Calculate the heat of formation of Ca(OH)_{2( | vedclass

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(B) The heat of formation of $Ca(OH)_{2(s)}$ is the enthalpy change for the reaction: $Ca_{(s)} + O_{2(g)} + H_{2(g)} 	o Ca(OH)_{2(s)}$.We are given:

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$-235.43$

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(B) The heat of formation of $Ca(OH)_{2(s)}$ is the enthalpy change for the reaction: $Ca_{(s)} + O_{2(g)} + H_{2(g)} 	o Ca(OH)_{2(s)}$.We are given:$(i) CaO_{(s)} + H_2O_{(l)} 	o Ca(OH)_{2(s)}$; $\Delta H_1 = -15.26\,K\,cal$$(ii) H_{2(g)} + \frac{1}{2}O_{2(g)} 	o H_2O_{(l)}$; $\Delta H_2 = -68.37\,K\,cal$$(iii) Ca_{(s)} + \frac{1}{2}O_{2(g)} 	o CaO_{(s)}$; $\Delta H_3 = -151.80\,K\,cal$Adding equations $(i)$,$(ii)$,and $(iii)$:$CaO_{(s)} + H_2O_{(l)} + H_{2(g)} + \frac{1}{2}O_{2(g)} + Ca_{(s)} + \frac{1}{2}O_{2(g)} 	o Ca(OH)_{2(s)} + H_2O_{(l)} + CaO_{(s)}$Canceling common terms on both sides:$Ca_{(s)} + O_{2(g)} + H_{2(g)} 	o Ca(OH)_{2(s)}$The total enthalpy change is $\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = (-15.26) + (-68.37) + (-151.80) = -235.43\,K\,cal$.

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