For a reaction,Delta H = +3 , kJ and Delta S | vedclass

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Question

(A) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative $(\Delta G < 0)$.Given the equation $\Delta G = \Delta H

Vedclass · Accepted Answer

$300$

Vedclass · Answer

(A) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative $(\Delta G Given the equation $\Delta G = \Delta H - T\Delta S$,for $\Delta G This implies $\Delta H  \frac{\Delta H}{\Delta S}$.Given $\Delta H = 3 \, kJ = 3000 \, J$ and $\Delta S = 10 \, J/K$.Substituting these values: $T > \frac{3000 \, J}{10 \, J/K} = 300 \, K$.Thus,the reaction becomes spontaneous at temperatures above $300 \, K$.

Process	$\Delta H \ (kJ \ mol^{-1})$	$\Delta S \ (J \ K^{-1} \ mol^{-1})$
$A$	$-25$	$-80$
$B$	$-22$	$40$
$C$	$25$	$-50$
$D$	$22$	$20$

For a reaction,$\Delta H = +3 \, kJ$ and $\Delta S = +10 \, J/K$. At what minimum temperature (in $K$) will the reaction become spontaneous?

Similar Questions

For a reaction,$\Delta H = -40 \, kJ$ and $\Delta S = -50 \, J/K$. At what temperature range will it change from spontaneous to non-spontaneous?

For independent processes at $300 \ K$,determine the number of non-spontaneous processes from the following table:

Process $\Delta H \ (kJ \ mol^{-1})$ $\Delta S \ (J \ K^{-1} \ mol^{-1})$

$A$ $-25$ $-80$

$B$ $-22$ $40$

$C$ $25$ $-50$

$D$ $22$ $20$

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