Thermodynamics

We have

$\Delta$G$^\circ$ = $\Delta$H$^\circ$ $-$ T$\Delta$S$^\circ$

$\Delta$G$^\circ$ = $-$ 29.8 $-$ 298 ($-$ 0.1) = 0

$\Delta$G$^\circ$ = $-$ 2.302 RT log K_eq

log K_eq = 0 $\Rightarrow$ K_eq = 1

Given

$C\left( s \right) + {O_2}\left( g \right) \to C{O_2}\left( g \right);$

$\Delta H = - 393.5\,\,kJ\,mo{l^{ - 1}}......\left( i \right)$

$CO\left( g \right) + {1 \over 2}{O_2}\left( g \right) \to C{O_2}\left( g \right);$

$\Delta H = - 283.5\,kJ\,mo{l^{ - 1}}\,\,\,\,\,....\left( {ii} \right)$

$\therefore$ Heat of formation of

$Co = eqn\left( i \right) - eqn\left( {ii} \right)$

$ = - 393.5 - \left( { - 283.5} \right)$

$ = - 110\,kJ$

$\Delta {G^ \circ }_{NO\left( g \right)}$

$\,\,\,\,\,\,\,\,\,\,\,\,\, = 86.6kJ/mol = 86600J/mol;$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$${G^ \circ }_{N{O_2}\left( g \right)} = x\,J/mol$

$T = 298,\,{K_p} = 1.6 \times {10^{12}}$

$\Delta {G^ \circ } = - RT\,\ln \,{K_p}$

Given equation,

$2NO\left( g \right) + {O_2}\left( g \right)\,\rightleftharpoons\,2N{O_2}\left( g \right)$

$2\Delta {G^ \circ }_{N{O_2}} - 2\Delta {G^ \circ }_{NO}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)$

$2\Delta {G^ \circ }_{N{O_2}} - 2 \times 86600$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)$

$2\Delta {G^ \circ }_{N{O_2}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ = 2 \times 86600 - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)$

$\Delta {G^ \circ }_{N{O_2}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ = {1 \over 2}\left[ {2 \times 86600 - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)} \right]$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ = 0.5\left[ {2 \times 86600 - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)} \right]$

${C_2}{H_5}OH\left( \ell \right) + 3{O_2}\left( g \right)$

$ \to 2C{O_2}\left( g \right) + 3{H_2}O\left( \ell \right)$

Bomb calorimeter gives $\Delta U$ of the reaction

Given, $\Delta U = - 1364.47\,kJ\,mo{l^{ - 1}}$

$\Delta {n_g} = - 1$

$\Delta H = \Delta U + \Delta {n_g}RT$

$ = - 1364.47 - {{1 \times 8.314 \times 298} \over {1000}}$

$ = - 1366.93\,kJ\,mo{l^{ - 1}}$

By I^st law of thermodynamics, q = $\Delta $E – W

At const T, ∆E = 0

q = – W

Heat absorbed = 208 J

$ \therefore $ q = +208 J

W = – 208 J

$\Delta {G^ \circ } = \Delta {H^ \circ } - T\Delta {S^ \circ };$

$\,\,\,\,\,\,\,\,\, - RT\,\ell nK = \Delta {H^ \circ } - T\Delta {S^ \circ }$

$\ell nK = - {{\Delta {H^ \circ } - T\Delta {S^ \circ }} \over {RT}}$

Entropy change for an isothermal reversible process is given by

$\Delta S = nR{\mkern 1mu} ln{{{V_2}} \over {{V_1}}}$

$ = 2 \times 8.314 \times 2.303log\,{{100} \over {10}}$

$ = 38.3\,J\,mo{l^{ - 1}}\,\,{K^{ - 1}}$

At equilibrium $\Delta G = 0$

Hence, $\Delta G = \Delta H - {T_e}\Delta S = 0$

$\therefore$ $\,\,\,\,\,\,\Delta H = {T_e}\Delta S\,\,\,$

or $\,\,\,\,{T_e} = {{\Delta H} \over {\Delta S}}\,\,$

For a spontaneous reaction

$\Delta G$ must be negative

which is possible only if $\Delta H < T\Delta S$

or $\,\,\,T > {{\Delta H} \over {\Delta S}};{T_e} < T$

${N_2} + 3{H_2} \to 2N{H_3}\,$

$\,\Delta H = 2 \times - 46.0\,\,kJ\,mo{l^{ - 1}}$

Let $x$ be the bond enthalpy of $N-H$ bond then

[Note : Enthalpy of formation or bond formation enthalpy is given which is negative but the given reaction involves bond breaking hence values should be taken as positive. ]

