Given that 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 :

Options

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

Answer

D

Explanation

`(1)/(2)H_(2(g))+(1)/(2)Cl_(2(g))rarrHCl_((g)) , Delta H_(f) =-90 kJ mol^(-1)`
`Delta H = Sigma (B.E.)_("Reactants")- Sigma (B.E.)_("Products")`
`-90=((1)/(2)xx430+(1)/(2)xx240)-(B.E.)_(HCl)`
`thereforer (B.E.)_(HCl)=425 kJ mol^(-1)`