Chemical Thermodynamics MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & T = \frac{{\Delta H}}{{\Delta S}} \cr & \,\,\,\,\, = \frac{{7.25}}{{0.007}} \cr & \,\,\,\,\, = 1035.7\,K \cr} $$

$$\Delta G = - 2.303\,RT\,\,\log \,K$$
$$ = - 2.303 \times 8.314 \times 300\,\log \,10 = - 5744.1\,J$$         $$ \Rightarrow - 5.74\,kJ$$

$$\eqalign{ & {\text{For reaction,}} \cr & C\left( s \right) + 2{H_2}\left( g \right) \to C{H_4}\left( g \right),\,\Delta {H^ \circ } = ? \cr & C + {O_2} \to C{O_2},\,\Delta H = - 94\,kcal\,\,...{\text{(i)}} \cr} $$
$$2{H_2} + {O_2} \to 2{H_2}O,\,\Delta H$$     $$ = - 68 \times 2\,kcal\,\,...{\text{(ii)}}$$
$$C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O,$$       $$\Delta H = - 213\,kcal\,\,...{\text{(iii)}}$$
On adding Eqs. (i) and (ii) and then subtracting Eq. (iii)
$$\eqalign{ & = \left( { - 94} \right) + \left( { - 2 \times 68} \right) - \left( { - 213} \right) \cr & = - 230 + 213 \cr & = - 17\,k\,cal/mol \cr} $$

For the reaction,
$${C_3}{H_8}\left( g \right) + 5{O_2}\left( g \right) \to $$     $$3C{O_2}\left( g \right) + 4{H_2}O\left( l \right)$$
$$\Delta {n_g} = $$  number of gaseous moles of products $$ - $$ number of gaseous moles of reactants
$$\eqalign{ & = 3 - 6 = - 3 \cr & \therefore \,\,\Delta H = \Delta E + \Delta n\,RT \cr & {\text{or}}\,\,\Delta H - \Delta E = \Delta nRT \cr & \therefore \,\,\Delta H - \Delta E = - 3RT \cr} $$

Reactions in which there is a decrease in the number of moles of the gaseous components, i.e. $$\Delta {n_g}$$  is negative, the enthalpy change $$\left( {\Delta H} \right)$$  is lesser than the internal energy change $$\left( {\Delta E} \right).$$
Reaction in which there is a increase in the number of moles of gaseous components i.e. $$\Delta ng$$  is positive, the enthalpy change is greater than the internal energy change.
$$\Delta H = \Delta E + \Delta {n_g}RT$$

$$\eqalign{ & {\text{Bond dissociation energy of}} \cr & P{H_3}\left( g \right) = 228\,kcal\,mo{l^{ - 1}} \cr & P - H\,\,{\text{bond energy}} = \frac{{228}}{3} = 76\,kcal\,mo{l^{ - 1}} \cr} $$

$$\eqalign{ & {\text{Bond energy of}}\,\,4\left( {P - H} \right) + \left( {P - P} \right) \cr & = 355\,kcal\,mo{l^{ - 1}} \cr & {\text{or}}\,\,4 \times 76 + \left( {P - P} \right) = 355\,kcal\,mo{l^{ - 1}} \cr & P - P\,\,{\text{bond energy}} = 51\,kcal\,mo{l^{ - 1}} \cr} $$

The spontaneity of a reaction is based upon the negative value of $$\Delta G$$  and $$\Delta G$$  is based upon $$T,\Delta S$$  and $$\Delta H$$  according to following equation ( Gibbs-Helmholtz equation )
$$\Delta G = \Delta H - T\Delta S$$
If the magnitude of $$\Delta H - T\Delta S$$   is negative, then the reaction is spontaneous.
when $$T\Delta S > \Delta H$$   or we can say that $$\Delta H$$  and $$\Delta S$$  are positive, then $$\Delta G$$  is negative.

Since the system is insulated, heat is not allowed to enter or leave the system.
Thus, $$q = 0,\Delta U = q + w \Rightarrow \Delta U = w$$

From enthalpy equation, $$\Delta H = \Delta E + {n_g}RT$$
For the reaction, $$PC{l_5}\left( g \right) \rightleftharpoons PC{l_3}\left( g \right) + C{l_2}\left( g \right)$$
$$\Delta {n_g} = $$  product mole - reactant mole $$\Delta n = 2 - 1 = 1$$
Thus, the value of $$\Delta H$$ is positive or $$ > 0.$$
$$\Delta G = \Delta H - T\Delta S$$
For a spontaneous reaction, $$\Delta G$$  must be negative. Since in this reaction $$\Delta H$$  is positive, so for the negative value of $$\Delta G,\Delta S$$   must be positive or $$ > 0.$$
Hence,   $$\Delta H > 0,\,\Delta S > 0$$

$$\eqalign{ & \Delta G = \Delta H - T\Delta S \cr & \therefore \,\,\Delta G = 9100 - 373.2 \times 25.5 \cr & = - 416.6\,cal\,mo{l^{ - 1}} \cr} $$