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. `245 kJ mol^(-1)`
B. `290 kJ mol^(-1)`
C. `380 kJ mol^(-1)`
D. `425 kJ mol^(-1)`
A. `245 kJ mol^(-1)`
B. `290 kJ mol^(-1)`
C. `380 kJ mol^(-1)`
D. `425 kJ mol^(-1)`
Correct Answer – D
The enthalpy of formation of `HCl` is the enthalpy change when 1 mol of `HCl`is the enthalpy from its elements in their stantard states:
`(1)/(2) H_(2) (g) + (1)/(2) Cl_(2) (g) rarr HCl (g) , Delta_(g) H^(@) = – 90 kJ mol^(-1)`
According to Eq. , we have
`Delta_(r) H^(@) = sum “Bond enthalpies”_(“reactants”) – sum “Bond enthalpies”_(“products”)`
`((1)/(2) Delta_(H – H) H^(@) + (1)/(2) Delta_(Cl – Cl) H^(@)) – Delta_(H – Cl) H^(@)`
`- 90 = (1)/(2) (430) + (1)/(2) (240) – Delta_(H – Cl) H^(@)`
`:. Delta_(H – Cl) H^(@) = (215) + (120) + (90) = 425 kJ mol^(-1)`