Chemical Thermodynamics MCQ Questions & Answers in Physical Chemistry | Chemistry

$$2{H_2}{O_2}\left( l \right) \to 2{H_2}O\left( l \right) + {O_2}\left( g \right)\Delta H = ?$$
$$\Delta H = \left[ {\left( {2 \times \Delta {H_f}\,\,{\text{of}}\,\,{H_2}O\left( l \right)} \right)} \right. + $$      $$\left( {\Delta {H_f}\,\,{\text{of}}\,\,\left. {{O_2}\left( g \right)} \right]} \right.$$   $$ - \left[ {\left( {2 \times \Delta {H_f}\,\,{\text{of}}\,\,{H_2}{O_2}\left( l \right)} \right)} \right]$$
        $$ = \left[ {\left( {2 \times - 286} \right) + \left( 0 \right) - \left( {2 \times - 188} \right)} \right]$$
        $$ = \left[ { - 572 + 376} \right]$$
        $$ = - 196\,kJ/mol$$

$$\eqalign{ & 1\,watt = 1\,J/\sec \cr & {\text{Total heat supplied for}}\,\,36\,mL\,\,{H_2}O \cr & = 806 \times 100 \cr & = 80600\,J \cr & \Delta {H_{vap}} = \frac{{80600}}{{36}} \times 18 \cr & = 40300\,J\,\,{\text{or}}\,\,40.3\,kJ/mol \cr} $$

$$\eqalign{ & 16\,g\,C{H_4}\,\,{\text{gives}}\,\,880\,kJ\,\,{\text{of heat}}{\text{.}} \cr & 3.2\,g\,C{H_4}\,\,{\text{will give}}\,\,\frac{{880}}{{16}} \times 3.2 = 176\,kJ \cr} $$

According to Gibbs-Helmholtz equation,
Gibbs energy $$\left( {\Delta G} \right) = \Delta H - T\Delta S$$
Where, $$\Delta H$$ = Enthalpy change
$$\Delta S$$  = Entropy change
$$T$$ = Temperature
For a reaction to be spontaneous
$$\Delta G < 0.$$
∴ Gibbs -Helmholtz equation becomes,
$$\eqalign{ & \Delta G = \Delta H - T\Delta S < 0 \cr & {\text{or,}}\,\Delta H < T\Delta S \cr & {\text{or}},\,\,T > \frac{{\Delta H}}{{\Delta S}} \cr & = \frac{{35.5\,kJ\,mo{l^{ - 1}}}}{{83.6\,J{K^{ - 1}}mo{l^{ - 1}}}} \cr & = \frac{{35.5 \times 1000}}{{83.6}} \cr & = 425K \cr & T > 425K \cr} $$

The thermodynamic parameters which depend only upon the initial and final states of system, are called state functions, such as enthalpy $$\left( {H = q + W} \right),$$   Gibbs free energy $$\left( {G = H - TS} \right),$$    etc. While those parameters which depend on the path by which the process is performed rather than on the initial and final states, are called path functions, such as work done, heat, etc.

$$\Delta {G^ \circ } = - RT\,\,{\text{ln}}K$$
$$ = - 8.314J{K^{ - 1}}\,mo{l^{ - 1}} \times $$     $$298\,K \times 2.303\,{\text{log}}\,{10^{ - 14}}$$
$$ = 7.98 \times {10^4} \simeq 8 \times {10^4}\,J\,mo{l^{ - 1}}$$

$$\eqalign{ & {\text{Given}} \cr & {H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to {H_2}O\left( l \right); \cr & \Delta {H^ \circ } = - 285.9\,kJ\,mo{l^{ - 1}}\,\,\,...\left( 1 \right) \cr & {H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to {H_2}O\left( g \right); \cr & \Delta {H^ \circ } = - 241.8\,kJ\,mo{l^{ - 1}}\,...\left( 2 \right) \cr & {\text{We have to calculate}} \cr & {H_2}O\left( l \right) \to {H_2}O\left( g \right)\,;\,\,\Delta {H^ \circ } = ? \cr & {\text{On substracting eqn}}{\text{. (2) from eqn}}{\text{. (1) we get}} \cr & {{\text{H}}_2}O\left( l \right) \to {H_2}O\left( g \right)\,;\Delta {H^ \circ } = - 241.8 - \left( { - 285.9} \right) \cr & = 44.1\,kJ\,mo{l^{ - 1}} \cr} $$

Hydration energy of $$C{l^ + }$$  is very less than $${H^ + },$$  hence it doesn’t form $$C{l^ + }\left( {aq} \right)ions.$$

For reaction,

$$\Delta {H_{{\text{reaction}}}}$$   $$ = \sum {B{E_{\left( {{\text{reactant}}} \right)}}} - \sum {B{E_{\left( {{\text{product}}} \right)}}} ,$$       $$\left[ {BE = {\text{bond energy}}} \right]$$
$$\Delta {H_r} = $$   $$\left[ {4 \times B{E_{\left( {C - H} \right)}} + 1 \times B{E_{\left( {C = C} \right)}} + 1 \times B{E_{\left( {H - H} \right)}}} \right]$$         $$ - \left[ {6 \times B{E_{\left( {C - H} \right)}} + 1 \times B{E_{\left( {C - C} \right)}}} \right]$$
$$ = \left( {4 \times 410.50 + 1 \times 606.10 + 1 \times 431.37} \right)$$         $$ - \left[ {\left( {6 \times 410.50} \right) + \left( {1 \times 336.49} \right)} \right]kJ\,mo{l^{ - 1}}$$
$$ = \left[ {1642 + 606.1 + 431.37} \right]$$      $$ - \left[ {2463 + 336.49} \right]kJ\,mo{l^{ - 1}}$$
$$ = \left[ {2679.47} \right] - \left[ {2799.49} \right]\,kJ\,mo{l^{ - 1}}$$
$$ = - 120.0\,kJ\,mo{l^{ - 1}}$$

$${\text{At equilibrium}}\,\,\Delta G = 0$$