Chemical Thermodynamics MCQ Questions & Answers in Physical Chemistry | Chemistry

$$S\left( s \right) + \frac{3}{2}{O_2}\left( g \right) \to S{O_3}\left( g \right) + 2x\,kcal\,\,\,...{\text{(i)}}$$
By inverting second equation we get,
$$S{O_3}\left( g \right) \to S{O_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) - y\,kcal\,...{\text{(ii)}}$$
On addition Eqs. (i) and (ii)
$$S\left( s \right) + {O_2}\left( g \right) \to S{O_2}\left( g \right) + \left( {2\,x - y} \right)\,kcal$$
Hence, heat of formation of $$S{O_2}$$  is $$\left( {2x - y} \right)\,kcal.$$

Since the reaction shows combustion of one mole of carbon, $$\Delta {H_{{\text{reaction}}}}$$   is enthalpy of combustion as $$1\,mole$$   of $$C{O_2}$$  is formed. $$\Delta {H_{{\text{reaction}}}}$$   is also the enthalpy of formation of $$C{O_2}.$$

Energy profile diagram for a reaction is as

From the figure, it is clear that
$${\left( {{E_a}} \right)_b} = {\left( {{E_a}} \right)_f} + \Delta H$$
[ Here $${\left( {{E_a}} \right)_b} = $$   activation energy of backward reaction
and $${\left( {{E_a}} \right)_f} = $$   activation energy of forward reaction ].
$$\eqalign{ & {\text{If}}\,{\left( {{E_a}} \right)_b} = {\left( {{E_a}} \right)_f} \cr & {\text{then}}\,\,\Delta H = 0 \cr} $$

The standard enthalpy of formation of $$C{H_4}$$ is given by the equation :
$$C\left( s \right) + 2{H_2}\left( g \right) \to C{H_4}\left( g \right)$$
Hence, dissociation energy of hydrogen and enthalpy of sublimation of carbon is required.

When $$q$$  amount of heat is given to the system, the system do $$w$$  work, i.e., utilise heat but does not exchange matter, thus system is dosed.

If the Gibbs free energy for a system $$\left( {\Delta {G_{{\text{system)}}}}} \right)$$   is equal to zero, then system is present in equilibrium at a constant temperature and pressure.
$$\eqalign{ & \Delta G = {\Delta _r}{G^ \circ } + RT\,\ln \,K;\,\,\Delta G = 0 \cr & {\Delta _r}{G^ \circ } = - RT\,{\text{ln}}\,K \cr} $$
$$K = $$  equilibrium constant
If $$\Delta {G_{{\text{system}}}} < 0,$$   then the process is spontaneous

$$\eqalign{ & \Delta H = \Delta U + \Delta nRT\,\,for\,\,{N_2} + 3{H_2} \to 2N{H_3} \cr & \Delta {n_g} = 2 - 4 = - 2 \cr & \therefore \,\,\Delta H = \Delta U - 2RT\,\,or\,\,\Delta U = \Delta H + 2RT \cr & \therefore \,\,\Delta U > \Delta H \cr} $$

In adiabatic process, heat exchange is constant, so $$q=0$$  and for free expanion, $$W = 0,\,\,\therefore \,\,\Delta T = 0.$$

$$\mathop {Pb}\limits^{ + 4} \,{O_2} \to \mathop {Pb}\limits^{ + 2} \,O,\,\,\Delta {G_{298}} < 0$$
For this reaction $$\Delta G$$  is negative, hence $$P{b^{2 + }}$$  is more stable than $$P{b^{4 + }}.$$
$$\mathop {Sn}\limits^{ + 4} \,{O_2} \to \mathop {Sn}\limits^{ + 2} \,O,\,\,\Delta {G_{298}} > 0$$
For this reaction $$\Delta G$$  is positive, hence $$S{n^{4 + }}$$  is more stable than $$S{n^{2 + }}$$  because for spontaneous change $$\Delta G$$  must be negative.