Chemical Thermodynamics MCQ Questions & Answers in Physical Chemistry | Chemistry

TIPS/Formulae:
Heat capacity at constant volume $$\left( {{q_v}} \right) = \Delta E$$
Heat capacity of constant pressure $$\left( {{q_p}} \right) = \Delta H$$
$$\eqalign{ & \Delta H = \Delta E + \Delta nRT\,\,\,\,or\,\,\,\Delta H - \Delta E = nRT \cr & \Delta n = {\text{no}}{\text{.}}\,{\text{of}}\,{\text{moles}}\,{\text{of}}\,{\text{gaseous}}\,{\text{products - no}}{\text{.}}\,{\text{of}}\,{\text{moles}}\,{\text{of}}\,{\text{gaseous}}\,{\text{reactants}} \cr & = 12 - 15 = - 3 \cr & \Delta H - \Delta E = - 3 \times 8.314 \times 298J = -7.43\,kJ. \cr} $$

The enthalpy of ionisation of weak acid is given by
$$\Delta {H_{ion}}\left( {HA} \right)$$
$$ = \Delta {H_N}$$   ( weak acid / strong base ) $$ -\, \,\Delta {H_N}$$   ( strong acid / strong base )
$$\eqalign{ & = - 56.1 - \left( { - 57.3} \right) = 1.2\,kJ\,mo{l^{ - 1}} \cr & \Delta {H_{{\text{(ionisation)}}}} = 1.5\,kJ\,mo{l^{ - 1}} \cr} $$
Hence % ionisation in $$1 M$$  solution
$$\eqalign{ & = \frac{{\left( {1.5 - 1.2} \right)}}{{1.5}} \times 100 \cr & = 20\% \cr} $$

$$\eqalign{ & \Delta S{\text{(entropy change)}} \cr & = 2.303\,\,nR\,\,{\text{lo}}{{\text{g}}_{10}}\frac{{{V_2}}}{{{V_1}}} \cr & = 2.303 \times 2 \times 2 \times {\log _{10}}\frac{{20}}{2} \cr & = 2.303 \times 2 \times 2 \times 1 \cr & = 9.212\,cal \cr} $$

A catalyst is a substance which alters the reaction rate but itself remains unchanged in amount and chemical composition at the end of the reaction. It provides a new reaction path with a lower energy barrier ( lowering activation energy ).

$$w = - {P_{ex}}\left( {{V_f} - {V_i}} \right)$$    $$ = - 3\left( {6 - 4} \right)L\,atm = - 6\,L\,atm$$
$$ = - 607.92\,J = - 608\,J$$

$$\eqalign{ & q = - {C_\nu } \times \Delta t \times \frac{M}{m} \cr & = - 30 \times 4 \times \frac{{12}}{4} \cr & = - 360\,kJ\,mo{l^{ - 1}} \cr & \Delta H = - 360\,kJ\,mo{l^{ - 1}} \cr} $$

$$\eqalign{ & \Delta S = 2.303nR\,\,\log \frac{{{V_2}}}{{{V_1}}} \cr & \,\,\,\,\,\,\,\,\,\, = 2.303 \times 1 \times 8.314\,\,\log \frac{{100}}{{10}} \cr & \,\,\,\,\,\,\,\,\,\, = 19.14\,J\,{K^{ - 1}} \cr} $$

For $$5\,moles$$  of gas at temperature $$T,$$
$$P{V_1} = 5RT$$
For $$5\,moles$$  of gas at temperature $$T - 2,$$
$$P{V_2} = 5R\left( {T - 2} \right)$$
$$\eqalign{ & \therefore \,\,P\left( {{V_2} - {V_1}} \right) \cr & = 5R\left( {T - 2 - T} \right); \cr & P\Delta V = - 10R, \cr & - P\Delta V = 10R \cr} $$
When $$\Delta V$$  is negative, $$W$$ is $$ + ve.$$

$$\eqalign{ & \frac{1}{2}{H_2} + \frac{1}{2}C{l_2} \to HCl \cr & \Delta H = \frac{1}{2} \times 104 + \frac{1}{2} \times 58 - 103 = - 22\,kcal \cr} $$

$$\eqalign{ & {\text{We know that,}} \cr & H = E + W \cr} $$
Enthalpy = internal energy + pressure × volume
$$\eqalign{ & \,\,\,\,\,H = E + pV \cr & \Delta H = \Delta E + \Delta \left( {pV} \right) \cr & \Delta H = \Delta E + \Delta \left( {{n_g}RT} \right)\,\,\left( {\because \,\,pV = nRT} \right) \cr} $$
For isothermal expansion of ideal gas, $$\Delta T = 0$$
$$\therefore \,\,\Delta H = \Delta E$$