Calculate ΔHrxn for the following reaction:CH4(g)+4Cl2(g)→CCl4(g)+4HC l(g)Use the following reactions and given ΔH′s. C(s)+2H2(g)→CH4(g)ΔH=−74.6kJC(s)+2C l2(g)→CCl4(g)ΔH=−95.7kJH2(g)+Cl2(g) →2HCl(g)ΔH=−184.6kJ

ΔH= -390.3kJ

Explanation:

Using Hess's law, we can obtain ΔHrxn from the sum of related reactions.

Using:

(1) C(s) + 2H2(g) → CH4(g) ΔH=−74.6kJ

(2) C(s) + 2Cl2(g) → CCl4(g) ΔH=−95.7kJ

(3) H2(g) + Cl2(g) → 2HCl(g) ΔH=−184.6kJ

(2) - (1):

CH4(g) + C(s) + 2Cl2(g) → CCl4(g) + C(s) + 2H2(g)

CH4(g) + 2Cl2(g) → CCl4(g) + 2H2(g)

ΔH= -95.7kJ - (-74.6kJ) = -21.1kJ

This reaction + 2*(3):

2H2(g) + 2Cl2(g) + CH4(g) + 2Cl2(g) → CCl4(g) + 2H2(g) +4HCl(g)

CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)

ΔH= -21.1kJ + 2*-184.6kJ

the answers are

c and c