ThermochemistryHard

Question

Calculate the enthalpy of formation (in kcal/mol) of HI(g) from the following data.

H2(g) + Cl2(g) → 2HCl(g): ΔH = −44.20 kcal

HCl(g) + aq → HCl(aq): ΔH = −17.31 kcal

HI(g) + aq → HI(aq): ΔH = −19.21 kcal

KOH(aq) + HCl(aq) → KCl(aq): ΔH = −13.74 kcal

KOH(aq) + HI(aq) → KI(aq): ΔH = −13.67 kcal

Cl2(g) + 2KI(aq) → 2KCl(aq) + I2(s): ΔH = −52.42 kcal

Options

A.+19.21
B.−32.87
C.+5.94
D.+7.82

Solution

For the reaction, $\frac{1}{2}H_{2}(g) + \frac{1}{2}I_{2}(s) \rightarrow HI(g);$

$\Delta H_{\text{Required}} = \frac{1}{2}( - 44.20) - \frac{1}{2}( - 52.42) + ( - 17.31) - ( - 19.21) + ( - 13.74) - ( - 13.67) = 5.94\text{ kcal}$

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