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(B) The relationship between Gibbs free energy change, enthalpy change, and entropy change is given by $\Delta G^\circ = \Delta H^\circ - T\Delta S^\c

Vedclass · Accepted Answer

-$50$

Vedclass · Answer

(B) The relationship between Gibbs free energy change, enthalpy change, and entropy change is given by $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$.Also, $\Delta G^\circ = -RT \ln K$.Given $T = 300 	ext{ K}$, $\Delta H^\circ = 28.40 	ext{ kJ mol}^{-1} = 28400 	ext{ J mol}^{-1}$, $K = 1.8 	imes 10^{-7}$, and $R = 8.3 	ext{ J K}^{-1} 	ext{ mol}^{-1}$.First, calculate $\ln K$: $\ln K = \ln(1.8 	imes 10^{-7}) = \ln(18 	imes 10^{-8}) = \ln(2 	imes 3^2) - 8 \ln 10 = \ln 2 + 2 \ln 3 - 8 \ln 10$.Using $\ln x = 2.3 \log x$, we have $\ln 2 = 2.3 	imes 0.30 = 0.69$, $\ln 3 = 2.3 	imes 0.48 = 1.104$, and $\ln 10 = 2.3$.So, $\ln K = 0.69 + 2(1.104) - 8(2.3) = 0.69 + 2.208 - 18.4 = -15.502$.Now, $\Delta G^\circ = -RT \ln K = -(8.3)(300)(-15.502) = 38599.98 	ext{ J mol}^{-1} \approx 38.60 	ext{ kJ mol}^{-1}$.Finally, $\Delta S^\circ = \frac{\Delta H^\circ - \Delta G^\circ}{T} = \frac{28400 - 38600}{300} = \frac{-10200}{300} = -34 	ext{ J K}^{-1} 	ext{ mol}^{-1}$.The nearest integer value is $-34$.

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