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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]\(\frac{4}{3} \, PCl_3(g) \rightarrow \frac{1}{3} \, P_4(s) + 2 \, Cl_2(g)\)[/tex] & [tex]\(\square \, \text{kJ}\)[/tex] \\
\hline
[tex]\(4 \, PCl_3(g) \rightarrow P_4(s) + 6 \, Cl_2(g)\)[/tex] & [tex]\(\square \, \text{kJ}\)[/tex] \\
\hline
[tex]\(\frac{1}{4} \, P_4(s) + \frac{3}{2} \, Cl_2(g) \rightarrow PCl_3(g)\)[/tex] & [tex]\(\square \, \text{kJ}\)[/tex] \\
\hline
\end{tabular}



Answer :

GinnyAnswer
Sure, let's break down the enthalpy changes for each reaction step-by-step:

1. First Reaction:

[tex]\[ \frac{4}{3} PCl_3(g) \rightarrow \frac{1}{3} P_4(s) + 2 Cl_2(g) \][/tex]

This reaction can be derived from the given reaction:

[tex]\[ P_4(s) + 6 Cl_2(g) \rightarrow 4 PCl_3(g) \quad \Delta H = -1148 \text{ kJ} \][/tex]

By reversing the given reaction and adjusting for the coefficients, we get:

[tex]\[ 4 PCl_3(g) \rightarrow P_4(s) + 6 Cl_2(g) \][/tex]

To balance the coefficients:

[tex]\[ \frac{4}{3} PCl_3(g) \rightarrow \frac{1}{3} P_4(s) + 2 Cl_2(g) \][/tex]

The enthalpy change for this reaction is:

[tex]\[ \Delta H = \left(-\frac{1}{3}\right) \times (-1148 \text{ kJ}) = 383 \text{ kJ} \][/tex]

2. Second Reaction:

[tex]\[ 4 PCl_3(g) \rightarrow P_4(s) + 6 Cl_2(g) \][/tex]

This can be derived by simply reversing the given reaction:

[tex]\[ 4 PCl_3(g) \rightarrow P_4(s) + 6 Cl_2(g) \][/tex]

The enthalpy change for this reaction is the negative of the given reaction:

[tex]\[ \Delta H = 1148 \text{ kJ} \][/tex]

3. Third Reaction:

[tex]\[ \frac{1}{4} P_4(s) + \frac{3}{2} Cl_2(g) \rightarrow PCl_3(g) \][/tex]

This reaction can be derived from the given reaction:

[tex]\[ P_4(s) + 6 Cl_2(g) \rightarrow 4 PCl_3(g) \quad \Delta H = -1148 \text{ kJ} \][/tex]

To balance the coefficients:

[tex]\[ \frac{1}{4} P_4(s) + \frac{3}{2} Cl_2(g) \rightarrow PCl_3(g) \][/tex]

The enthalpy change for this reaction is:

[tex]\[ \Delta H = \left(\frac{1}{4}\right) \times -1148 \text{ kJ} = -287 \text{ kJ} \][/tex]

Finally, we complete the table with these enthalpy changes:

[tex]\[ \begin{tabular}{|c|c|} \hline reaction & $\Delta H$ \\ \hline \frac{4}{3} PCl_3(g) \rightarrow \frac{1}{3} P_4(s) + 2 Cl_2(g) & 383 \text{ kJ} \\ \hline 4 PCl_3(g) \rightarrow P_4(s) + 6 Cl_2(g) & 1148 \text{ kJ} \\ \hline \frac{1}{4} P_4(s) + \frac{3}{2} Cl_2(g) \rightarrow PCl_3(g) & -287 \text{ kJ} \\ \hline \end{tabular} \][/tex]

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