Question

Given the reactions and their Δh values: CH₄(g) + 4Cl₂(g) → CCl₄(g) + 4HCl(g), the Δh values for the reactions are -74.6 kJ, -95.7 kJ, and -184.6 kJ. What are the products of the reactions?

1) CH₄(g), CCl₄(g), 2HCl(g)
2) C(s), H₂(g), 2H₂(g), 2Cl₂(g)
3) C(s), H₂(g), 2H₂(g), Cl₂(g)
4) C(s), H₂(g), 2H₂(g), 4Cl₂(g)

Answer 1

The products of the given reaction CH₄(g) + 4Cl₂(g) → CCl₄(g) + 4HCl(g) are carbon tetrachloride (CCl₄) and hydrogen chloride (HCl). The enthalpy values alone cannot determine the products for the mismatched reactions provided.

Step-by-step explanation:

The reactions provided in the question do not correspond directly to the Δh values given, so it seems there might be an error or misunderstanding in the question.

The products of the reactions cannot be determined from the Δh values alone. However, it is possible to infer some information from the given chemical equation: CH₄(g) + 4Cl₂(g) → CCl₄(g) + 4HCl(g). The reaction shows that when methane reacts with chlorine gas, the products formed are carbon tetrachloride and hydrogen chloride gas.

Furthermore, the formation of HCl(g) from hydrogen gas and chlorine gas is known to be exothermic, with a standard molar enthalpy of formation of -92.307 kJ/mol for HCl(g), which means that the formation of two moles of HCl would have a Δh of -184.6 kJ as stated in the appendix and the provided material.