ThermochemistryHard

Question

The enthalpy of formation of HCl(g) from the following reaction

H2(g) + Cl2(g) → 2HCl(g) + 44 kcal is

Options

A.−44 kcal mol–1
B.−22 kcal mol–1
C.22 kcal mol–1
D.−88 kcal mol–1

Solution

$\Delta_{f}H_{HCl(g)} = - \frac{44}{2} = - 22\text{ kcal/mol}$

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