Answer :
To find the enthalpy change for the overall reaction \( CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl(g) \), we'll combine the given intermediate reactions using Hess's Law. Here's a step-by-step solution:
1. Given intermediate reactions:
[tex]\[ \begin{array}{ll} CH_4(g) \rightarrow C(s) + 2H_2(g) & \Delta H_1 = 74.6 \text{ kJ} \\ CCl_4(g) \rightarrow C(s) + 2Cl_2(g) & \Delta H_2 = 95.7 \text{ kJ} \\ H_2(g) + Cl_2(g) \rightarrow 2HCl(g) & \Delta H_3 = -92.3 \text{ kJ} \end{array} \][/tex]
2. Rewrite the target overall reaction:
[tex]\[ CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl(g) \][/tex]
3. Manipulate the given reactions to align with the target reaction:
- The first reaction is already in terms of \( CH_4 \):
[tex]\[ CH_4(g) \rightarrow C(s) + 2H_2(g) \quad \Delta H_1 = 74.6 \text{ kJ} \][/tex]
- The second reaction is reversed to fit \( CCl_4 \) becoming a product:
[tex]\[ C(s) + 2Cl_2(g) \rightarrow CCl_4(g) \quad \Delta H_2' = -95.7 \text{ kJ} \][/tex]
- The third reaction needs to be multiplied by 2 to balance the 4HCl in the final equation:
[tex]\[ 2H_2(g) + 2Cl_2(g) \rightarrow 4HCl(g) \quad \Delta H_3' = 2 \times (-92.3 \text{ kJ}) = -184.6 \text{ kJ} \][/tex]
4. Add the enthalpy changes of these manipulated reactions:
[tex]\[ \Delta H_{\text{overall}} = \Delta H_1 + \Delta H_2' + \Delta H_3' \][/tex]
[tex]\[ \Delta H_{\text{overall}} = 74.6 \text{ kJ} + (-95.7 \text{ kJ}) + (-184.6 \text{ kJ}) \][/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]
5. Sum the enthalpies:
[tex]\[ \Delta H_{\text{overall}} = -14.3 \text{ kJ} \][/tex]
Thus, the enthalpy of the overall chemical reaction \( CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl(g) \) is:
[tex]\[ \boxed{-14.3 \text{ kJ}} \][/tex]
1. Given intermediate reactions:
[tex]\[ \begin{array}{ll} CH_4(g) \rightarrow C(s) + 2H_2(g) & \Delta H_1 = 74.6 \text{ kJ} \\ CCl_4(g) \rightarrow C(s) + 2Cl_2(g) & \Delta H_2 = 95.7 \text{ kJ} \\ H_2(g) + Cl_2(g) \rightarrow 2HCl(g) & \Delta H_3 = -92.3 \text{ kJ} \end{array} \][/tex]
2. Rewrite the target overall reaction:
[tex]\[ CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl(g) \][/tex]
3. Manipulate the given reactions to align with the target reaction:
- The first reaction is already in terms of \( CH_4 \):
[tex]\[ CH_4(g) \rightarrow C(s) + 2H_2(g) \quad \Delta H_1 = 74.6 \text{ kJ} \][/tex]
- The second reaction is reversed to fit \( CCl_4 \) becoming a product:
[tex]\[ C(s) + 2Cl_2(g) \rightarrow CCl_4(g) \quad \Delta H_2' = -95.7 \text{ kJ} \][/tex]
- The third reaction needs to be multiplied by 2 to balance the 4HCl in the final equation:
[tex]\[ 2H_2(g) + 2Cl_2(g) \rightarrow 4HCl(g) \quad \Delta H_3' = 2 \times (-92.3 \text{ kJ}) = -184.6 \text{ kJ} \][/tex]
4. Add the enthalpy changes of these manipulated reactions:
[tex]\[ \Delta H_{\text{overall}} = \Delta H_1 + \Delta H_2' + \Delta H_3' \][/tex]
[tex]\[ \Delta H_{\text{overall}} = 74.6 \text{ kJ} + (-95.7 \text{ kJ}) + (-184.6 \text{ kJ}) \][/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]
5. Sum the enthalpies:
[tex]\[ \Delta H_{\text{overall}} = -14.3 \text{ kJ} \][/tex]
Thus, the enthalpy of the overall chemical reaction \( CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl(g) \) is:
[tex]\[ \boxed{-14.3 \text{ kJ}} \][/tex]
