# Problem: Calculate ΔHrxn Delta  H_{ m rxn} for the following reaction: CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)Given these reactions and their ΔH values:C(s) + 2 H2(g) → CH4(g); ΔH = –74.6 kJC(s) + 2 Cl2(g) → CCl4(g); ΔH = –95.7 kJH2(g) + Cl2(g) → 2 HCl(g); ΔH = –92.3 kJ﻿

###### FREE Expert Solution

We are asked to find ΔHrxn for the following reaction: CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)

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

Reaction                                                                                                Modification

CH4(g) → C(s) + 2 H2(g)                                                                    -Eqn1   [reversed]
C(s) + 2 Cl2(g) → CCl4(g)                                                                 +Eqn2   [no change]
H2(g) + Cl2(g) → 2 HCl(g)                                                                +2Eqn3   [multiplied by 2]

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CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)

C(s) + 2 H2(g) → CH4(g); ΔH = –74.6 kJ
C(s) + 2 Cl2(g) → CCl4(g); ΔH = –95.7 kJ
H2(g) + Cl2(g) → 2 HCl(g); ΔH = –92.3 kJ ###### Problem Details

Calculate ΔHrxn for the following reaction: CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)

Given these reactions and their ΔH values:
C(s) + 2 H2(g) → CH4(g); ΔH = –74.6 kJ
C(s) + 2 Cl2(g) → CCl4(g); ΔH = –95.7 kJ
H2(g) + Cl2(g) → 2 HCl(g); ΔH = –92.3 kJ

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