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A chemist measures the enthalpy change [tex]$\Delta H$[/tex] during the following reaction:
[tex]\[
P_4(s) + 6 Cl_2(g) \rightarrow 4 PCl_3(g) \quad \Delta H = -1148 \text{ kJ}
\][/tex]

Use this information to complete the table below. Round each of your answers to the nearest kJ.

\begin{tabular}{|c|c|}
\hline
Reaction & [tex]$\Delta H$[/tex] \\
\hline
[tex]$\frac{1}{6} P_4(s) + Cl_2(g) \rightarrow \frac{2}{3} PCl_3(g)$[/tex] & [tex]$\square$[/tex] kJ \\
\hline
[tex]$\frac{4}{3} PCl_3(g) \rightarrow \frac{1}{3} P_4(s) + 2 Cl_2(g)$[/tex] & [tex]$\square$[/tex] kJ \\
\hline
[tex]$4 PCl_3(g) \rightarrow P_4(s) + 6 Cl_2(g)$[/tex] & [tex]$\square$[/tex] kJ \\
\hline
\end{tabular}



Answer :

GinnyAnswer
To determine the enthalpy changes for the given reactions, we can start from the provided enthalpy change of the reaction:

[tex]\[ P_4(s) + 6 \text{Cl}_2(g) \rightarrow 4 \text{PCl}_3(g) \quad \Delta H = -1148 \text{ kJ} \][/tex]

We'll proceed step-by-step to calculate the enthalpy changes for the specified reactions.

### 1. For the reaction:
[tex]\[ \frac{1}{6} P_4(s) + \text{Cl}_2(g) \rightarrow \frac{2}{3} \text{PCl}_3(g) \][/tex]

This reaction is essentially the original reaction divided by 6. To find the enthalpy change, we divide the enthalpy change of the original reaction by 6:

[tex]\[ \Delta H_1 = \frac{-1148 \text{ kJ}}{6} = -191 \text{ kJ} \][/tex]

So, the enthalpy change for this reaction is:
[tex]\[ \boxed{-191 \text{ kJ}} \][/tex]

### 2. For the reaction:
[tex]\[ \frac{4}{3} \text{PCl}_3(g) \rightarrow \frac{1}{3} P_4(s) + 2 \text{Cl}_2(g) \][/tex]

This reaction is the reverse of the reaction in step 1. In the reversal of a reaction, the enthalpy change is the opposite (negated) of the forward reaction's enthalpy change:

[tex]\[ \Delta H_2 = -(-191 \text{ kJ}) = 191 \text{ kJ} \][/tex]

So, the enthalpy change for this reaction is:
[tex]\[ \boxed{191 \text{ kJ}} \][/tex]

### 3. For the reaction:
[tex]\[ 4 \text{PCl}_3(g) \rightarrow P_4(s) + 6 \text{Cl}_2(g) \][/tex]

This reaction is the reverse of the original reaction. Therefore, the enthalpy change will be the opposite (negated) of the original reaction's enthalpy change:

[tex]\[ \Delta H_3 = -(-1148 \text{ kJ}) = 1148 \text{ kJ} \][/tex]

So, the enthalpy change for this reaction is:
[tex]\[ \boxed{1148 \text{ kJ}} \][/tex]

### Summary

Here's the complete table with the calculated enthalpy changes:

[tex]\[ \begin{array}{|c|c|} \hline \text{reaction} & \Delta H \\ \hline \frac{1}{6} P_4(s) + \text{Cl}_2(g) \rightarrow \frac{2}{3} \text{PCl}_3(g) & -191 \text{ kJ} \\ \hline \frac{4}{3} \text{PCl}_3(g) \rightarrow \frac{1}{3} P_4(s) + 2 \text{Cl}_2(g) & 191 \text{ kJ} \\ \hline 4 \text{PCl}_3(g) \rightarrow P_4(s) + 6 \text{Cl}_2(g) & 1148 \text{ kJ} \\ \hline \end{array} \][/tex]

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