Delta H and Delta S for a reaction are +30.0 | vedclass

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(B) Given: $\Delta H = +30.0 \ kJ \ mol^{-1}$, $\Delta S = 0.06 \ kJ \ K^{-1} \ mol^{-1}$.For the reaction to be at equilibrium, $\Delta G = 0$.Using

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$227^{\circ} C$ and non-spontaneous

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(B) Given: $\Delta H = +30.0 \ kJ \ mol^{-1}$, $\Delta S = 0.06 \ kJ \ K^{-1} \ mol^{-1}$.For the reaction to be at equilibrium, $\Delta G = 0$.Using the relation $\Delta G = \Delta H - T \Delta S$, we set $\Delta G = 0$:$0 = 30.0 - T 	imes 0.06$$T = \frac{30.0}{0.06} = 500 \ K$.Converting to Celsius: $T(^{\circ} C) = 500 - 273 = 227^{\circ} C$.Since $\Delta H$ is positive (endothermic) and $\Delta S$ is positive, the reaction is spontaneous only at temperatures above $500 \ K$ (or $227^{\circ} C$).Therefore, below $227^{\circ} C$, the reaction is non-spontaneous.

	$\Delta_r H^{\circ}$	$\Delta_r S^{\circ}$	$\Delta_r G^{\circ}$	Spontaneity of the reaction
$A$	$+$	$+$	$+$	Spontaneous at low $T$
$B$	$+$	$+$	$-$	Spontaneous at high $T$
$C$	$-$	$-$	$-$	Spontaneous at low $T$
$D$	$+$	$-$	$+$	Spontaneous at high $T$

$\Delta H$ and $\Delta S$ for a reaction are $+30.0 \ kJ \ mol^{-1}$ and $0.06 \ kJ \ K^{-1} \ mol^{-1}$ at $1 \ atm$ pressure. The temperature at which free energy change is equal to zero and the nature of the reaction below this temperature are:

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