Chemical Thermodynamics MCQ Questions & Answers in Physical Chemistry | Chemistry

\[\underset{\begin{smallmatrix} 98 \\ 49 \end{smallmatrix}}{\mathop{{{H}_{2}}S{{O}_{4}}}}\,+\underset{\begin{smallmatrix} 112 \\ 56 \end{smallmatrix}}{\mathop{2KOH}}\,\to \underset{\begin{smallmatrix} 174 \\ 87 \end{smallmatrix}}{\mathop{{{K}_{2}}S{{O}_{4}}}}\,+\underset{\begin{smallmatrix} 2\,mole \\ 1\,mole \end{smallmatrix}}{\mathop{2{{H}_{2}}O}}\,\]
$$13.7\,kcal$$  is the heat evolved when $$1\,gev$$  of strong acid is neutralised by $$1\,gev$$  of strong base.

$$\Delta {H^ \circ } = \Delta {U^ \circ } + \Delta {n_g}RT$$
$$\Delta {H^ \circ } = - 310\,kJ,\Delta {n_g} = 3 - 2 = 1,$$       $$R = 8.314 \times {10^{ - 3}}\,kJ\,{k^{ - 1}}\,mo{l^{ - 1}},$$      $$T = 298\,K$$
$$\eqalign{ & \Delta {U^ \circ } = - 310 - \left( {1 \times 8.314 \times {{10}^{ - 3}} \times 298} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = - 312.47\,kJ \cr} $$

$$\eqalign{ & \Delta {S_{{\text{vap}}}} = \frac{{\Delta {H_{{\text{vap}}}}}}{{{T_b}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{40.98 \times 1000}}{{373}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 109.8\,J\,{K^{ - 1}}\,mo{l^{ - 1}} \cr} $$

The equilibrium constant at two different temperatures for a thermodynamic process is given by
$$\log \frac{{{K_2}}}{{{K_1}}} = \frac{{\Delta {H^ \circ }}}{{2.303R}}\left[ {\frac{1}{{{T_1}}} - \frac{1}{{{T_2}}}} \right]$$
Here, $${K_1}$$  and $${K_2}$$  are replaced by $${K_p}$$  and \[K_{p}^{'}.\]
Therefore, \[\log \frac{{{K}^{'}}_{p}}{{{K}_{p}}}=\frac{\Delta {{H}^{\circ }}}{2.303R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\]
For exothermic reaction,
\[\begin{align} & {{T}_{2}}>{{T}_{1}}\,\,\text{and}\,\,H=-ve \\ & \therefore \,\,{{K}_{p}}>{{K}^{'}}_{p} \\ \end{align}\]

$$\eqalign{ & \Delta H = \Delta E + \Delta nRT\,\,{\text{For}}\,\Delta H \ne \Delta E,\Delta n \ne 0 \cr & {\text{Where}}\,\Delta n = {\text{no}}{\text{.}}\,{\text{of}}\,{\text{moles}}\,{\text{of}}\,{\text{gaseous}}\,{\text{products}} - {\text{no}}{\text{.}}\,{\text{of}}\,{\text{moles}}\,{\text{of}}\,{\text{gaseous}}\,{\text{reactants}} \cr & \left( {\text{a}} \right)\,\,\Delta n = 2 - 2 = 0 \cr & \left( {\text{b}} \right)\,\,\Delta n = 0\,\,\,\left( {\because \,\,{\text{they}}\,{\text{are either in solid or liquid state}}} \right) \cr & \left( {\text{c}} \right)\,\;\Delta n = 1 - 1 = 0\,\,\,\left( {\because \,\,C\,{\text{is}}\,{\text{in}}\,{\text{solid}}\,{\text{state}}} \right) \cr & \left( {\text{d}} \right)\,\,\Delta n = 2 - 4 = - 2 \cr} $$

$$\eqalign{ & \Delta {S_{{\text{fusion}}}} = \frac{{\Delta {H_{{\text{fusion}}}}}}{{{T_{{\text{fusion}}}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{6.01 \times 1000}}{{273}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 22\,J\,{K^{ - 1}}mo{l^{ - 1}} \cr} $$

The balanced equation for combustion of methane is $$C{H_{4\left( g \right)}} + 2{O_{2\left( g \right)}} \to C{O_{2\left( g \right)}} + 2{H_2}{O_{\left( l \right)}}$$
$${\text{Here}},\Delta {n_g} = 1 - 3 = - 2$$
$${\Delta _f}{H^ \circ } = {\Delta _f}{U^ \circ } + \Delta {n_g}RT;$$       $${\Delta _f}{H^ \circ } = - 393 - 2RT$$
$$\therefore \,\,{\Delta _f}{H^ \circ } < {\Delta _f}{U^ \circ }$$

$$\eqalign{ & {\text{Given}}\,\,\Delta H = 41\,kJ\,mo{l^{ - 1}} = 41000\,J\,mo{l^{ - 1}} \cr & T = {100^ \circ }C = 273 + 100 = 373\,K \cr & \Delta n = 1 \cr & \Delta U = \Delta H - \Delta nRT = 41000 - \left( {1 \times 8.314 \times 373} \right) \cr & = 37898.88\,J\,mo{l^{ - 1}} \simeq 37.9\,kJ\,mo{l^{ - 1}} \cr} $$

$$C + {O_2} \to C{O_2},\,\,\Delta {H^ \circ } = - x\,kJ...{\text{(i)}}$$
On reversing given second equation we get,
$$\eqalign{ & 2C{O_2} \to 2CO + {O_2},\,\Delta {H^ \circ } = + y\,kJ \cr & {\text{or}}\,\,C{O_2} \to CO + \frac{1}{{2{O_2}}},\,\Delta {H^ \circ } = + \frac{y}{2}kJ...{\text{(ii)}} \cr} $$
From Eqs. (i) and (ii) ( by addition )
$$\eqalign{ & C + \frac{1}{2}{O_2} \to CO, \cr & \Delta {H^ \circ } = \frac{y}{2} - x \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{y - 2x}}{2}kJ \cr} $$

The change in Gibbs free energy is given by
$$\Delta G = \Delta H - T\Delta S$$
where, $$\Delta H = $$  change enthalpy of the reaction
            $$\Delta S = $$  change entropy of the reaction
Thus, in order to determine $$\Delta G,$$ the values of $$\Delta H$$ must be known. The value of $$\Delta H$$ can be calculated by using equation
$$\Delta H = \Delta U + \Delta {n_g}RT\,\,\,...{\text{(i)}}$$
where, $$\Delta U = $$  change in internal energy using
$$\Delta {n_g} =$$  number of moles of gaseous products $$ - $$ number of moles of gaseous reactants
$$= 2 - 0 = 2$$
$$R = {\text{gas constant}} = 2\,cal$$
given, $$\Delta U = 2.1\,kcal$$
                  $$ = 2.1 \times {10^3}cal\,\left[ {\because \,\,1\,kcal = {{10}^3}\,cal} \right]$$
By putting the values in eq. (i) we get,
$$\eqalign{ & \therefore \,\,\Delta H = \left( {2.1 \times {{10}^3}} \right) + \left( {2 \times 2 \times 300} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3300\,cal \cr & {\text{Hence,}}\,\Delta G = \Delta H - T\Delta S \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta G = \left( {3300} \right) - \left( {300 \times 20} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta G = - 2700\,cal \cr & \therefore \,\,\,\,\,\,\,\,\,\,\Delta G = - 2.7\,kcal \cr} $$