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NEET CHEMISTRYMedium

The bond energy of H—H and Cl—Cl is 430 kJ mol−1430 \text{ kJ mol}^{-1} and 240 kJ mol−1240 \text{ kJ mol}^{-1} respectively and ΔfH\Delta_f H for HCl is −90 kJ mol−1-90 \text{ kJ mol}^{-1}. The bond enthalpy of HCl is:

A

290 kJ mol−1290 \text{ kJ mol}^{-1}

B

380 kJ mol−1380 \text{ kJ mol}^{-1}

C

425 kJ mol−1425 \text{ kJ mol}^{-1}

D

245 kJ mol−1245 \text{ kJ mol}^{-1}

Step-by-Step Solution

The chemical equation for the standard enthalpy of formation of HCl is: 12H2(g)+12Cl2(g)→HCl(g)\frac{1}{2}\text{H}_2\text{(g)} + \frac{1}{2}\text{Cl}_2\text{(g)} \rightarrow \text{HCl(g)}

The enthalpy of formation (ΔfH\Delta_f H) is related to bond enthalpies by the equation: ΔfH=∑B.E.(reactants)−∑B.E.(products)\Delta_f H = \sum \text{B.E.(reactants)} - \sum \text{B.E.(products)} ΔfH=[12B.E.(H−H)+12B.E.(Cl−Cl)]−B.E.(H−Cl)\Delta_f H = \left[ \frac{1}{2}\text{B.E.}(\text{H}-\text{H}) + \frac{1}{2}\text{B.E.}(\text{Cl}-\text{Cl}) \right] - \text{B.E.}(\text{H}-\text{Cl})

Given: ΔfH=−90 kJ mol−1\Delta_f H = -90 \text{ kJ mol}^{-1} B.E.(H−H)=430 kJ mol−1\text{B.E.}(\text{H}-\text{H}) = 430 \text{ kJ mol}^{-1} B.E.(Cl−Cl)=240 kJ mol−1\text{B.E.}(\text{Cl}-\text{Cl}) = 240 \text{ kJ mol}^{-1}

Substituting the values: −90=[12(430)+12(240)]−B.E.(H−Cl)-90 = \left[ \frac{1}{2}(430) + \frac{1}{2}(240) \right] - \text{B.E.}(\text{H}-\text{Cl}) −90=(215+120)−B.E.(H−Cl)-90 = (215 + 120) - \text{B.E.}(\text{H}-\text{Cl}) −90=335−B.E.(H−Cl)-90 = 335 - \text{B.E.}(\text{H}-\text{Cl}) B.E.(H−Cl)=335+90=425 kJ mol−1\text{B.E.}(\text{H}-\text{Cl}) = 335 + 90 = 425 \text{ kJ mol}^{-1}

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