We are asked to find ΔH_{rxn} for the following reaction: CH_{4}(g) + 4Cl_{2}(g) → CCl_{4}(g) + 4HCl(g)

Modify the equation in such a way that it adds up to be the target reaction.

Reaction Modification

CH_{4}(g) → C(s) + 2 H_{2}(g) -Eqn1 [reversed]

C(s) + 2 Cl_{2}(g) → CCl_{4}(g) +Eqn2 [no change]

H_{2}(g) + Cl_{2}(g) → 2 HCl(g) +2Eqn3 [multiplied by 2]

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CH_{4}(g) + 4Cl_{2}(g) → CCl_{4}(g) + 4HCl(g)

C(s) + 2 H_{2}(g) → CH_{4}(g); ΔH = –74.6 kJ

C(s) + 2 Cl_{2}(g) → CCl_{4}(g); ΔH = –95.7 kJ

H_{2}(g) + Cl_{2}(g) → 2 HCl(g); ΔH = –92.3 kJ

Calculate ΔH_{rxn} for the following reaction: CH_{4}(g) + 4Cl_{2}(g) → CCl_{4}(g) + 4HCl(g)

Given these reactions and their ΔH values:

C(s) + 2 H_{2}(g) → CH_{4}(g); ΔH = –74.6 kJ

C(s) + 2 Cl_{2}(g) → CCl_{4}(g); ΔH = –95.7 kJ

H_{2}(g) + Cl_{2}(g) → 2 HCl(g); ΔH = –92.3 kJ

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