Determine enthalpy change for the reaction,
`CH_(4(g))+Cl_(2(g))rarr CH_(3)Cl_((g))+HCl_((g))`
Bond energies for `C-H,C-Cl,Cl-Cl,H-Cl` are `412,338,242,431k J mol^(-1)` respectively.

Explanation

`Delta_(r)H^(Theta)` = [Total energy used in dissociation of bonds]-[Total energy used in formation of bonds]
`=[4 DeltaH_(C-H)+Delta H_(Cl-Cl)]-[3 Delta H_(C-H)+Delta H_(C-Cl)+Delta H_(H-Cl)]`
= [4(412) + 1(242)] - [ 3(412) + 1(338) + 1(431)]
= [1648 + 242] - [1236 + 338 + 431]
= [1890] - [2005]
= -115 kJ `mol^(-1)`