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 :

Options

A. `380 kJ mol^(-1)`
B. `425 kJ mol^(-1)`
C. `245 kJ mol^(-1)`
D. `290 kJ mol^(-1)`

Answer

B

Explanation

`(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)`