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To find the enthalpy change ([tex]\(\Delta H\)[/tex]) for the overall reaction:
[tex]\[ CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl(g) \][/tex]
we need to manipulate the given intermediate reactions. Here's a detailed step-by-step solution:
Given intermediate reactions:
1. [tex]\( CH_4(g) \rightarrow C(s) + 2H_2(g) \quad \Delta H_1 = 74.6 \text{ kJ} \)[/tex]
2. [tex]\( CCl_4(g) \rightarrow C(s) + 2Cl_2(g) \quad \Delta H_2 = 95.7 \text{ kJ} \)[/tex]
3. [tex]\( H_2(g) + Cl_2(g) \rightarrow 2HCl(g) \quad \Delta H_3 = -92.3 \text{ kJ} \)[/tex]
We need to construct the overall desired reaction by reversing and combining these intermediate reactions correctly.
1. Reverse the second reaction to match the desired products:
[tex]\[ C(s) + 2Cl_2(g) \rightarrow CCl_4(g) \quad \Delta H = -\Delta H_2 = -95.7 \text{ kJ} \][/tex]
2. Double the third reaction since we need [tex]\(4HCl(g)\)[/tex] in the overall reaction:
[tex]\[ 2H_2(g) + 2Cl_2(g) \rightarrow 4HCl(g) \quad \Delta H = 2 \times \Delta H_3 = 2 \times -92.3 \text{ kJ} = -184.6 \text{ kJ} \][/tex]
3. Now, add up the modified reactions:
[tex]\[ CH_4(g) \rightarrow C(s) + 2H_2(g) \quad \Delta H_1 = 74.6 \text{ kJ} \][/tex]
[tex]\[ C(s) + 2Cl_2(g) \rightarrow CCl_4(g) \quad \Delta H = -95.7 \text{ kJ} \][/tex]
[tex]\[ 2H_2(g) + 2Cl_2(g) \rightarrow 4HCl(g) \quad \Delta H = -184.6 \text{ kJ} \][/tex]
When we add these reactions together, we need to ensure intermediates cancel out correctly to achieve the overall equation:
- [tex]\(C(s)\)[/tex] cancels out.
- [tex]\(2H_2(g)\)[/tex] cancels out.
- [tex]\(2Cl_2(g)\)[/tex] remains:
[tex]\[ CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl(g) \][/tex]
Finally, sum the enthalpies:
[tex]\[ \Delta H_{\text{overall}} = \Delta H_1 + (-\Delta H_2) + 2 \times \Delta H_3 \][/tex]
[tex]\[ \Delta H_{\text{overall}} = 74.6 \text{ kJ} + (-95.7 \text{ kJ}) + (-184.6 \text{ kJ}) \][/tex]
[tex]\[ \Delta H_{\text{overall}} = 74.6 - 95.7 - 184.6 \][/tex]
[tex]\[ \Delta H_{\text{overall}} = -205.7 \text{ kJ} \][/tex]
Therefore, the enthalpy change for the overall reaction is [tex]\(-205.7 \text{ kJ}\)[/tex].
The correct answer is:
[tex]\[ -205.7 \text{ kJ} \][/tex]
[tex]\[ CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl(g) \][/tex]
we need to manipulate the given intermediate reactions. Here's a detailed step-by-step solution:
Given intermediate reactions:
1. [tex]\( CH_4(g) \rightarrow C(s) + 2H_2(g) \quad \Delta H_1 = 74.6 \text{ kJ} \)[/tex]
2. [tex]\( CCl_4(g) \rightarrow C(s) + 2Cl_2(g) \quad \Delta H_2 = 95.7 \text{ kJ} \)[/tex]
3. [tex]\( H_2(g) + Cl_2(g) \rightarrow 2HCl(g) \quad \Delta H_3 = -92.3 \text{ kJ} \)[/tex]
We need to construct the overall desired reaction by reversing and combining these intermediate reactions correctly.
1. Reverse the second reaction to match the desired products:
[tex]\[ C(s) + 2Cl_2(g) \rightarrow CCl_4(g) \quad \Delta H = -\Delta H_2 = -95.7 \text{ kJ} \][/tex]
2. Double the third reaction since we need [tex]\(4HCl(g)\)[/tex] in the overall reaction:
[tex]\[ 2H_2(g) + 2Cl_2(g) \rightarrow 4HCl(g) \quad \Delta H = 2 \times \Delta H_3 = 2 \times -92.3 \text{ kJ} = -184.6 \text{ kJ} \][/tex]
3. Now, add up the modified reactions:
[tex]\[ CH_4(g) \rightarrow C(s) + 2H_2(g) \quad \Delta H_1 = 74.6 \text{ kJ} \][/tex]
[tex]\[ C(s) + 2Cl_2(g) \rightarrow CCl_4(g) \quad \Delta H = -95.7 \text{ kJ} \][/tex]
[tex]\[ 2H_2(g) + 2Cl_2(g) \rightarrow 4HCl(g) \quad \Delta H = -184.6 \text{ kJ} \][/tex]
When we add these reactions together, we need to ensure intermediates cancel out correctly to achieve the overall equation:
- [tex]\(C(s)\)[/tex] cancels out.
- [tex]\(2H_2(g)\)[/tex] cancels out.
- [tex]\(2Cl_2(g)\)[/tex] remains:
[tex]\[ CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl(g) \][/tex]
Finally, sum the enthalpies:
[tex]\[ \Delta H_{\text{overall}} = \Delta H_1 + (-\Delta H_2) + 2 \times \Delta H_3 \][/tex]
[tex]\[ \Delta H_{\text{overall}} = 74.6 \text{ kJ} + (-95.7 \text{ kJ}) + (-184.6 \text{ kJ}) \][/tex]
[tex]\[ \Delta H_{\text{overall}} = 74.6 - 95.7 - 184.6 \][/tex]
[tex]\[ \Delta H_{\text{overall}} = -205.7 \text{ kJ} \][/tex]
Therefore, the enthalpy change for the overall reaction is [tex]\(-205.7 \text{ kJ}\)[/tex].
The correct answer is:
[tex]\[ -205.7 \text{ kJ} \][/tex]
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