$\Delta H = \sum \, $ Bond energies of products

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \sum \, $ Bond energies of reactants

$2 \times - 46 = 712 + 3 \times \left( {436} \right) - 6x;$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\, - 92 = 2020 - 6x$

$6x = 2020 + 92$

$ \Rightarrow 6x = 2112$

$ \Rightarrow x = + 352\,kJ/mol$

Given, for reaction

$(i)$ $\,\,\,\,\,\,{H_2}O\left( \ell \right) \to {H^ + }\left( {aq.} \right) + O{H^ - }\left( {aq.} \right);$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H_r} = 57.32\,kJ$

$(ii)$ $\,\,\,\,\,\,{H_2}\left( g \right) + {1 \over 2}{O_2}\left( g \right) \to {H_2}O\left( \ell \right);$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H_r} = - 286.20\,kJ$

For reaction $(i)$

$\Delta {H_r} = \Delta {H^ \circ }_f\left( {{H^ + }.aq} \right) + $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H^ \circ }_f\left( {O{H^ - }.aq} \right) - \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right)$

$57.32 = 0 + \Delta {H^ \circ }_f\left( {O{H^{ - 1}},aq} \right) - $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right)\,\,\,\,\,\,...\left( {iii} \right)$

For reaction $(ii)$

$\Delta {H_r} = \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right) - \Delta {H^ \circ }_f\left( {{H_2},g} \right) - $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{1 \over 2}\Delta {H^ \circ }_f\left( {{O_2},g} \right)$

$ - 286.20 = \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right)$

On replacing this value in equ. $(iii)$ we have

$57.32 = \Delta {H^ \circ }_f\left( {O{H^ - }.aq} \right) - \left( { - 286.20} \right)$

$\Delta {H^ \circ }_f = - 286.20 + 57.32$

$ = - 228.88\,kJ$

For a reaction to be at equilibrium $\Delta G = 0.$

Since $\Delta G = \Delta H - T\Delta S$

so at equilibrium $\Delta H - T\Delta S = 0$

or $\,\,\,\,\Delta H = T\Delta S$

For the reaction

${1 \over 2}{X_2} + {3 \over 2}{Y_2} \to X{Y_3};\,\,\,$

$\Delta H = - 30kJ\,\,\left( {given} \right)$

Calculating $\Delta S$ for the above reaction, we get

$\Delta S = 50 - \left[ {{1 \over 2} \times 60 + {3 \over 2} \times 40} \right]J{K^{ - 1}}$

$ = 50 - \left( {30 + 60} \right)J{K^{ - 1}}$

$ = - 40\,J{K^{ - 1}}$

At equilibrium, $\,\,\,T\Delta S = \Delta H$

[ as $\,\,\,\Delta G = 0$ ]

$\therefore$ $\,\,\,\,\,$ $T \times \left( { - 40} \right) = - 30 \times 1000$

[ as $1kJ=1000J$ ]

or $\,\,\,\,\,$ $T = {{ - 30 \times 1000} \over { - 40}}$

or $\,\,\,\,\,$ = $750$ $K$

The energy involved in the conversion of

${1 \over 2}C{{\rm}_2}\left( g \right)\,\,$ to ${\mkern 1mu} {\mkern 1mu} {\mkern 1mu} C{l^{ - 1}}\left( {aq} \right)$ is given by

