The ΔHf of AgCl is -120 kJ mol-1.

If Ag^+_{(aq)}+Cl^-_{(aq)}\rightarrow AgCl_{(s)} \: \: \: \Delta H:-127kJ \: mol^-^1, find the enthalpy change for the below equation:

Ag_{(s)}+\frac{1}{2}Cl_{2(aq)}\rightarrow Ag^+_{(aq)}+Cl^-_{(aq)} \: \: \: \Delta H:\: ?kJ \: mol^-^1.

clockwise = anticlockwise

{\color{Red} x}{\color{Blue} -127} = {\color{Magenta} -120}

x=7kJ