Given that bond energies of H - H and Cl - Cl are 430 KJ mol- and 240 KJ mol- respectively and ΔfH for HCI is −90kJmol-1. Bond enthalpy of HCI is

A
290 kJ $m o l^{- 1}$
B
380 kJ $m o l^{- 1}$
C
425 kJ $m o l^{- 1}$
D
245 kJ $m o l^{- 1}$

Solution:

$\begin{array}{l} \frac{1}{2} H_{2} + \frac{1}{2} {Cl}_{2} \to HCl \\ Δ H_{f} = [\frac{1}{2} Δ H_{H_{2}} + \frac{1}{2} Δ H_{{Cl}_{2}}] - [{ΔH}_{HCl}] \\ - 90 = [\frac{1}{2} \times 430 + \frac{1}{2} \times 240] - Δ H_{HCl} \\ - 90 = 215 + 120 - Δ H_{HCl} \\ {ΔH}_{HCl} = 335 + 90 = 425 kJ / mol \end{array}$

Given that bond energies of H - H and Cl - Cl are 430 ${KJ mol}^{-}$ and $240 KJ {mol}^{-}$ respectively and $Δ_{f} H$ for HCI is $- 90 kJ {mol}^{- 1}$ . Bond enthalpy of HCI is

Solution:

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