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(D) The enthalpy of formation of $PCl_5(s)$ is the enthalpy change for the reaction: $\frac{1}{4} P_{4(s)} + \frac{5}{2} Cl_{2(g)} 	o PCl_{5(s)}$.Ste

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$-454.5 \ kJ$

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(D) The enthalpy of formation of $PCl_5(s)$ is the enthalpy change for the reaction: $\frac{1}{4} P_{4(s)} + \frac{5}{2} Cl_{2(g)} 	o PCl_{5(s)}$.Step $1$: Divide the first equation by $2$:$\frac{1}{4} P_{4(s)} + \frac{3}{2} Cl_{2(g)} 	o PCl_3(\ell) ; \Delta H = \frac{-635}{2} \ kJ = -317.5 \ kJ$Step $2$: Add the second equation:$PCl_3(\ell) + Cl_{2(g)} 	o PCl_{5(s)} ; \Delta H = -137 \ kJ$Step $3$: Add the two resulting equations:$\frac{1}{4} P_{4(s)} + (\frac{3}{2} + 1) Cl_{2(g)} 	o PCl_{5(s)} ; \Delta H = -317.5 + (-137) = -454.5 \ kJ$.

Find the value of enthalpy of formation of $PCl_5(s)$ given the following thermochemical equations:
$1) \frac{1}{2} P_{4(s)} + 3Cl_{2(g)} \to 2PCl_3(\ell) ; \Delta H = -635 \ kJ$
$2) PCl_3(\ell) + Cl_{2(g)} \to PCl_{5(s)} ; \Delta H = -137 \ kJ$

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Find the value of enthalpy of formation of $PCl_5(s)$ given the following thermochemical equations:$1) \frac{1}{2} P_{4(s)} + 3Cl_{2(g)} \to 2PCl_3(\ell) ; \Delta H = -635 \ kJ$$2) PCl_3(\ell) + Cl_{2(g)} \to PCl_{5(s)} ; \Delta H = -137 \ kJ$