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To determine the enthalpy change for the overall reaction [tex]\( CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl(g) \)[/tex], we can use Hess's Law. Hess's Law states that the total enthalpy change for a reaction is the sum of the enthalpy changes of the individual steps that lead to the overall reaction.
Let's consider the given intermediate equations and their enthalpy changes:
1. [tex]\( CH_4(g) \rightarrow C(s) + 2H_2(g) \)[/tex] with [tex]\( \Delta H_1 = 74.6 \, \text{kJ} \)[/tex]
2. [tex]\( CCl_4(g) \rightarrow C(s) + 2Cl_2(g) \)[/tex] with [tex]\( \Delta H_2 = 95.7 \, \text{kJ} \)[/tex]
3. [tex]\( H_2(g) + Cl_2(g) \rightarrow 2HCl(g) \)[/tex] with [tex]\( \Delta H_3 = -92.3 \, \text{kJ} \)[/tex]
We need to rearrange and combine these equations to match the overall reaction [tex]\( CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl(g) \)[/tex].
First, let's reverse reaction 2 so that [tex]\( CCl_4(g) \)[/tex] is on the product side:
[tex]\[ C(s) + 2Cl_2(g) \rightarrow CCl_4(g) \quad \Delta H = -95.7 \, \text{kJ} \][/tex]
Next, we need 4 moles of [tex]\( HCl(g) \)[/tex] in the product, so we multiply reaction 3 by 2:
[tex]\[ 2H_2(g) + 2Cl_2(g) \rightarrow 4HCl(g) \quad \Delta H = 2 \times (-92.3) \, \text{kJ} = -184.6 \, \text{kJ} \][/tex]
Now combine these rearranged reactions with reaction 1:
[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 steps together, the intermediates [tex]\( C(s) \)[/tex] and [tex]\( 2H_2(g) \)[/tex] cancel out:
[tex]\[ CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl(g) \][/tex]
The overall enthalpy change [tex]\( \Delta H_{\text{overall}} \)[/tex] is the sum of the enthalpy changes of these steps:
[tex]\[ \Delta H_{\text{overall}} = \Delta H_1 + (-95.7) \, \text{kJ} + (-184.6) \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{\text{overall}} = 74.6 \, \text{kJ} - 95.7 \, \text{kJ} - 184.6 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{\text{overall}} = 74.6 - 280.3 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{\text{overall}} = -205.7 \, \text{kJ} \][/tex]
Therefore, the enthalpy of the overall chemical reaction [tex]\( CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl (g) \)[/tex] is [tex]\( \boxed{-14.3 \, \text{kJ}} \)[/tex].
Let's consider the given intermediate equations and their enthalpy changes:
1. [tex]\( CH_4(g) \rightarrow C(s) + 2H_2(g) \)[/tex] with [tex]\( \Delta H_1 = 74.6 \, \text{kJ} \)[/tex]
2. [tex]\( CCl_4(g) \rightarrow C(s) + 2Cl_2(g) \)[/tex] with [tex]\( \Delta H_2 = 95.7 \, \text{kJ} \)[/tex]
3. [tex]\( H_2(g) + Cl_2(g) \rightarrow 2HCl(g) \)[/tex] with [tex]\( \Delta H_3 = -92.3 \, \text{kJ} \)[/tex]
We need to rearrange and combine these equations to match the overall reaction [tex]\( CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl(g) \)[/tex].
First, let's reverse reaction 2 so that [tex]\( CCl_4(g) \)[/tex] is on the product side:
[tex]\[ C(s) + 2Cl_2(g) \rightarrow CCl_4(g) \quad \Delta H = -95.7 \, \text{kJ} \][/tex]
Next, we need 4 moles of [tex]\( HCl(g) \)[/tex] in the product, so we multiply reaction 3 by 2:
[tex]\[ 2H_2(g) + 2Cl_2(g) \rightarrow 4HCl(g) \quad \Delta H = 2 \times (-92.3) \, \text{kJ} = -184.6 \, \text{kJ} \][/tex]
Now combine these rearranged reactions with reaction 1:
[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 steps together, the intermediates [tex]\( C(s) \)[/tex] and [tex]\( 2H_2(g) \)[/tex] cancel out:
[tex]\[ CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl(g) \][/tex]
The overall enthalpy change [tex]\( \Delta H_{\text{overall}} \)[/tex] is the sum of the enthalpy changes of these steps:
[tex]\[ \Delta H_{\text{overall}} = \Delta H_1 + (-95.7) \, \text{kJ} + (-184.6) \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{\text{overall}} = 74.6 \, \text{kJ} - 95.7 \, \text{kJ} - 184.6 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{\text{overall}} = 74.6 - 280.3 \, \text{kJ} \][/tex]
[tex]\[ \Delta H_{\text{overall}} = -205.7 \, \text{kJ} \][/tex]
Therefore, the enthalpy of the overall chemical reaction [tex]\( CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl (g) \)[/tex] is [tex]\( \boxed{-14.3 \, \text{kJ}} \)[/tex].
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