graphite}}} \right) + \frac{1}{2}{O_2}\left( g \right) \to CO\left( g \right)$$
entropy increases because randomness (disorder) increases. Thus, standard entropy change $$\left( {\Delta {S^ \circ }} \right)$$  is positive. Moreover, it is a combustion reaction and all the combustion reactions are generally exothermic, i.e. $$\Delta {H^ \circ } = - ve$$
We know that,
$$\eqalign{ & \Delta {G^ \circ } = \Delta {H^ \circ } - T\Delta {S^ \circ } \cr & \Delta {G^ \circ } = - ve - T\left( { + ve} \right) \cr} $$
Thus, as the temperature increases, the value of $$\Delta {G^ \circ }$$ decreases.

As the heat is absorbed.
$$\therefore q = + 208\,J$$
Now for reversible isothermal process,
$$\eqalign{ & q = - W \cr & \therefore \,W = - q = - 208\,J. \cr} $$
Hence the values of $$q$$ and $$W$$ for the process will be $$ + 208\,J$$  and $$ - 208\,J$$  respectively.