Question

See the standard enthalpy of formation, in KJ.mol-1 and at 25°F , of ​​some substances: CH4(g) = -74.8 CHCl3(l) = - 134.5 HCl(g) = - 92, 3If we carry out the chlorination reaction of methane, what will be the value of the change in the enthalpy of the process? CH4(g) + 3Cl2(g) → CHCl3(l) + 3HCl(g) *





a) -115.9 KJ.mol-1
b) 186.3 KJ.mol-1
c) -376.2 KJ.mol-1
d) -336.6 KJ.mol-1
e) 148.5 KJ.mol-1

Answer 1

Answer:

bro it's is so easy so the answer was 336.6kj.mol-1