Certainly! Let's go through the problem step-by-step.
### Step 1: Original Reaction and Enthalpy
The given original reaction and its enthalpy change (ΔH) is:
[tex]\[
\begin{array}{r}
CH_4(g) + 4 \, Cl_2(g) \rightarrow CCl_4(g) + 4 \, HCl(g) \\
\Delta H = -390.3 \, \text{kJ}
\end{array}
\][/tex]
### Step 2: Reverse the Reaction
To find the enthalpy change for the altered reaction, note that the altered reaction appears to be a reverse of the original with some modifications. The reverse of the original reaction would be:
[tex]\[
CCl_4(g) + 4 \, HCl(g) \rightarrow CH_4(g) + 4 \, Cl_2(g)
\][/tex]
For the reverse reaction, the enthalpy change (ΔH) will be the opposite in sign to that of the original reaction. Therefore:
[tex]\[
\Delta H_{\text{reverse}} = +390.3 \, \text{kJ}
\][/tex]
### Step 3: Formulate the Altered Reaction
The altered reaction given is:
[tex]\[
8 \, HCl(g) + 2 \, CCl_4(g) \rightarrow 2 \, CH_4(g) + 8 \, Cl_2(g)
\][/tex]
This altered reaction is exactly the double of the reversed original reaction. Let's see how:
1. Each coefficient in the reversed reaction is doubled:
[tex]\[
2 \left( CCl_4(g) + 4 \, HCl(g) \rightarrow CH_4(g) + 4 \, Cl_2(g) \right)
\][/tex]
2. This results in the altered reaction:
[tex]\[
\begin{array}{c}
2 \, CCl_4(g) + 8 \, HCl(g) \rightarrow 2 \, CH_4(g) + 8 \, Cl_2(g)
\end{array}
\][/tex]
### Step 4: Calculate the Enthalpy for the Altered Reaction
Since the reaction is exactly double the reversed reaction, the enthalpy change for the altered reaction will be double that of the reversed reaction. Therefore:
[tex]\[
\Delta H_{\text{altered}} = 2 \times +390.3 \, \text{kJ}
\][/tex]
[tex]\[
\Delta H_{\text{altered}} = +780.6 \, \text{kJ}
\][/tex]
### Step 5: Final Answer
The enthalpy change (ΔH) for the altered reaction:
[tex]\[
8 \, HCl(g) + 2 \, CCl_4(g) \rightarrow 2 \, CH_4(g) + 8 \, Cl_2(g)
\][/tex]
is:
[tex]\[
\boxed{+780.6 \, \text{kJ}}
\][/tex]
