Answer :
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]
[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]
