Chemical Thermodynamics MCQ Questions & Answers in Physical Chemistry | Chemistry

$${\text{I - II,}}$$  gives, $$2Al + F{e_2}{O_3} \to 2Fe + A{l_2}{O_3}$$
we have $$ - 400 - \left( { - 200} \right) = - 200\,;$$      For one $$mole$$  of iron the value is $$ - 100\,cal$$

$$\eqalign{ & 2S{O_{2\left( g \right)}} + {O_{2\left( g \right)}} \to 2S{O_{3\left( g \right)}} \cr & \Delta {n_g} = 2 - 3 = - 1,\Delta S = - ve \cr} $$

$$\Delta {G^ \circ } = - RT\,\ln {K_c}\,or\, - \Delta {G^ \circ } = RT\,\ln {k_c}$$

$$\eqalign{ & Hg\left( l \right) \rightleftharpoons Hg\left( g \right), \cr & {\Delta _R}{S^ \circ } = 174.4 - 77.4 = 97\,J/K{\text{ - }}mol \cr & \because \,\,\Delta {G^ \circ } = \Delta {H^ \circ } - T\Delta {S^ \circ } = 0 \cr & T = \frac{{\Delta {H^ \circ }}}{{\Delta {S^ \circ }}} \cr & = \frac{{60.8 \times 1000}}{{97}} \cr & = 626.8\,K \cr} $$

$$\eqalign{ & \Delta {H_{hyd.}} = \Delta {H_{sol.}} - \Delta {H_{lattice}} \cr & = 1 - 180 = - 179\,kcal\,mo{l^{ - 1}} \cr & {\text{Then}}\,\,\Delta {H_{hyd.}}\left( {N{a^ + }} \right) + \Delta {H_{hyd.}}\left( {C{l^ - }} \right) = - 179 \cr & {\text{or}}\,\,\Delta {H_{hyd.}}\left( {N{a^ + }} \right) + \frac{2}{3}\Delta {H_{hyd}} = - 179 \cr & {\text{or}}\,\,\Delta {H_{hyd.}}\left( {N{a^ + }} \right) = - 107.4\,kcal\,mo{l^{ - 1}} \cr} $$

The given phase equilibria is $${\text{Liquid}} \rightleftharpoons {\text{Vapour}}.$$    This equilibrium states that, when liquid is heated, it converts into vapour but on cooling, it further converts into liquid, which is derived by Clausius Clapeyron and the relationship is written as,
$$\frac{{d\ln P}}{{dT}} = - \frac{{\Delta {H_v}}}{{R{T^2}}}$$
where, $$\Delta {H_v} = $$  Heat of vaporisation

TIPS/Formulae :
$$\Delta H_f^ \circ $$ is the enthalpy change when 1 mole of the substance is formed from its elements in their standard states.
In (A) carbon is present in diamond however standard state of carbon is graphite . Again, in (D) $$CO$$ $$(g)$$ is involved so it can't be the right option. Further in (C) 2 moles of $$N{H_3}$$ are generated. Hence the correct option is (B).

$${H^ + }\left( {aq} \right) + O{H^ - }\left( {aq} \right) \rightleftharpoons {H_2}O\left( l \right);$$       $$\Delta H = - 57.3\,kJ\,\,....\left( {\text{i}} \right)$$
$${H_2}{C_2}{O_4} + 2O{H^ - } \rightleftharpoons $$       $${C_2}O_4^{ - - } + 2{H_2}O;\Delta H = - 106\,kJ\,\,....\left( {{\text{ii}}} \right)$$
For the second reaction the value should have been
$$\eqalign{ & 2 \times \left( { - 57.3} \right) = - 114.6\,kJ \cr & {\text{The difference}} \cr} $$
$$\left( {114.6 - 106} \right) = 8.6\,kJ\,mo{l^{ - 1}}$$      is used to effect of the ionisation of oxalic acid.

For spontaneous process, $$\Delta S$$  must be positive. In reversible process
$$\Delta {S_{{\text{system}}}} + \Delta {S_{{\text{surrounding}}}} = 0$$
Hence, system is present in equilibrium.
( i.e. it is not spontaneous process )
While in irreversible process
$$\Delta {S_{{\text{system}}}} + \Delta {S_{{\text{surrounding}}}} > 0$$
Hence, in the process $$\Delta S$$  is positive.

$$\eqalign{ & {\text{For a reaction to be at equilibrium}}\,\,\Delta G = 0.\,\,Since \cr & \Delta G = \Delta H - T\Delta S\,\,{\text{so}}\,{\text{at}}\,{\text{equilibrium}}\,\,\Delta H - T\Delta S = 0 \cr & or\,\,\Delta H = T\Delta S \cr & {\text{For}}\,{\text{the}}\,{\text{reaction}} \cr & \frac{1}{2}{X_2} + \frac{3}{2}{Y_2} \to X{Y_3}\,;\,\Delta H = - 30\,kJ\,\left( {{\text{given}}} \right) \cr & {\text{Calculating}}\,\Delta S\,{\text{for}}\,{\text{the}}\,{\text{above}}\,{\text{reaction,}}\,{\text{we}}\,{\text{get}} \cr & \Delta S = 50 - \left[ {\frac{1}{2} \times 60 + \frac{3}{2} \times 40} \right]J{K^{ - 1}} \cr & = 50 - \left( {30 + 60} \right)J{K^{ - 1}} \cr & = - 40J{K^{ - 1}} \cr & {\text{At}}\,{\text{equilibrium,}}\,\,\,T\Delta S = \Delta H\,\,\,\left[ {\,\because \Delta G = 0} \right] \cr & \therefore \,\,T \times \left( { - 40} \right) = - 30 \times 1000\,\,\left[ {\,\because 1kJ = 1000J} \right] \cr & or\,\,\,T = \frac{{ - 30 \times 1000}}{{ - 40}}\,or\,750K \cr} $$