Calculate the enthalpy change for the reaction `H_(2)(g) + Cl_(2)(g) to 2HCl(g)`. The bond enthalpy of `H - H, Cl - Cl and H - Cl` ar 437, 244 and 433 kj `mol^(-1)` respectively.

Explanation

We are given that
Bond enthalpy of H - H 437 kj/mol
Bond enthalpy of Cl - Cl = 244 kj/mol
Bond enthalpy of H - Cl = 433 kj/mol
The given chemical reaction is : `H_(2)(g) +Cl_(2)(g) to 2HCl (g)`
we know, that
`Delta H_(r)` = (Sum of bond enthalpies of reactants) - (Sum of bond enthalpies of products).
We are to find out the enthalpy change `(Delta H)`.
Using the relation
`Delta H_(f) = [Delta H_(H - H) + Delta H_(Cl-Cl)]- 2 [Delta H_(H-Cl)]=[437 + 244]-2 [433] = 681` kj/mol - 866 kj/mol So, enthalpy of reaction is = - 185 kj/mol