Chemical Thermodynamics MCQ Questions & Answers in Physical Chemistry | Chemistry

TIPS/Formulae:
$$\Delta G = \Delta H - T\Delta S$$
Since $$\Delta G = \Delta H - T\Delta S$$     for an endothermic reaction,
$$\Delta H = + ve$$   and at low temperature $$\Delta S = + ve$$
Hence $$\Delta G = \left( + \right)\Delta H - T\left( + \right)\Delta S$$
and if $$T\Delta S < \Delta H$$         (at low temp)
$$\Delta G = + ve$$         (non spontaneous)
But at high temperature, reaction becomes spontaneous i.e. $$\Delta G = - ve.$$
because at higher temperature $$T\Delta S > \Delta H.$$

$$\eqalign{ & \Delta {G^ \circ } = \Delta {H^ \circ } - T\Delta {S^ \circ }\,\,\,...\left( {\text{i}} \right) \cr & {\text{Given that,}}\,\,\Delta {H^ \circ } = - 382.64\,kJ\,mo{l^{ - 1}} \cr & \Delta {S^ \circ } = 145.6\,J\,{K^{ - 1}}\,mo{l^{ - 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = - 145.6 \times {10^{ - 3}}\,kJ\,{K^{ - 1}} \cr & T = 298\,K \cr} $$
On putting the given values in $$eq.$$  (i) we get,
or $$\Delta {G^ \circ } = - 382.64 - $$   $$\left[ {298 \times \left( { - 145.6 \times {{10}^{ - 3}}} \right)} \right]$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 339.3\,kJ\,mo{l^{ - 1}}$$

If the process is carried out adiabatically and isochorically,
$$\eqalign{ & \Delta U = \Delta {U_{{\text{heating}}}} + \Delta {U_{298\,K}} = 0\,\,{\text{or}}\,\,\Delta {U_{{\text{heating}}}} = - \Delta {U_{298\,K}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \int\limits_{298\,K}^{{T_f}} {n\sum {{C_\nu }dT} = + 240.60\,kJ\,mo{l^{ - 1}}} \cr & \sum {n{C_\nu } = n \cdot {C_{\nu \left( {{H_2}{O_{\left( v \right)}}} \right)}} + n{C_{\nu \left( {{N_{2\left( g \right)}}} \right)}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {39.06 + 2 \times 26.40} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 91.86\,J{K^{ - 1}}\,mo{l^{ - 1}} \cr} $$
by using the value of $$\sum {n{C_\nu }} $$   in the above equation
$$\eqalign{ & \left( {91.86} \right)\left( {{T_f} - 298} \right) = 240600\,J\,mo{l^{ - 1}} \cr & {T_f} - 298 = \frac{{240600}}{{91.86}} = 2619\,K \cr & {T_f} = 2619 + 298 = 2917\,K \cr} $$

$$\eqalign{ & \Delta S = \frac{{\Delta {Q_{rev.}}}}{T}\,\,;\,\,\,75 = \frac{{30 \times {{10}^3}}}{T} \cr & \therefore \,\,T = 400\,K \cr} $$

The species in its elemental form has zero standard molar enthalpy of formation at $$298\,K.$$   At $$298\,K,C{l_2}$$   is gas while $$B{r_2}$$  is liquid.

No. of moles of $${O_2}$$ required to supplied $$30\,kJ$$  heat to second reaction
$$\eqalign{ & = \frac{{30}}{{1260}} \times \frac{3}{2} = \frac{1}{{28}} \cr & {n_{{O_2}}}:{n_{{H_2}}} = \frac{1}{{28}}:3\,\,\,{\text{or}}\,\,1:84 \cr} $$

Since vessel is thermally insulated, i.e., the process is adiabatic hence, $$q = 0.$$  Also, $${P_{{\text{ext}}}} = 0,$$   hence $$w = 0$$
From 1^st law of thermodynamics, $$\Delta U = q + w$$
$$\therefore \,\,\Delta U = 0$$    ( for ideal gas )
$$\therefore \,\,\Delta T = 0$$    or $${T_2} = {T_1}$$
[ $$\because $$  Internal energy of an ideal gas is a function of temperature. ]
Applying ideal gas equation, $$PV = nRT$$
where $$n, R$$  and $$T$$  are constant. Then $${P_1}{V_1} = {P_2}{V_2}$$

$$2C{l_2} + 2{H_2}O \to 4HCl + {O_2}$$
$$\Delta H = $$   $$B.E.\,\,{\text{of}}$$   $$\left( {2 \times Cl - Cl} \right) + \left( {2 \times 2 \times O - H} \right) - \left[ {4 \times H - Cl) + (O = O} \right]$$
$$\eqalign{ & \,\,\,\,\,\,\,\, = \left( {2 \times 242.8 + 4 \times 464} \right) - \left( {4 \times 431.8 + 442} \right) \cr & \,\,\,\,\,\,\,\, = 172.4\,kJ\,mo{l^{ - 1}} \cr} $$

$$\eqalign{ & \Delta H = \Delta U + \Delta {n_g}RT \cr & {\text{For the reaction}}\,\,\Delta {n_g} = 12 - 15 = - 3 \cr & \Delta H - \Delta U = - 3 \times 8.314 \times 300 \cr & = - 7482\,J\,mo{l^{ - 1}} \cr} $$

$$\because \,\,\Delta G = \Delta H - T \times \Delta S$$
For a spontaneous reaction, $$\Delta G$$  should be negative
$$\Delta H = - 238\,kJ,\,\Delta S = - 87\,J{K^{ - 1}}$$
Hence, reaction will be spontaneous when $$\Delta H > T \times \Delta S.$$   Therefore, at 1000, 1500 and 3000 $$K$$ the reaction would be spontaneous.