If for a certain reaction Delta_{r} H is 30 ; | vedclass

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(B) For a reaction to be spontaneous,the change in Gibbs free energy,$\Delta_{r} G$,must be negative $(\Delta_{r} G < 0)$.The relationship is given by

Vedclass · Accepted Answer

$> 66.67$

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(B) For a reaction to be spontaneous,the change in Gibbs free energy,$\Delta_{r} G$,must be negative $(\Delta_{r} G The relationship is given by the equation: $\Delta_{r} G = \Delta_{r} H - T \Delta_{r} S$.Substituting the condition for spontaneity: $\Delta_{r} H - T \Delta_{r} S Rearranging the inequality: $\Delta_{r} H  \frac{\Delta_{r} H}{T}$.Given $\Delta_{r} H = 30 \; kJ \; mol^{-1} = 30,000 \; J \; mol^{-1}$ and $T = 450 \; K$.$\Delta_{r} S > \frac{30,000 \; J \; mol^{-1}}{450 \; K} = 66.67 \; J \; K^{-1} \; mol^{-1}$.Therefore,the reaction will be spontaneous if $\Delta_{r} S > 66.67 \; J \; K^{-1} \; mol^{-1}$.

If for a certain reaction $\Delta_{r} H$ is $30 \; kJ \; mol^{-1}$ at $450 \; K$,the value of $\Delta_{r} S$ (in $J \; K^{-1} \; mol^{-1}$) for which the same reaction will be spontaneous at the same temperature is

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