Chemical Thermodynamics MCQ Questions & Answers in Physical Chemistry | Chemistry

$${N_2} + 3{H_2} \to 2N{H_3}\,\,\,\,\Delta H = 2 \times - 46.0\,kJ\,mo{l^{ - 1}}$$
Let $$x$$ be the bond enthalpy of $$N - H$$  bond then
[Note : Enthalpy of formation or bond formation enthalpy is given which is negative but the given reaction involves bond breaking hence values should be taken as positive. ]
$$\eqalign{ & \Delta H = \sum {{\text{Bond energies of products}} - \sum {{\text{Bond energies of reactants}}} } \cr & 2 \times - 46 = 712 + 3 \times \left( {436} \right) - 6x; - 92 = 2020 - 6x \cr & 6x = 2020 + 92 \Rightarrow 6x = 2112 \Rightarrow x = + 352\,kJ/mol \cr} $$

$$\eqalign{ & {C_{\left( {graphite} \right)}} \to {C_{\left( {diamond} \right)}}\,\,\left( {{\text{Isothermally}}} \right) \cr & {\Delta _r}{G^ \circ } = \Delta {G^ \circ }_{\left( {diamond} \right)} - \Delta {G^ \circ }_{\left( {graphite} \right)} \cr & = 2.9 - 0 \cr & = 2.9\,kJ\,mo{l^{ - 1}} \cr} $$
Gibbs free energy is the maximum useful work, then
$$\eqalign{ & - \Delta G = {w_{\max }} = P\Delta V \cr & - 2.9 \times {10^3} = - P \times 2 \times {10^{ - 6}} \cr & P = \frac{{2.9 \times {{10}^3}}}{{2 \times {{10}^{ - 6}}}} \cr & = 1.45 \times {10^9}Pa \cr & = 1.45 \times {10^9} \times {10^{ - 5}}bar \cr & = 1.45 \times {10^4}bar \cr & = 14500\,bar \cr} $$

For spontaneous reaction, $$dS > 0$$  and $$dG$$ should be negative ie. $$<0.$$

Mass independent properties ( molar conductivity and electromotive force ) are intensive properties. Resistance and heat capacity are mass dependent, hence extensive properties.

$$\Delta {H_r} = \sum {H_{f{\text{(product}}s)}^ \circ - \sum {H_{f{\text{(reactants)}}}^ \circ } } $$

$$\eqalign{ & \Delta {G^ \circ } = \Delta {H^ \circ } - T\Delta {S^ \circ };\, - RT\,\ell nK = \Delta {H^ \circ } - T\Delta {S^ \circ } \cr & \ell nK = \frac{{\Delta {H^ \circ } - T\Delta {S^ \circ }}}{{RT}} \cr} $$

\[\begin{align} & {{N}_{2}}\left( g \right)+\frac{1}{2}{{O}_{2}}\to {{N}_{2}}O\left( g \right) \\ & N\equiv N\left( g \right)+\frac{1}{2}\left( O=O \right)\to \underset{\scriptscriptstyle\centerdot\centerdot}{\ddot{N}}=\overset{+}{\mathop{N}}\,=\underset{\scriptscriptstyle\centerdot\centerdot}{\ddot{O}}\left( g \right) \\ \end{align}\]
$$\Delta {H_f}^ \circ = $$   [ Energy required for breaking of bonds ] – [ Energy released for forming of bonds ]
$$\eqalign{ & = (\Delta {H_{N \equiv N}} + \frac{1}{2}\Delta {H_{O = O}} - (\Delta {H_{N = N}} + \Delta {H_{N = O}}) \cr & = (946 + \frac{1}{2} \times 498) - (418 + 607) \cr & = 170\,kJ\,mo{l^{ - 1}} \cr & {\text{Resonance energy}} = 82 - 170 \cr & = - 88\,kJ\,mo{l^{ - 1}} \cr} $$

$$\eqalign{ & {\text{At constant volume,}}\,\Delta V = 0 \cr & w = - P\Delta V = 0\,\,\,\,\,\Delta U = q = 200\,J \cr} $$

$$\eqalign{ & \Delta H = \sum {\Delta {H_P} - \sum {\Delta {H_R}} } \cr & \,\,\,\,\,\,\,\,\,\, = - 1670 - \left( { - 834} \right) \cr & \,\,\,\,\,\,\,\,\,\, = - 836\,kJ\,mo{l^{ - 1}} \cr} $$

As $$\Delta S$$  does not depend on path and only depends on initial and final stages i.e., it is a state function thus
$$\Delta {S_{X \to Z}} = \Delta {S_{X \to Y}} + \Delta {S_{Y \to Z}}$$
As we know that work is not a state function and depends on path,
Thus,
$$\eqalign{ & {w_{X \to Z}} \ne {w_{X \to Y}} + {w_{Y \to Z}} \cr & {w_{X \to Y}} = PdV\left( {P\,\,{\text{is}}\,\,{\text{constant}}} \right) \cr & {w_{Y \to Z}} = 0\,\,\,\,\left( {V\,\,{\text{is}}\,\,{\text{constant}}} \right) \cr & {w_{X \to Z}} = 2.303nRT\,\,\log \frac{{{V_2}}}{{{V_1}}} \cr & {w_{X \to Y \to Z}} = {w_{X \to Y}} + {w_{Y \to Z}} \cr} $$
As $${w_{Y \to Z}} = 0,$$   hence $${w_{X \to Y \to Z}} = {w_{X \to Y}}$$