madey21
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An original reaction and enthalpy is given below:
[tex]\[
\begin{array}{r}
CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl(g) \\
\Delta H = -390.3 \, \text{kJ}
\end{array}
\][/tex]

What is the enthalpy for the altered reaction below?
[tex]\[
\begin{array}{c}
8HCl(g) + 2CCl_4(g) \rightarrow 2CH_4(g) + 8Cl_2(g) \\
\Delta H = [?] \, \text{kJ}
\end{array}
\][/tex]

Enter either a "+" or "-" sign and the magnitude. Use significant figures.

Sagot :

To determine the enthalpy change for the altered reaction, we need to carefully analyze how the original reaction has been transformed and accordingly adjust the enthalpy change. Let's break it down step-by-step:

### Original Reaction:
[tex]\[ CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(g) + 4HCl(g) \][/tex]
[tex]\[ \Delta H = -390.3 \, \text{kJ} \][/tex]

### Altered Reaction:
[tex]\[ 8HCl(g) + 2CCl_4(g) \rightarrow 2CH_4(g) + 8Cl_2(g) \][/tex]

### Step-by-Step Solution:

1. Reverse the Original Reaction:

Reversing the reaction means that the products become reactants and the reactants become products. For the reverse reaction, the sign of the enthalpy change also reverses. Therefore:
[tex]\[ CCl_4(g) + 4HCl(g) \rightarrow CH_4(g) + 4Cl_2(g) \][/tex]
The enthalpy change for the reverse reaction is:
[tex]\[ \Delta H = +390.3 \, \text{kJ} \][/tex]

2. Scale the Reaction Coefficients:

The coefficients in the altered reaction are doubled compared to the reverse reaction. When the entire reaction is scaled by a factor, the enthalpy change is also scaled by the same factor. Here, we are scaling by a factor of 2:
[tex]\[ 2 \times (CCl_4(g) + 4HCl(g) \rightarrow CH_4(g) + 4Cl_2(g)) \][/tex]
This gives:
[tex]\[ 2CCl_4(g) + 8HCl(g) \rightarrow 2CH_4(g) + 8Cl_2(g) \][/tex]
And the enthalpy change is:
[tex]\[ \Delta H = 2 \times +390.3 \, \text{kJ} = +780.6 \, \text{kJ} \][/tex]

### Final Answer:

The enthalpy change for the altered reaction
[tex]\[ 8HCl(g) + 2CCl_4(g) \rightarrow 2CH_4(g) + 8Cl_2(g) \][/tex]
is [tex]\(\boxed{+780.6 \, \text{kJ}}\)[/tex].
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