For a reaction at 25 , ^oC,the enthalpy chang | vedclass

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(B) The Gibbs free energy change is given by the formula: $\Delta G = \Delta H - T\Delta S$. Given: $\Delta H = -11.7 	imes 10^3 \, J \, mol^{-1}$,$\

Vedclass · Accepted Answer

$19.59$

Vedclass · Answer

(B) The Gibbs free energy change is given by the formula: $\Delta G = \Delta H - T\Delta S$. Given: $\Delta H = -11.7 	imes 10^3 \, J \, mol^{-1}$,$\Delta S = -105 \, J \, mol^{-1} \, K^{-1}$,and $T = 25 + 273 = 298 \, K$. Substituting the values: $\Delta G = -11.7 	imes 10^3 - (298 	imes -105)$. $\Delta G = -11700 + 31290 = 19590 \, J \, mol^{-1}$. Converting to $kJ$: $\Delta G = 19590 / 1000 = 19.59 \, kJ \, mol^{-1}$.

For a reaction at $25 \, ^oC$,the enthalpy change and entropy change are $-11.7 \times 10^3 \, J \, mol^{-1}$ and $-105 \, J \, mol^{-1} \, K^{-1}$ respectively. What is the Gibbs free energy in $kJ$?

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