1.

Given that bond energies H-H and Cl-Cl are 430 kJ mol^(-1) and 240 kJ mol^(-1) respecively and Delta_(f) H for HCl is - 90 kJ mol^(-1). Bond enthalpy of HCl is :

Answer»

`380 KJ mol^(-1)`
`425 kJ mol^(-1)`
`245 kJ mol^(-1)`
`290 kJ mol^(-1)`

Solution :`(1)/(2) H_(2) + (1)/(2) Cl_(2) rarr 2HCl`
`DELTA H = (1)/(2) B. E (H-H) + (1)/(2) B. E (Cl - Cl) - B. E (H - Cl)`
`- 90 = (1)/(2) (430) + (1)/(2) (240) - B. E. (H - Cl)`
`B. E. (H - Cl) = 215 + 120 + 90 = 425 kJ mol^(-1)`


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