Standard free energies of formation (in kJ/mo | vedclass

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(A) The standard Gibbs free energy of reaction is calculated as: $\Delta G^{\circ}_{rxn} = \Sigma \Delta G^{\circ}_f (	ext{products}) - \Sigma \Delta

Vedclass · Accepted Answer

$1.0968$

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(A) The standard Gibbs free energy of reaction is calculated as: $\Delta G^{\circ}_{rxn} = \Sigma \Delta G^{\circ}_f (	ext{products}) - \Sigma \Delta G^{\circ}_f (	ext{reactants})$.For the combustion of pentane: $C_5H_{12(g)} + 8O_{2(g)} ightarrow 5CO_{2(g)} + 6H_2O_{(l)}$.$\Delta G^{\circ}_{rxn} = [5 	imes \Delta G^{\circ}_f(CO_2) + 6 	imes \Delta G^{\circ}_f(H_2O)] - [\Delta G^{\circ}_f(C_5H_{12}) + 8 	imes \Delta G^{\circ}_f(O_2)]$.Given $\Delta G^{\circ}_f(O_2) = 0 \ kJ/mol$,$\Delta G^{\circ}_f(CO_2) = -394.4 \ kJ/mol$,$\Delta G^{\circ}_f(H_2O) = -237.2 \ kJ/mol$,and $\Delta G^{\circ}_f(C_5H_{12}) = -8.2 \ kJ/mol$.$\Delta G^{\circ}_{rxn} = [5(-394.4) + 6(-237.2)] - [-8.2 + 0] = -1972 - 1423.2 + 8.2 = -3387 \ kJ/mol = -3387000 \ J/mol$.In the fuel cell reaction,the number of electrons transferred $(n)$ is $32$.Using $\Delta G^{\circ} = -nFE^{\circ}_{cell}$,where $F = 96500 \ C/mol$:$E^{\circ}_{cell} = \frac{-\Delta G^{\circ}}{nF} = \frac{3387000}{32 	imes 96500} = 1.0968 \ V$.

Standard free energies of formation (in $kJ/mol$) at $298 \ K$ are $-237.2$,$-394.4$,and $-8.2$ for $H_2O_{(l)}$,$CO_{2_{(g)}}$,and pentane$_{(g)}$ respectively. The value of $E^o_{cell}$ for the pentane-oxygen fuel cell is .......... $V$.

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