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Q.

${PCl}_{5} (s)$ reacts with $H_{2} O (l)$ according to the equation:
${PCl}_{5} (s) + 4 H_{2} O (l) ⟶ H_{3} {PO}_{4} (a q) + 5 HCl (a q)$ What is $Δ H^{\circ}$ for this reaction in $kJ / {mol}^{- 1} ?$
Substance $Δ H_{f}^{\circ}, kJ / mol$
${PCl}_{5} (s)$ -443.5
$H_{2} O (l)$ -285.8
$H_{3} {PO}_{4} (a q)$ -1284.4
$HCM (a q)$ -167.1

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a

-722.2

b

533.2

c

-533.2

d

722.2

answer is B.

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Detailed Solution

For the given values, change in enthalpy,

$\begin{array}{l} Δ H^{\circ} = - 167.1 \times 5 - 1284.4 + 285.8 \times 4 + 443.5 \\ = - 533.2 \end{array}$

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PCl5(s) reacts with H2O(l) according to the equation:PCl5(s)+4H2O(l)⟶H3PO4(aq)+5HCl(aq) What is ΔH∘ for this reaction in kJ/mol−1?SubstanceΔHf∘,kJ/molPCl5(s)-443.5H2O(l)-285.8H3PO4(aq)-1284.4HCM(aq)-167.1