Given the following information $$Cl(g) \rightleftharpoons 2Cl(g); \quad K = 1.13 \times 10^{-6} M \text{ at } 1100^{\circ}C$$ what would be the numerical value of the equilibrium constant for the reaction $$2Cl(g) \rightleftharpoons Cl(g)$$ at the same temperature? A - VCE - SSCE Chemistry - Question 11 - 2006 - Paper 1

To find the equilibrium constant for the reaction $2Cl(g) \rightleftharpoons Cl(g)$, we can use the relationship between the equilibrium constants of the forward and reverse reactions.

The given equilibrium constant for the reaction $Cl(g) \rightleftharpoons 2Cl(g)$ is: $K = 1.13 \times 10^{-6}$.

For the reverse reaction, the equilibrium constant is the reciprocal of the forward reaction: $K' = \frac{1}{K} = \frac{1}{1.13 \times 10^{-6}}.$

Now, calculating that,

$K' = 8.85 \times 10^{5}.$

Thus, the numerical value of the equilibrium constant for the reaction $2Cl(g) \rightleftharpoons Cl(g)$ at 1100°C is $8.85 \times 10^{5}$, which corresponds to option A.