Calculate ΔH o rxn for the following: CH4(g) + Cl2(g) → CCl4(l) + HCl(g)[unbalanced] ΔH o f [CH4(g)] = −74.87 kJ/mol ΔH o f [CCl4(g)] = −96.0 kJ/mol ΔH o f [CCl4(l)] = −139 kJ/mol ΔH o f [HCl(g)] = −92.31 kJ/mol ΔH o f [HCl(aq)] = −167.46 kJ/mol ΔH o f [Cl(g)] = 121.0 kJ/mol

 ΔH for the reaction is -277.4 kJ

Explanation:

The balanced chemical reaction is,

The expression for enthalpy change is,

where,

n = number of moles

Now put all the given values in this expression, we get

Therefore, the enthalpy change for this reaction is, -277.4 kJ

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