Calculate the enthalpy of formation of HCl if bond enthalpies of H - H , Cl - Cl and H - Cl bonds are 434 kJ mol^ -1 , 242 kJ mol^ -1 and 431 kJ mol^ -1 , respectively.

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Quantrex Academy · Accepted Answer

Correct answer: B. -93 kJ mol^ -1

The chemical equation for the formation of

\text{HCl}

is:

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

The enthalpy of formation is given by:

\Delta_f H = \sum \text{Bond enthalpies of reactants} - \sum \text{Bond enthalpies of products}

\Delta_f H = \left[ \dfrac{1}{2} \Delta H_{\text{H-H}} + \dfrac{1}{2} \Delta H_{\text{Cl-Cl}} \right] - \Delta H_{\text{H-Cl}}

Substituting the given values:

\Delta_f H = \left[ \dfrac{1}{2}(434) + \dfrac{1}{2}(242) \right] - 431

\Delta_f H = (217 + 121) - 431

\Delta_f H = 338 - 431 = -93 \text{ kJ mol}^{-1}

Answer:

-93

kJ mol^{-1}

Calculate the enthalpy of formation of HCl if bond enthalpies of H...

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