For a certain reaction,Delta H = -225 kJ mo | vedclass

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(A) The relationship between Gibbs free energy change $(\Delta G)$,enthalpy change $(\Delta H)$,and entropy change $(\Delta S)$ is given by the equati

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$1500$

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(A) The relationship between Gibbs free energy change $(\Delta G)$,enthalpy change $(\Delta H)$,and entropy change $(\Delta S)$ is given by the equation: $\Delta G = \Delta H - T \Delta S$. For $\Delta G = 0$,the equation becomes: $0 = \Delta H - T \Delta S$,which implies $T = \frac{\Delta H}{\Delta S}$. Given: $\Delta H = -225 \ kJ \ mol^{-1} = -225000 \ J \ mol^{-1}$ and $\Delta S = -150 \ J \ K^{-1} \ mol^{-1}$. Substituting the values: $T = \frac{-225000 \ J \ mol^{-1}}{-150 \ J \ K^{-1} \ mol^{-1}} = 1500 \ K$. Therefore,the temperature at which $\Delta G$ is zero is $1500 \ K$.

For a certain reaction,$\Delta H = -225 \ kJ \ mol^{-1}$ and $\Delta S = -150 \ J \ K^{-1} \ mol^{-1}$. Find the temperature at which $\Delta G$ is zero. (in $K$)

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