Given two processes:frac{1}{2} P_{4(s)} + 3 C | vedclass

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(B) The heat of formation is the enthalpy change when $1 \ mol$ of a substance is formed from its constituent elements in their standard states.The ta

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$-454.5$

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(B) The heat of formation is the enthalpy change when $1 \ mol$ of a substance is formed from its constituent elements in their standard states.The target equation is: $\frac{1}{4} P_{4(s)} + \frac{5}{2} Cl_{2(g)} 	o PCl_{5(s)}$.From the first given equation:$\frac{1}{2} P_{4(s)} + 3 Cl_{2(g)} 	o 2 PCl_{3(l)} \;; \Delta H = -635 \ kJ$Dividing by $2$ gives: $\frac{1}{4} P_{4(s)} + \frac{3}{2} Cl_{2(g)} 	o PCl_{3(l)} \;; \Delta H = -317.5 \ kJ$.From the second given equation:$PCl_{3(l)} + Cl_{2(g)} 	o PCl_{5(s)} \;; \Delta H = -137 \ kJ$.Adding these two equations:$(\frac{1}{4} P_{4(s)} + \frac{3}{2} Cl_{2(g)}) + (PCl_{3(l)} + Cl_{2(g)}) 	o PCl_{3(l)} + PCl_{5(s)}$$\frac{1}{4} P_{4(s)} + \frac{5}{2} Cl_{2(g)} 	o PCl_{5(s)}$$\Delta H_{f} = -317.5 \ kJ + (-137 \ kJ) = -454.5 \ kJ \ mol^{-1}$.

Given two processes:
$\frac{1}{2} P_{4(s)} + 3 Cl_{2(g)} \to 2 PCl_{3(l)} \;; \Delta H = -635 \ kJ$
$PCl_{3(l)} + Cl_{2(g)} \to PCl_{5(s)} \;; \Delta H = -137 \ kJ$
The value of the heat of formation of $PCl_{5(s)}$ is ...... $kJ \ mol^{-1}$.

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Given two processes:$\frac{1}{2} P_{4(s)} + 3 Cl_{2(g)} \to 2 PCl_{3(l)} \;; \Delta H = -635 \ kJ$$PCl_{3(l)} + Cl_{2(g)} \to PCl_{5(s)} \;; \Delta H = -137 \ kJ$The value of the heat of formation of $PCl_{5(s)}$ is ...... $kJ \ mol^{-1}$.