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To determine the enthalpy change for the overall reaction [tex]\( \text{CH}_4(g) + 4\text{Cl}_2(g) \rightarrow \text{CCl}_4(g) + 4\text{HCl}(g) \)[/tex], we need to utilize the given intermediate reactions and their associated enthalpy changes. This can be done by manipulating these intermediate reactions to match the overall reaction.
The given intermediate reactions are as follows:
1. [tex]\( \text{CH}_4(g) \rightarrow \text{C}(s) + 2\text{H}_2(g) \quad \Delta H_1 = 74.6 \, \text{kJ} \)[/tex]
2. [tex]\( \text{CCl}_4(g) \rightarrow \text{C}(s) + 2\text{Cl}_2(g) \quad \Delta H_2 = 95.7 \, \text{kJ} \)[/tex]
3. [tex]\( \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g) \quad \Delta H_3 = -92.3 \, \text{kJ} \)[/tex]
Let's adjust these reactions to derive the desired overall reaction:
1. The first reaction [tex]\( \text{CH}_4(g) \rightarrow \text{C}(s) + 2\text{H}_2(g) \)[/tex] does not need to be altered.
2. The second reaction [tex]\( \text{CCl}_4(g) \rightarrow \text{C}(s) + 2\text{Cl}_2(g) \)[/tex] needs to be reversed to produce [tex]\( \text{CCl}_4(g) \)[/tex] on the right side.
[tex]\[ \text{C}(s) + 2\text{Cl}_2(g) \rightarrow \text{CCl}_4(g) \quad \Delta H_2' = -95.7 \, \text{kJ} \][/tex]
3. For the third reaction, we need 4 HCl molecules in the product side. Since the given equation produces 2 HCl molecules, we'll multiply the entire equation by 2.
[tex]\[ 2\left( \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g) \right) \quad \Delta H_3' = 2 \times -92.3 \, \text{kJ} = -184.6 \, \text{kJ} \][/tex]
Now we combine these adjusted reactions:
1. [tex]\( \text{CH}_4(g) \rightarrow \text{C}(s) + 2\text{H}_2(g) \quad \Delta H_1 = 74.6 \, \text{kJ} \)[/tex]
2. [tex]\( \text{C}(s) + 2\text{Cl}_2(g) \rightarrow \text{CCl}_4(g) \quad \Delta H_2' = -95.7 \, \text{kJ} \)[/tex]
3. [tex]\( 2\left( \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g) \right) \quad \Delta H_3' = -184.6 \, \text{kJ} \)[/tex]
Let's sum these reactions to find the overall enthalpy change:
[tex]\[ \text{CH}_4(g) + 2\text{H}_2(g) + 2\text{Cl}_2(g) + 2\text{H}_2(g) + 2\text{Cl}_2(g) \rightarrow \text{C}(s) + 2\text{H}_2(g) + \text{C}(s) + \text{CCl}_4(g) + 2\text{H}_2(g) + 4\text{HCl}(g) \][/tex]
Notice that intermediate species ([tex]\(\text{C}(s)\)[/tex], [tex]\(\text{H}_2(g)\)[/tex]) cancel out on both sides, leaving:
[tex]\[ \text{CH}_4(g) + 4\text{Cl}_2(g) \rightarrow \text{CCl}_4(g) + 4\text{HCl}(g) \][/tex]
Summing the enthalpies:
[tex]\[ \Delta H_{\text{overall}} = 74.6 \, \text{kJ} + (-95.7 \, \text{kJ}) + (-184.6 \, \text{kJ}) = -205.7 \, \text{kJ} \][/tex]
Therefore, the enthalpy change for 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]\(\Delta H_{\text{overall}} = -205.7 \, \text{kJ}\)[/tex].
So, the correct answer is [tex]\( \boxed{-205.7 \, \text{kJ}} \)[/tex].
The given intermediate reactions are as follows:
1. [tex]\( \text{CH}_4(g) \rightarrow \text{C}(s) + 2\text{H}_2(g) \quad \Delta H_1 = 74.6 \, \text{kJ} \)[/tex]
2. [tex]\( \text{CCl}_4(g) \rightarrow \text{C}(s) + 2\text{Cl}_2(g) \quad \Delta H_2 = 95.7 \, \text{kJ} \)[/tex]
3. [tex]\( \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g) \quad \Delta H_3 = -92.3 \, \text{kJ} \)[/tex]
Let's adjust these reactions to derive the desired overall reaction:
1. The first reaction [tex]\( \text{CH}_4(g) \rightarrow \text{C}(s) + 2\text{H}_2(g) \)[/tex] does not need to be altered.
2. The second reaction [tex]\( \text{CCl}_4(g) \rightarrow \text{C}(s) + 2\text{Cl}_2(g) \)[/tex] needs to be reversed to produce [tex]\( \text{CCl}_4(g) \)[/tex] on the right side.
[tex]\[ \text{C}(s) + 2\text{Cl}_2(g) \rightarrow \text{CCl}_4(g) \quad \Delta H_2' = -95.7 \, \text{kJ} \][/tex]
3. For the third reaction, we need 4 HCl molecules in the product side. Since the given equation produces 2 HCl molecules, we'll multiply the entire equation by 2.
[tex]\[ 2\left( \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g) \right) \quad \Delta H_3' = 2 \times -92.3 \, \text{kJ} = -184.6 \, \text{kJ} \][/tex]
Now we combine these adjusted reactions:
1. [tex]\( \text{CH}_4(g) \rightarrow \text{C}(s) + 2\text{H}_2(g) \quad \Delta H_1 = 74.6 \, \text{kJ} \)[/tex]
2. [tex]\( \text{C}(s) + 2\text{Cl}_2(g) \rightarrow \text{CCl}_4(g) \quad \Delta H_2' = -95.7 \, \text{kJ} \)[/tex]
3. [tex]\( 2\left( \text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g) \right) \quad \Delta H_3' = -184.6 \, \text{kJ} \)[/tex]
Let's sum these reactions to find the overall enthalpy change:
[tex]\[ \text{CH}_4(g) + 2\text{H}_2(g) + 2\text{Cl}_2(g) + 2\text{H}_2(g) + 2\text{Cl}_2(g) \rightarrow \text{C}(s) + 2\text{H}_2(g) + \text{C}(s) + \text{CCl}_4(g) + 2\text{H}_2(g) + 4\text{HCl}(g) \][/tex]
Notice that intermediate species ([tex]\(\text{C}(s)\)[/tex], [tex]\(\text{H}_2(g)\)[/tex]) cancel out on both sides, leaving:
[tex]\[ \text{CH}_4(g) + 4\text{Cl}_2(g) \rightarrow \text{CCl}_4(g) + 4\text{HCl}(g) \][/tex]
Summing the enthalpies:
[tex]\[ \Delta H_{\text{overall}} = 74.6 \, \text{kJ} + (-95.7 \, \text{kJ}) + (-184.6 \, \text{kJ}) = -205.7 \, \text{kJ} \][/tex]
Therefore, the enthalpy change for 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]\(\Delta H_{\text{overall}} = -205.7 \, \text{kJ}\)[/tex].
So, the correct answer is [tex]\( \boxed{-205.7 \, \text{kJ}} \)[/tex].
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