Question

Given the following chemical reactions and their respective enthalpy changes:

CH₄(g) → C(s) + 2H₂(g) ΔH₁ = 74.6 kJ
CCl₂(g) → C(s) + 2Cl₂(g) ΔH₂ = 2012 kJ
H₂(g) + Cl₂(g) → 2HCl(g) ΔH₃ = -92.3 kJ
What is the enthalpy change for the overall chemical reaction:

CH₄(g) + 4Cl₂(g) → CCl₄(g) + 4HCl(g)?

Answer 1

The enthalpy change for the reaction CH₄(g) + 4Cl₂(g) → CCl₄(g) + 4HCl(g) is calculated by summing the changes given for intermediate reactions, resulting in an enthalpy change of 1902.0 kJ.

Step-by-step explanation:

The enthalpy change for the overall chemical reaction is calculated using Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction. Therefore, we must manipulate the given reactions to derive the overall reaction:

CH₄(g) + 4Cl₂(g) → CCl₄(g) + 4HCl(g).

The steps involve breaking the bonds of CH₄ and creating C(s) and H₂(g), then breaking bonds in CCl₂(g) to create C(s) and Cl₂(g), and finally forming HCl from H₂(g) and Cl₂(g). By summing up the given enthalpies changes for the reactions (ΔH₁ = 74.6 kJ, ΔH₂ = 2012 kJ, and ΔH₃ = -92.3 kJ for each 2 moles of HCl) and considering the stoichiometry, we get:

ΔH = ΔH₁ + ΔH₂ + (2 * ΔH₃)

Plugging in the values:

ΔH = 74.6 kJ + 2012 kJ + (2 * (-92.3 kJ))

ΔH = 74.6 kJ + 2012 kJ - 184.6 kJ

ΔH = 1902.0 kJ

Therefore, the enthalpy change for the overall reaction is 1902.0 kJ.