The bond energies of H-H and Cl-Cl are 430 , | vedclass

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(D) The chemical equation for the formation of $HCl$ is: $\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \longrightarrow HCl(g)$ The enthalpy of formation i

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$425$

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(D) The chemical equation for the formation of $HCl$ is: $\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \longrightarrow HCl(g)$ The enthalpy of formation is given by: ${\Delta _f}H = \sum 	ext{Bond Energy of Reactants} - \sum 	ext{Bond Energy of Products}$ $-90 = [\frac{1}{2} 	imes BE(H-H) + \frac{1}{2} 	imes BE(Cl-Cl)] - BE(H-Cl)$ $-90 = [\frac{1}{2}(430) + \frac{1}{2}(240)] - BE(H-Cl)$ $-90 = [215 + 120] - BE(H-Cl)$ $-90 = 335 - BE(H-Cl)$ $BE(H-Cl) = 335 + 90 = 425 \, kJ \, mol^{-1}$

The bond energies of $H-H$ and $Cl-Cl$ are $430 \, kJ \, mol^{-1}$ and $240 \, kJ \, mol^{-1}$ respectively. If the value of ${\Delta _f}H$ for $HCl$ is $-90 \, kJ \, mol^{-1}$,then the $H-Cl$ bond energy is ..... $kJ \, mol^{-1}$.

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