${\mkern 1mu} \Delta H = {1 \over 2}{\Delta _{diss}}H_{C{l_2}}^{\left( - \right)} + {\Delta _{eg}}H_{Cl}^{\left( - \right)} + {\Delta _{hy{\rm{I}}}}H_{Cl}^{\left( - \right)}$

Substituting various values from given data, we get

$\Delta H = \left( {{1 \over 2} \times 240} \right) + \left( { - 349} \right) + \left( { - 381} \right)kJ\,mo{l^{ - 1}}$

$ = \left( {120 - 349 - 381} \right)kJmo{l^{ - 1}}$

$ = - 610\,kJ\,mo{l^{ - 1}}$

i.e., the correct answer is $(b)$

Given

$\Delta H = 41\,kJ\,mo{l^{ - 1}}$

$ = 41000\,J\,mo{l^{ - 1}}$

$T = {100^ \circ }C = 273 + 100$

$ = 373\,K,n = 1$

$\Delta U = \Delta H - \Delta nRT$

$ = 41000 - \left( {2 \times 8.314 \times 373} \right)$

$ = 37898.88\,J\,mo{l^{ - 1}}$

$ = 37.9\,kJ\,mo{l^{ - 1}}$

$\Delta {G^ \circ } = \Delta {H^ \circ } - T\Delta {S^ \circ }$

For a spontaneous reaction $\Delta {G^ \circ } < 0$

or $\,\,\,\Delta {H^ \circ } - T\Delta {S^ \circ } < 0$

$ \Rightarrow T > {{\Delta {H^ \circ }} \over {\Delta {S^ \circ }}}$

$ \Rightarrow T > {{179.3 \times {{10}^3}} \over {160.2}} > 1117.9K \approx 1118K$

Spontaneity of reaction depends on tendency to acquire minimum energy state and maximum randomness. For a spontaneous process in an isolated system the change in entropy is positive.

${{\rm{I}}_2}\left( s \right) + C{l_2}\left( g \right) \to 2{\rm{I}}Cl\left( g \right)$

$\Delta A = \left[ {\Delta {{\rm{I}}_2}\left( s \right) \to {l_2}\left( g \right) + \Delta {H_{{\rm I} - l}} + \Delta {H_{C{\rm I} - Cl}}} \right] - $

${\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 2\left[ {\Delta {H_{{\rm{I}} - Cl}}} \right]$

$ = 151.0 + 242.3 + 62.76 - 2 \times 211.3$

$ = 33.46$

$\Delta H_f^0\left( {{\rm{I}}Cl} \right) = {{33.46} \over 2}$

$ = 16.73\,\,kJ/mol$

NOTE : In a reversible process the work done is greater than in irreversible process. Hence the heat absorbed in reversible process would be greater than in the latter case. So,

$\,\,\,\,\,\,\,\,\,\,\,$ ${T_f}\left( {rev.} \right) < {T_f}\left( {irr.} \right)$

The standard enthalpy of formation of $C{H_4}$ is given by the equation :

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $C\left( s \right) + 2{H_2}\left( g \right) \to C{H_4}\left( g \right)$

In order to calculate average energy for C – H bond formation we should know the following data.

C_(graphite) $ \to $ C_(g) ; $\Delta H_f^o$ = enthalpy of sublimation of carbon

H_2(g) $ \to $ 2H(g) ; $\Delta $H = bond dissociation energy of H₂

For the reaction, ${C_{\left( s \right)}} + {1 \over 2}{O_{2\left( g \right)}} \to CO$

