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(D) The standard free energy change is calculated using the Gibbs-Helmholtz equation: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. Given: $\De

Vedclass · Accepted Answer

$5.39$

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(D) The standard free energy change is calculated using the Gibbs-Helmholtz equation: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. Given: $\Delta H^\circ = 58.04 \, kJ = 58040 \, J$,$\Delta S^\circ = 176.7 \, J/K$,and $T = 25 + 273 = 298 \, K$. Substituting the values: $\Delta G^\circ = 58040 \, J - (298 \, K 	imes 176.7 \, J/K)$. $\Delta G^\circ = 58040 \, J - 52656.6 \, J = 5383.4 \, J$. Converting to $kJ$: $\Delta G^\circ = 5.3834 \, kJ \approx 5.39 \, kJ$.

The standard enthalpy of the decomposition of $N_2O_4$ to $NO_2$ is $58.04 \, kJ$ and standard entropy of this reaction is $176.7 \, J/K$. The standard free energy change for this reaction at $25 \, ^oC$ is $..... \, kJ$

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