Question

the equilibrium constant, ( k_{c} ), for the following reaction is 5.49 at 389 k.

( 2ch_{2}cl_{2}(g)
ightleftharpoons ch_{4}(g)+ccl_{4}(g) )

calculate ( k_{c} ) at this temperature for:

( ch_{4}(g)+ccl_{4}(g)
ightleftharpoons 2ch_{2}cl_{2}(g) )

( k_{c}= )

Explanation:

Step1: Write the expression for \(K_{c}\) of the first reaction

For the reaction \(2CH_{2}Cl_{2}(g)
ightleftharpoons CH_{4}(g)+CCl_{4}(g)\), \(K_{c1}=\frac{[CH_{4}][CCl_{4}]}{[CH_{2}Cl_{2}]^{2}} = 5.49\)

Step2: Write the expression for \(K_{c}\) of the second reaction

For the reaction \(CH_{4}(g)+CCl_{4}(g)
ightleftharpoons 2CH_{2}Cl_{2}(g)\), \(K_{c2}=\frac{[CH_{2}Cl_{2}]^{2}}{[CH_{4}][CCl_{4}]}\)

Step3: Relate \(K_{c2}\) to \(K_{c1}\)

Since \(K_{c2}=\frac{1}{K_{c1}}\), substituting \(K_{c1} = 5.49\) gives \(K_{c2}=\frac{1}{5.49}\approx0.182\)

Answer:

\(0.182\)