To determine the enthalpy change (ΔH) of the overall reaction [tex]\( \text{CH}_4(g) + 4 \text{Cl}_2(g) \rightarrow \text{CCl}_4(g) + 4 \text{HCl}(g) \)[/tex], we will follow a step-by-step approach using the given intermediate reactions and their enthalpy changes.
Step 1: List the intermediate reactions with their enthalpy changes.
1. [tex]\( \text{CH}_4(g) \rightarrow \text{C}(s) + 2 \text{H}_2(g) \)[/tex], ΔH = 74.6 kJ
2. [tex]\( \text{CCl}_4(g) \rightarrow \text{C}(s) + 2 \text{Cl}_2(g) \)[/tex], ΔH = 95.7 kJ
3. [tex]\( \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2 \text{HCl}(g) \)[/tex], ΔH = -92.3 kJ
Step 2: Reverse the second reaction to match the overall reaction's direction.
[tex]\[ \text{C}(s) + 2 \text{Cl}_2(g) \rightarrow \text{CCl}_4(g), \Delta H = -95.7 \text{ kJ} \][/tex]
Reversing the reaction changes the sign of ΔH.
Step 3: Determine the number of times each intermediate reaction contributes to the overall reaction.
For the production of [tex]\( 4 \, \text{HCl}(g) \)[/tex]:
[tex]\[ 2 \left(\text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2 \text{HCl}(g) \right), \Delta H = 2 \times (-92.3)\][/tex]
Combine this with the reversed second reaction and the first reaction to match the overall reaction.
Step 4: Add the enthalpy changes of the individual steps.
[tex]\[ \Delta H_{\text{overall}} = \Delta H_1 + (-\Delta H_2) + 2 \times \Delta H_3 \][/tex]
Given the values:
[tex]\[ \Delta H_{\text{overall}} = 74.6 \text{ kJ} + (-95.7 \text{ kJ}) + 2 \times (-92.3 \text{ kJ}) \][/tex]
[tex]\[ \Delta H_{\text{overall}} = 74.6 \text{ kJ} - 95.7 \text{ kJ} - 184.6 \text{ kJ} \][/tex]
[tex]\[ \Delta H_{\text{overall}} = -205.7 \text{ kJ} \][/tex]
Hence, the enthalpy change of the overall reaction [tex]\( \text{CH}_4(g) + 4 \text{Cl}_2(g) \rightarrow \text{CCl}_4(g) + 4 \text{HCl}(g) \)[/tex] is [tex]\( -205.7 \text{ kJ} \)[/tex].
Answer: The enthalpy of the overall chemical reaction is [tex]\( -205.7 \text{ kJ} \)[/tex].
So the correct answer is:
[tex]\[ \boxed{-205.7 \text{ kJ}} \][/tex]
