Question

Calculate ΔrH for the following reaction: CH₄(g)+4Cl₂(g)→CCl₄(g)+4HCl(g)

Use the following reactions and given
a. ΔrH's. C(s)+2H₂(g)→CH₄(g) ΔrH=−74.6kJmol⁻¹
b. C(s)+2Cl₂(g)→CCl₄(g) ΔrH=−95.7kJmol⁻¹
c. H₂(g)+Cl₂(g)→2HCl(g) ΔrH=−92.3kJmol⁻¹

Answer 1

The enthalpy change for the reaction is
`\(-539.5 \, \text{kJ/mol}\)`.

To calculate the enthalpy change
`(\(\Delta_rH\))` for the given reaction:

`\[ \text{CH}_4(g) + 4\text{Cl}_2(g) \rightarrow \text{CCl}_4(g) + 4\text{HCl}(g) \]`

We can use the enthalpy changes provided for the given reactions:

a.
`\( C(s) + 2H_2(g) \rightarrow \text{CH}_4(g) \) with \( \Delta_rH = -74.6 \, \text{kJ/mol} \)`

b.
`\( C(s) + 2Cl_2(g) \rightarrow \text{CCl}_4(g) \) with \( \Delta_rH = -95.7 \, \text{kJ/mol} \)`

c.
`\( H_2(g) + Cl_2(g) \rightarrow 2\text{HCl}(g) \) with \( \Delta_rH = -92.3 \, \text{kJ/mol} \)`

The enthalpy change for the reaction can be calculated using Hess’s Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step in the reaction process.

The target reaction is a combination of the above reactions, and we can use them to find the enthalpy change for the target reaction.

`\[ \text{CH}_4(g) + 4\text{Cl}_2(g) \rightarrow \text{CCl}_4(g) + 4\text{HCl}(g) \]`

`\[ \Delta_rH = \Delta_rH_a + \Delta_rH_b + \Delta_rH_c \]`

`\[ \Delta_rH = (-74.6 \, \text{kJ/mol}) + (-95.7 \, \text{kJ/mol}) + 4(-92.3 \, \text{kJ/mol}) \]`

`\[ \Delta_rH = -74.6 - 95.7 - 369.2 \]`

`\[ \Delta_rH = -539.5 \, \text{kJ/mol} \]`

Therefore, the enthalpy change for the reaction is
`\(-539.5 \, \text{kJ/mol}\)`.