Thermodynamics

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.