Bond dissociation enthalpy of `H_(2)` , `Cl_(2)` and `HCl` are `434, 242` and `431KJmol^(-1)` respectively. Enthalpy of formation of `HCl` is

Options

A. `93 kJ mol^(-1)`
B. `- 245 kJ mol^(-1)`
C. `- 93 kJ mol^(-1)`
D. `245 kJ mol^(-1)`

Answer

C

Explanation

The enthalpy of formation of `HCl` is the enthalpy change for the following reaction:
`(1)/(2) H_(2) (g) + (1)/(2) Cl_(2) (g) rarr HCl (g),` `Delta_(r) H^(@) = ?`
That is, one mole of `HCl` is synthesized in its standard state from tis elements in their standard states.
According to Eq., we have
`Delta_(r) H^(@) = sum "Bond enthalpies"_("reactants") - sum "Bond enthalies"_("products")`
`((1)/(2) Delta_(H - H) H^(@) + (1)/(2) Delta_(Cl - Cl) H^(@)) - Delta_(H - Cl) H^(@)`
`= (1)/(2) (434) + (1)/(2) (242) - (431)`
`= (271) + (121) - (431) = - 93 kJ mol^(-1)`