Calculate the bond enthalpy of Cl-Cl bond from the following data
CH4+Cl2 --------> Ch3Cl+HCl
ΔH= -100.3kJ. Also the bond enthalpies of C-H, C-Cl, H-Cl bonds are 413, 326 & 431 kJ mol -1 respectively.
ΔH = (Total enthalpy of product) - (Total enthalpy of reactant)
so..
-100.3= 3(413)+326+431-4(413)-H(Cl-Cl)
on solving...
H(Cl-Cl)= 444.3 KJ/mol