Question
Determine whether the following reaction is spontaneous under standard state conditions.
$2 H_2O_{(l)} + O_{2(g)} \rightarrow 2H_2O_{2(l)}$
if $\triangle H^0 = 196 kJ, \triangle S^0 = -126 J/K$
Does it have a cross-over temperature?

Answer

Given : $2H_2O_{(l)} + O_{2(g)} \rightarrow 2H_2O_{2(l)}$
$\triangle H^0 = +196 kJ$
$\triangle S^0 = -126 JK-1 =0.126 kj K-1$
$T= 298 K$
$\triangle G^0 = ?$
Cross over temperature $= T = ?$
$\triangle G^0 = \triangle H^0 – T\triangle S^0$
$= 196 – 298 (-0.126)$
$= 196 + 37.55$
$= + 233.55 kJ$
$\because \triangle G^0 > 0$, the reaction is non-spontaneous.
$\triangle H^0 > 0, \triangle S^0 < 0,$
Since at all temperatures,$ \triangle G^0 > 0$, there is no cross over temperature.
Ans. The reaction is non-spontaneous.
There is no cross-over temperature for the reaction.

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