Chemical Thermodynamics MCQ Questions & Answers in Physical Chemistry | Chemistry

From 1^st law of thermodynamics
$$\Delta U = q + w$$
For adiabatic process :
$$q = 0$$
$$\therefore \,\Delta U = W$$
i.e. change in internal energy $$\left( {\Delta U} \right)$$   is equal to adiabatic work.

$$\eqalign{ & {\text{According to heat capacity rule,}} \cr & q = mc\Delta T,\,\,c = \frac{q}{{m\left( {{T_2} - {T_1}} \right)}} \cr & {\text{Given that,}}\,\,c = 75\,J{K^{ - 1}}\,mo{l^{ - 1}} \cr & q = 1.0\,kJ = 1000J \cr & {\text{Mass}} = 100\,g\,{\text{water}} \cr & {\text{Molar mass of water}} = 18g \cr} $$
$$75 = \frac{{1000}}{{5.55 \times \Delta T}}$$    $$\left( {{\text{number of moles}} = \frac{{100}}{{18}} = 5.55} \right)$$
$$\eqalign{ & \therefore \,\,\Delta T = \frac{{1000}}{{5.55 \times 75}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2.4\,K \cr} $$

Measure of disorder of a system is nothing but entropy. For a spontaneous reaction, $$\Delta G < 0.$$   As per Gibbs Helmholtz equation, $$\Delta G = \Delta H - T\Delta S$$
Thus $$\Delta G$$  is $$ - ve$$  only
when $$\Delta H = - ve$$   (exothermic)
and $$\Delta S = + ve$$   (increasing disorder)

$$\Delta G = \Delta H - T\Delta S$$
For a spontaneous reaction if $$\Delta H$$  is $$ + ve$$  and $$\Delta S = + ve,\Delta G$$    will be negative when $$T\Delta S > \Delta H.$$

In isothermal process, temperature remains constant.

We have the following relationship.
$$\ln \,{K_p} = - \frac{{\Delta H}}{{RT}} + {\text{constant}}$$
Now, if $$\Delta H$$  is $$ - ve,$$  Then the relation comes out to be :
$${\text{ln}}\,{{\text{K}}_p} = \frac{{\Delta H}}{{RT}} + {\text{constant}}$$
$$\therefore \,\,\ln \,{K_p}\,\,{\text{vs}}\,\frac{1}{T}$$    graph is similar to $$y = mx + c$$    where, $$m$$  and $$c$$  both are positive. So, the graph will be as shown.

$${\text{The}}\,\,{\text{value of}}\,\,{\Delta _c}H\,\,{\text{is}}\,\, - 2658\,kJ/mol.$$

Heat of neutralisation of strong acid and strong base is $$ - 57.33\,kJ.$$   $$MgO$$  is weak base while $$HCl$$  is strong acid, so the heat of neutralisation of $$MgO$$  and $$HCl$$  is lower than $$ - 57.33\,kJ$$   because $$MgO$$  requires some heat for ionisation, therefore the net released amount of heat is decreased.

$$\eqalign{ & \Delta G = \Delta H - T\Delta S \cr & {\text{At equilibrium}},\Delta G = 0,\Delta H = T\Delta S \cr} $$
$$\Delta S = \frac{{\Delta H}}{T} = \frac{{42600\,J\,mo{l^{ - 1}}}}{{373\,K}}$$       $$ = 114.2\,J\,{K^{ - 1}}\,mo{l^{ - 1}}$$

\[\begin{align} & {{H}_{2}}O\left( l \right)\xrightarrow{{{100}^{\circ }}C}{{H}_{2}}O\left( g \right) \\ & {{\Delta }_{vap}}{{H}^{\circ }}={{\Delta }_{vap}}{{E}^{\circ }}+\Delta {{n}_{g}}RT \\ & {{\Delta }_{vap}}{{H}^{\circ }}=\text{enthalpy of vaporisation} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=40.66\,kJ\,mo{{l}^{-1}} \\ & \text{For the above reaction,} \\ & \Delta {{n}_{g}}={{n}_{p}}-{{n}_{r}} \\ & \,\,\,\,\,\,\,\,\,\,=1-0 \\ & \,\,\,\,\,\,\,\,\,\,=1 \\ & R=8.314 \\ & T={{100}^{\circ }}C \\ & \,\,\,\,=273+100 \\ & \,\,\,\,=373K \\ \end{align}\]
$$\therefore 40.66\,kJ\,mo{l^{ - 1}} = $$     $${\Delta _{vap}}{E^ \circ } + 1 \times 8.314 \times {10^{ - 3}} \times 373$$
$$\eqalign{ & {\Delta _{vap}}{E^ \circ } = 40.66\,kJ\,mo{l^{ - 1}} - 3.1\,kJ\,mo{l^{ - 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = + 37.56\,kJ\,mo{l^{ - 1}} \cr} $$