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Consider the reaction below:

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

At 500 K, the reaction is at equilibrium with the following concentrations:

[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]

What is the equilibrium constant for the given reaction?

A. 0.042
B. 0.42
C. 2.4
D. 24


Sagot :

To find the equilibrium constant ([tex]\( K_c \)[/tex]) for the given reaction at equilibrium, we use the expression for the equilibrium constant involving the concentrations of the reactants and products. The reaction is:

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

The equilibrium constant expression for this reaction, [tex]\( K_c \)[/tex], is given by:

[tex]\[ K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \][/tex]

Given the equilibrium concentrations:
[tex]\[ [\text{PCl}_5] = 0.0095 \, M \][/tex]
[tex]\[ [\text{PCl}_3] = 0.020 \, M \][/tex]
[tex]\[ [\text{Cl}_2] = 0.020 \, M \][/tex]

We substitute these values into the equilibrium constant expression:

[tex]\[ K_c = \frac{(0.020)(0.020)}{0.0095} \][/tex]

First, calculate the product of the concentrations of [tex]\(\text{PCl}_3\)[/tex] and [tex]\(\text{Cl}_2\)[/tex]:

[tex]\[ 0.020 \times 0.020 = 0.0004 \][/tex]

Next, divide this product by the concentration of [tex]\(\text{PCl}_5\)[/tex]:

[tex]\[ \frac{0.0004}{0.0095} = 0.04210526315789474 \][/tex]

So, the equilibrium constant [tex]\( K_c \)[/tex] for the reaction at [tex]\( 500\, K \)[/tex] is approximately:

[tex]\[ K_c \approx 0.042 \][/tex]

Therefore, the correct answer is:

[tex]\[ 0.042 \][/tex]