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(A) For the equilibrium $N \rightleftharpoons D$, the equilibrium constant $K_{eq}$ is given by $[D]/[N]$.Since the concentrations of $N$ and $D$ are

Vedclass · Accepted Answer

$2.0$

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(A) For the equilibrium $N ightleftharpoons D$, the equilibrium constant $K_{eq}$ is given by $[D]/[N]$.Since the concentrations of $N$ and $D$ are equal at equilibrium, $K_{eq} = 1$.The relationship between the standard Gibbs free energy change and the equilibrium constant is $\Delta G^\circ = -RT \ln K_{eq}$.Since $K_{eq} = 1$, $\ln(1) = 0$, therefore $\Delta G^\circ = 0$.We also know that $\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$.Given $\Delta G^\circ = 0$, we have $\Delta H^\circ = T \Delta S^\circ$, which implies $\Delta S^\circ = \Delta H^\circ / T$.The temperature $T = 60 ^\circ C = 60 + 273.15 = 333.15 	ext{ K}$.Given $\Delta H^\circ = 666 	ext{ kJ mol}^{-1}$.Calculating $\Delta S^\circ = 666 / 333.15 \approx 1.999 	ext{ kJ K}^{-1} 	ext{mol}^{-1}$.Rounding to the nearest value, we get $2.0 	ext{ kJ K}^{-1} 	ext{mol}^{-1}$.

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