Question

Calculate ΔH_rxn for the following reaction:

CH_4(g) + 4Cl_2(g) → CCl_4(g) + 4HCl(g)
Given these reactions and their ΔH given:
C(s) + H2(g) → 2H_2(g), ΔH=−74.6kJ
2Cl_2(g) → Cl_4(g), ΔH=−95.7kJ
2H_2(g) + Cl_4(g) → CH_4(g), ΔH=−184.6kJ
a. -421.6 kJ
b. -351.3 kJ
c. -265.7 kJ
d. -184.6 kJ

Answer 1

By manipulating the given reactions using Hess's Law and adding them up to get the desired overall reaction, the calculated ΔH_rxn for the reaction CH₄(g) + 4Cl₂(g) → CCl₄(g) + 4HCl(g) is -486.0 kJ, which is not in the list of options provided.

Step-by-step explanation:

To calculate the enthalpy change (ΔH_rxn) for the given reaction, we can use Hess's Law, which states that the total enthalpy change for a reaction is the same no matter how many steps it takes. We are given the following reactions:

• C(s) + 2H₂(g) → CH₄(g), ΔH = −74.6 kJ
• 2Cl₂(g) → Cl₄(g), ΔH = −95.7 kJ
• 2H₂(g) + Cl₄(g) → CH₄(g), ΔH = −184.6 kJ

We need to arrange these reactions to yield our target reaction:

CH₄(g) + 4Cl₂(g) → CCl₄(g) + 4HCl(g)

We reverse the first equation and multiply it by 1, the second equation is fine as is but we multiply it by 2, and the third equation we reverse and multiply by 2:

1. CH₄(g) → C(s) + 2H₂(g), ΔH = +74.6 kJ
2. 4Cl₂(g) → 2Cl₄(g), ΔH = −191.4 kJ (2 x −95.7 kJ)
3. CH₄(g) + 2Cl₄(g) → 2H₂(g) + CCl₄(g), ΔH = −369.2 kJ (2 x −184.6 kJ)

By adding these adjusted reactions, we cancel out the H₂(g) and Cl₄(g) from both sides, leaving us with our target reaction. The summation of the ΔH values would give us ΔH_rxn:

ΔH_rxn = +74.6 kJ - 191.4 kJ - 369.2 kJ = -486.0 kJ

Therefore, the enthalpy change for the reaction is -486.0 kJ, which is not listed in the given options, indicating a potential error in the question or provided data.