Calculate the enthalpy change for the following reaction.
4 CHCl3 (l) + 5 O2 (g) ---> 4 CO2 (g) + 2 H2O (l) + 6 Cl2 (g)
Substance ----> H°f (kJ / mol)
CHCl3 (l) -132
CO2 (g) -393.5
H2O (l) -285.8
a. -8.11 x 102 kJ
b. -5.48 x 102 kJ
c. 1.62 x 103 kJ
d. 5.48 x 102kJ
e. -1.62 x 103 kJ