The bond enthalpies of H-H, Cl-Cl, and H-Cl bonds are 437 kJ/mol, 239 kJ/mol, and 433 kJ/mol respectively. Calculate the enthalpy of formation of HCl.

Question

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Correct answer: C. -95 kJ/mol

The reaction for the formation of HCl is:

\dfrac{1}{2} \text{H}_2(g) + \dfrac{1}{2} \text{Cl}_2(g) \rightarrow \text{HCl}(g)

The enthalpy of formation

\Delta H_f

is given by the difference between the bond enthalpies of reactants and products:

\Delta H_f = \sum \text{BE}(\text{reactants}) - \sum \text{BE}(\text{products})

\Delta H_f = \left( \dfrac{1}{2} \text{BE}(\text{H-H}) + \dfrac{1}{2} \text{BE}(\text{Cl-Cl}) \right) - \text{BE}(\text{H-Cl})

Substituting the given values:

\Delta H_f = \left( \dfrac{1}{2} \times 437 + \dfrac{1}{2} \times 239 \right) - 433

\Delta H_f = (218.5 + 119.5) - 433

\Delta H_f = 338 - 433 = -95 \text{ kJ/mol}

Answer:

-95

kJ/mol

The bond enthalpies of H-H, Cl-Cl, and H-Cl bonds are 437 kJ/mol,...

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