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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