kess473629
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Consider the reaction below.

[tex]\[ PCl_5(g) \longleftrightarrow PCl_3(g) + Cl_2(g) \][/tex]

At [tex]\(500 \, \text{K}\)[/tex], the reaction is at equilibrium with the following concentrations:

[tex]\[
\begin{array}{l}
\left[ PCl_5 \right] = 0.0095 \, M \\
\left[ PCl_3 \right] = 0.020 \, M \\
\left[ Cl_2 \right] = 0.020 \, M
\end{array}
\][/tex]

What is the equilibrium constant for the given reaction?

A. 0.042

B. 0.42

C. 2.4

D. 24



Answer :

GinnyAnswer
Certainly! Let's solve the problem step-by-step to find the equilibrium constant for the given chemical reaction.

Given the reaction:
[tex]\[ \text{PCl}_5(g) \longleftrightarrow \text{PCl}_3(g) + \text{Cl}_2(g) \][/tex]

The equilibrium concentrations at 500 K are:
[tex]\[ \left[ \text{PCl}_5 \right] = 0.0095 \, M \][/tex]
[tex]\[ \left[ \text{PCl}_3 \right] = 0.020 \, M \][/tex]
[tex]\[ \left[ \text{Cl}_2 \right] = 0.020 \, M \][/tex]

The equilibrium constant [tex]\( K_{eq} \)[/tex] is given by the expression:
[tex]\[ K_{eq} = \frac{\left[ \text{PCl}_3 \right] \left[ \text{Cl}_2 \right]}{\left[ \text{PCl}_5 \right]} \][/tex]

Now we substitute the values into this expression:
[tex]\[ K_{eq} = \frac{(0.020) \times (0.020)}{0.0095} \][/tex]

Simplifying the expression:
[tex]\[ K_{eq} = \frac{0.020 \times 0.020}{0.0095} \][/tex]
[tex]\[ K_{eq} = \frac{0.0004}{0.0095} \][/tex]
[tex]\[ K_{eq} \approx 0.042 \][/tex]

Therefore, the equilibrium constant [tex]\( K_{eq} \)[/tex] for the given reaction at 500 K is approximately [tex]\( 0.042 \)[/tex].

The correct answer is:
[tex]\[ 0.042 \][/tex]

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