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Given the bond energies of H - H and Cl ...

Given the bond energies of `H - H` and `Cl - Cl` are `430 kJ mol^(-1)` and `240 kJ mol^(-1)`, respectively, and `Delta_(f) H^(@)` for `HCl` is `- 90 kJ mol^(-1)`. Bond enthalpy of `HCl` is

A

290 kJ `mol^(-1)`

B

380 kJ `mol^(-1)`

C

425 kJ `mol^(-1)`

D

245 kJ `mol^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
`(1)/(2) H_(2) + (1)/(2) Cl_(2) rarr HCl`
`Delta H_(f)= [(1)/(2) Delta H_(H-H) + (1)/(2) Delta H_(Cl- Cl)]- [Delta H_(H-Cl)]`
`-90 = [(1)/(2) xx 430 + (1)/(2) xx 240]- Delta H_(H-Cl)`
`-90 = 215 + 120- Delta H_(H-Cl)`
`Delta H_(H-Cl) = 335 + 90= 425`kJ/mol
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