Question

Calculate ΔH⁰298 (in kJ) for the process P(s) + 5/2 Cl2(g) → PCl5(g) from the following information.

Answer 1

Answer:

$\Delta H_{298}^{0}$ for the process is -375 kJ

Explanation:

• Given reaction is a combination of the two given elementary steps.

• Summation of change in standard enthalpy ($\Delta H_{298}^{0}$)of the two elementary reactions give $\Delta H_{298}^{0}$ of the reaction $P(s)+\frac{5}{2}Cl_{2}(g)\rightarrow PCl_{5}(g)$.

$P(s)+\frac{3}{2}Cl_{2}(g)\rightarrow PCl_{3}(g)$......$\Delta H_{1}=-287kJ$

$PCl_{3}(g)+Cl_{2}(g)\rightarrow PCl_{5}(g)$......$\Delta H_{2}=-88kJ$

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$P(s)+\frac{5}{2}Cl_{2}(g)\rightarrow PCl_{5}(g)$

$\Delta H_{298}^{0}$ = $\Delta H_{1}+\Delta H_{2}=-287kJ-88kJ=-375kJ$