H_2 + frac{1}{2} O_2 to H_2O; Delta H = -68.3 | vedclass

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(B) The target reaction for the heat of formation of $KOH_{(s)}$ is:$K_{(s)} + \frac{1}{2} O_{2(g)} + \frac{1}{2} H_{2(g)} 	o KOH_{(s)}$Given equatio

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$-68.39 - 48 + 14$

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(B) The target reaction for the heat of formation of $KOH_{(s)}$ is:$K_{(s)} + \frac{1}{2} O_{2(g)} + \frac{1}{2} H_{2(g)} 	o KOH_{(s)}$Given equations:$(I)$ $H_2 + \frac{1}{2} O_2 	o H_2O; \Delta H_1 = -68.39 \ kcal$$(II)$ $K + H_2O + 	ext{water} 	o KOH_{(aq)} + \frac{1}{2} H_2; \Delta H_2 = -48 \ kcal$$(III)$ $KOH + 	ext{water} 	o KOH_{(aq)}; \Delta H_3 = -14 \ kcal$To obtain the target reaction,we perform: $(II)$ + $(I)$ - $(III)$:$K + H_2O + \frac{1}{2} O_2 + H_2 - KOH 	o KOH_{(aq)} + \frac{1}{2} H_2 + H_2O - KOH_{(aq)}$$K + \frac{1}{2} O_2 + \frac{1}{2} H_2 	o KOH$Therefore,$\Delta H_f = \Delta H_2 + \Delta H_1 - \Delta H_3$$\Delta H_f = -48 + (-68.39) - (-14) = -68.39 - 48 + 14 \ kcal$.

$H_2 + \frac{1}{2} O_2 \to H_2O; \Delta H = -68.39 \ kcal$
$K + H_2O + \text{water} \to KOH_{(aq)} + \frac{1}{2} H_2; \Delta H = -48 \ kcal$
$KOH + \text{water} \to KOH_{(aq)}; \Delta H = -14 \ kcal$
The heat of formation of $KOH$ is (in $kcal$):

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$H_2 + \frac{1}{2} O_2 \to H_2O; \Delta H = -68.39 \ kcal$$K + H_2O + \text{water} \to KOH_{(aq)} + \frac{1}{2} H_2; \Delta H = -48 \ kcal$$KOH + \text{water} \to KOH_{(aq)}; \Delta H = -14 \ kcal$The heat of formation of $KOH$ is (in $kcal$):