$\Delta H = \Delta U + \Delta nRT$

or $\,\,\,\,\,\Delta H - \Delta U = \Delta nRT$

$\Delta n = 1 - {1 \over 2} = {1 \over 2};\,\,$

$\Delta H - \Delta U = {1 \over 2} \times 8.314 \times 298$

$ = 1238.78\,Jmo{l^{ - 1}}$

${X_2} + {Y_2} \to 2XY,$ $\Delta H = 2\left( { - 200} \right).$

Let $x$ be the bond dissociation energy of ${X_2}.$

Then $\Delta H = - 400$

$ = {\xi _{x - x}} + {\xi _{y - y}} - 2{\xi _{x - y}}$

$ = x + 0.5x - 2x$

$ = - 0.5x$

or $\,\,\,x = {{400} \over {0.5}} = 800\,kJ\,mo{l^{ - 1}}$

$\Delta H = \Delta U + \Delta nRT$

for $\,\,\,{N_2} + 3{H_2} \to 2N{H_3}$

$\,\,\,\Delta {n_g} = 2 - 4 = - 2$

$\therefore$ $\,\,\,\Delta H = \Delta U - 2RT\,\,\,$

or $\,\,\,\Delta U = \Delta H + 2RT$

$\therefore$ $\,\,\,\Delta U > \Delta H$

Enthalpy of reaction $\left( {\Delta H} \right) = {E_{{a_{\left( f \right)}}}} - {E_{{a_{\left( b \right)}}}}$

for an endothermic reaction $\Delta H = + ve$

Hence for $\Delta H$ to be negative

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ ${E_{{a_{\left( b \right)}}}} < {E_{{a_{\left( f \right)}}}}$

$w = - P\Delta V = - {10^{ - 5}}\left( {1 \times {{10}^{ - 2}} - 1 \times {{10}^{ - 3}}} \right)$

$ = - 900J$

$\left( i \right)\,\,\,C + {O_2}\rightleftharpoons C{O_2},$

$\,\,\Delta H = - 393.5\,kJmo{l^{ - 1}}$

$\left( {ii} \right)\,\,\,CO + {1 \over 2}{O_2}\rightleftharpoons\,C{O_2},$

$\,\,\,\,\,\,\,\,\Delta H = - 283.0\,kJmo{l^{ - 1}}$

Operating $(i)-(ii),$ we have

$C + {1 \over 2}{O_2} \to CO$

$\,\,\,\,\,\,\,\,\Delta H = - 110.5\,kJmo{l^{ - 1}}$

For a cyclic process the net change in the internal energy is zero because the change in internal energy does not depend on the path.

$C{H_2} = C{H_2}\left( g \right) + {H_2}\left( g \right) \to C{H_3} - C{H_3}$

Enthalpy change $=$ Bond energy of reactants $-$ Bond energy of products.

$\Delta H = 1\left( {C = C} \right) + 4\left( {C - H} \right) + 1\left( {H - H} \right) - $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\left( {C - C} \right) - 6\left( {C - H} \right)$

$ = 1\left( {C = C} \right) + 1\left( {H - H} \right) - 1\left( {C - C} \right) - 2\left( {C - H} \right)$

$ = 615 + 435 - 347 - 2 \times 414$

$ = 1050 - 1175$

$ = - 125\,kJ.$

$\Delta {G^ \circ } = - RT\,1n{K_c}\,\,\,$ or $\,\,\, - \Delta {G^ \circ } = RT\,1n{K_c}$

For spontaneous reaction, $dS > 0$ and $dG$ should be negative i.e. $ < 0.$

Enthalpy change for a reaction does not depend upon the nature of intermediate reaction steps.

TIPS/Formulae :

$\Delta G = \Delta H - T\Delta S$

Since $\Delta G = \Delta H - T\Delta S$ for an endothermic reaction,

$\Delta H = + ve$ and at low temperature $\Delta S = + ve$

Hence $\Delta G = \left( + \right)\Delta H - T\left( + \right)\Delta S$

and if $T\Delta S < \Delta H$ (at low temp)

$\Delta G = + ve$ (non spontaneous)

But at high temperature, reaction becomes

spontaneous i.e. $\Delta G = - ve.$

because at higher temperature $T\Delta S > \Delta H.$

The heat required to raise the temperature of body by $1K$ is called thermal capacity or heat capacity.

According to first law of thermodynamics energy can neither be created nor destroyed although it can be converted from one form to another.

NOTE : Carnot cycle is based upon this principle but during the conversion of heat into work some mechanical energy is always converted to other form of energy hence this data violates $1$ st. law of thermodynamics.

$\Delta H$ negative shows that the reaction is spontaneous. Higher negative value for $Zn$ shows that the reaction is more feasible.