Given the following thermochemical equations: | vedclass

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(B) The heat of formation of $KOH_{(s)}$ corresponds to the reaction: $K_{(s)} + \frac{1}{2}H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow KOH_{(s)}$.To o

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$-68.39 - 48.0 + 14.0$

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(B) The heat of formation of $KOH_{(s)}$ corresponds to the reaction: $K_{(s)} + \frac{1}{2}H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow KOH_{(s)}$.To obtain this,we manipulate the given equations:$(i) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)} ; \Delta H_1 = -68.39 \, kcal$$(ii) \ K_{(s)} + H_2O_{(l)} \rightarrow KOH_{(aq)} + \frac{1}{2}H_{2(g)} ; \Delta H_2 = -48.0 \, kcal$$(iii) \ KOH_{(s)} \rightarrow KOH_{(aq)} ; \Delta H_3 = -14.0 \, kcal$Adding $(i)$ and $(ii)$ gives: $K_{(s)} + \frac{1}{2}H_{2(g)} + \frac{1}{2}O_{2(g)}$ $\rightarrow KOH_{(aq)} ; \Delta H = \Delta H_1 + \Delta H_2 = -68.39 - 48.0 = -116.39 \, kcal$.Subtracting $(iii)$ from this result: $K_{(s)} + \frac{1}{2}H_{2(g)} + \frac{1}{2}O_{2(g)}$ $\rightarrow KOH_{(s)} ; \Delta H = (-116.39) - (-14.0) = -68.39 - 48.0 + 14.0 \, kcal$.

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