Question

using enthalpy diagrams consider these equations ch₄(g) → c(s) + 2h₂(g) δh₁ = 74.6kj c(s) + 2cl₂(g) → ccl₄(g) δh₂ = -95.7kj 2h₂(g) + 2cl₂(g) → 4hcl(g) δh₃ = -184.6kj ch₄(g) + 4cl₂(g) → ccl₄(g) + 4hcl(g) δh₄ = -205.7kj complete this based on the diagram. arrow a:

Explanation:

Step1: Analyze the first equation

The first equation $\text{CH}_4(\text{g})
ightarrow\text{C}(\text{s}) + 2\text{H}_2(\text{g})$ has $\Delta H_1 = 74.6\text{ kJ}$. This is an endothermic reaction as $\Delta H>0$. Arrow A in the enthalpy - diagram likely represents this positive enthalpy change.

Answer:

$74.6\text{ kJ}$