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Consider the following intermediate chemical equations:

[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]

What is the enthalpy of the overall chemical reaction [tex]\( CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl(g) \)[/tex]?

A. [tex]\(-205.7 \, \text{kJ}\)[/tex]
B. [tex]\(-113.4 \, \text{kJ}\)[/tex]
C. [tex]\(-14.3 \, \text{kJ}\)[/tex]
D. [tex]\(78.0 \, \text{kJ}\)[/tex]

Sagot :

GinnyAnswer
To determine the enthalpy change for the overall chemical reaction given by
[tex]\[ \text{CH}_4(g) + 4 \text{Cl}_2(g) \rightarrow \text{CCl}_4(g) + 4 \text{HCl}(g), \][/tex]
we will need to manipulate the given intermediate reactions and their enthalpies. Here are the provided intermediate reactions with their respective enthalpy changes:

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]

First, we reverse the first reaction to get:
[tex]\[ \text{C}(s) + 2 \text{H}_2(g) \rightarrow \text{CH}_4(g) \][/tex]
Reversing a reaction flips the sign of the enthalpy change:
[tex]\[ \Delta H_1' = -74.6 \text{ kJ} \][/tex]

Next, we reverse the second reaction:
[tex]\[ \text{C}(s) + 2 \text{Cl}_2(g) \rightarrow \text{CCl}_4(g) \][/tex]
Reversing this reaction also flips the sign of the enthalpy change:
[tex]\[ \Delta H_2' = -95.7 \text{ kJ} \][/tex]

The third reaction will be used as it is. To match the overall reaction, we need to multiply it by 2:
[tex]\[ 2(\text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2 \text{HCl}(g)) \][/tex]
This means the enthalpy change for this step will be doubled:
[tex]\[ 2 \times \Delta H_3 = 2 \times -92.3 \text{ kJ} = -184.6 \text{ kJ} \][/tex]

Now we add up the enthalpy changes of these manipulated steps to find the enthalpy change of the overall reaction:

[tex]\[ \Delta H_{\text{overall}} = \Delta H_1' + \Delta H_2' + 2 \times \Delta H_3 \][/tex]
Substituting in the values:

[tex]\[ \Delta H_{\text{overall}} = -74.6 \text{ kJ} + -95.7 \text{ kJ} + -184.6 \text{ kJ} \][/tex]
[tex]\[ \Delta H_{\text{overall}} = -354.9 \text{ kJ} \][/tex]

Thus, 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]\(-354.9 \text{ kJ}\)[/tex].

Since none of the options provided match the calculated enthalpy change, it seems there's a discrepancy. The final answer is [tex]\(-354.9 \text{ kJ}\)[/tex], none of the given options.
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