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 c:

Explanation:

Step1: Analyze enthalpy change signs

Positive $\Delta H$ means endothermic (energy absorbed), negative $\Delta H$ means exothermic (energy released). For the given reactions, we need to relate them to the enthalpy - diagram.

Step2: Consider magnitude comparison

We are comparing the magnitudes of enthalpy changes represented by arrows in the diagram. From the equations, we can see that the reactions have different $\Delta H$ values. We assume that the lengths of the arrows in the diagram represent the magnitudes of the enthalpy changes.
If we consider the overall reaction $CH_4(g)+4Cl_2(g)
ightarrow CCl_4(g) + 4HCl(g)$ with $\Delta H_4=- 205.7kJ$ and other component - reactions with their respective $\Delta H$ values, we can infer the following about the arrows in the diagram. If arrow B represents a certain enthalpy change and we compare it with arrow C. Since $\Delta H_4$ is a combination of multiple steps and we assume the diagram represents the overall and intermediate enthalpy changes, and considering the signs and magnitudes of the $\Delta H$ values of the given reactions, we know that if B represents an intermediate step and C is part of the overall change or related to it in a way that the magnitude of the overall change (or relevant part) is compared.
If we assume that the reactions are set up in a way that the overall reaction's enthalpy change is composed of the intermediate steps, and if B is an intermediate endothermic step (positive $\Delta H$) and C is related to the overall exothermic change, we can say that C has a magnitude that is greater than that of B.

Answer:

has a magnitude that is greater than that of B