Which of the following conditions will oppose | vedclass

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(D) The spontaneity of a reaction is determined by the Gibbs free energy equation: $\Delta G = \Delta H - T \Delta S$. For a reaction to be non-sponta

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$\Delta H = +ve, \Delta S = -ve$

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(D) The spontaneity of a reaction is determined by the Gibbs free energy equation: $\Delta G = \Delta H - T \Delta S$. For a reaction to be non-spontaneous (opposed),$\Delta G$ must be positive $(> 0)$. When $\Delta H = +ve$ (endothermic) and $\Delta S = -ve$ (decrease in entropy),the term $\Delta G = (+ve) - T(-ve) = (+ve) + T(+ve)$ will always be positive regardless of the temperature $T$. Therefore,the condition $\Delta H = +ve$ and $\Delta S = -ve$ will always oppose the reaction.

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

Which of the following conditions will oppose or prevent a reaction from occurring?

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

$A$ $-25$ $-80$

$B$ $-22$ $40$

$C$ $25$ $-50$

$D$ $22$ $20$

Which of the following conditions will always lead to a non-spontaneous change?

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