Chemical Thermodynamics MCQ Questions & Answers in Physical Chemistry | Chemistry

Isothermal process means temperature remains constant. At constant temperature, internal energy $$\left( {\Delta E} \right)$$  also remains constant. So, $$\Delta E = 0$$

$$\eqalign{ & {\text{Atomisation of methane}} \cr & C{H_4}\left( g \right) \to C\left( g \right) + 4H\left( g \right); \cr & \Delta H = 360\,kcal \cr & \therefore \,\,C - H\,{\text{bond energy}} = \frac{{360}}{4} \cr & = 90\,kcal\,mo{l^{ - 1}} \cr & {C_2}{H_6}\left( g \right) \to 2C\left( g \right) + 6H\left( g \right); \cr & \Delta H = 620\,kcal \cr & {\text{or}}\,\,{H_{C - C}} + 6\,{H_{C - H}} = 620 \cr & \therefore \,\,{H_{C - C}} = 620 - 6\,{H_{C - H}} \cr & = 620 - 6 \times 90 = 80\,kcal\,mo{l^{ - 1}} \cr} $$

$$\eqalign{ & {C_6}{H_6}\left( l \right) + \frac{{15}}{2}{O_2}\left( g \right) \to 6C{O_2}\left( g \right) + 3{H_2}O\left( l \right) \cr & \Delta n = 6 - \frac{{15}}{2} \cr & = - \frac{3}{2} \cr} $$

A reversible process involves a slow change during operations and surrounding is always in equilibrium with system.

The first reaction is exothermic and the second reaction is endothermic. If on passing the mixture of $${O_2}$$  and $${H_2}O$$  (steam) over coke while keeping the temperature constant $$\Delta H$$  of both the reactions must be same.
Moles of $${O_2}$$ needed to evolve $$132\,kJ$$
$$\eqalign{ & = \frac{{0.5 \times 132}}{{110}} \cr & = 0.6 \cr} $$
Hence steam : $${O_2}$$ ratio must be 1 : 0.6

Entropy is the degree of randomness or disorder of the system. When the temperature of the system is zero kelvin, then all the motion of molecules ceases. According to third law of thermodynamics “At absolute zero the entropy of a perfectly crystalline substance is taken as zero.”

$$\eqalign{ & \Delta {S_{\left( {A\, \to \,B} \right)}} = \Delta {S_{\left( {A\, \to \,C} \right)}} + \Delta {S_{\left( {C\, \to \,D} \right)}} - \Delta {S_{\left( {B\, \to \,D} \right)}} \cr & = 50 + 30 - 20 \cr & = 60\,e.u. \cr} $$

$$\eqalign{ & {\text{Given,}}\,\,\Delta {H_{H - H}} = 434\,kJ/mol \cr & \Delta {H_{Cl - Cl}} = 242\,kJ/mol \cr & \Delta {H_{H - Cl}} = 431\,kJ/mol \cr & \frac{1}{2}{H_2} + \frac{1}{2}C{l_2} \to HCl,\Delta {H_r} = ? \cr & \Delta {H_r} = \frac{1}{2} \times \Delta {H_{H - H}} + \frac{1}{2} \times \Delta {H_{Cl - Cl}} - \Delta {H_{H - Cl}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \times 434 + \frac{1}{2} \times 242 - 431 \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 217 + 121 - 431 \cr & \,\,\,\,\,\,\,\,\,\,\,\, = - 93\,kJ/mol \cr} $$

$$\eqalign{ & 0.50\,mol\,\,{\text{of}}\,\,HCl \equiv 0.50\,mol\,\,{\text{of}}\,\,{H^ + } \cr & 0.30\,mol\,\,{\text{of}}\,\,NaOH \equiv 0.30\,mol\,\,{\text{of}}\,\,O{H^ - } \cr & {H^ + } + O{H^ - } \to {H_2}O{\text{(Neutralisation)}} \cr & 0.30\,mol\,\,{\text{of}}\,\,HCl \equiv 0.30\,mol\,\,{\text{of}}\,\,{H^ + } \cr & {\text{or}}\,\,0.30\,mol\,\,{\text{of}}\,\,{H^ + } \equiv 0.30\,mol\,\,{\text{of}}\,\,O{H^ - } \cr & {\text{Heat}}\,\,{\text{evolved with}}\,\,{\text{1}}\,{\text{mol}}\,\,{H^ + } = 57.1\,kJ \cr} $$
$${\text{Heat evolved with}}\,\,0.30\,mol\,\,{H^ + }$$      $$ = 57.1 \times 0.30 = 17.13\,kJ$$

For endothermic reactions, $${H_R} < {H_P}.$$