Question

Consider the following elementary reaction: CFCl3(g)→CFCl2(g)+Cl(g).Suppose we let k1 stand for the rate constant of this reaction, and k−1 stand for the rate constant of the reverse reaction.
Write an expression that gives the equilibrium concentration of CFCl3 in terms of k1,k−1, and the equilibrium concentrations of CFCl2 and Cl.

Answer 1

Answer:

       $[CFCl_3(g)]=\dfrac{k_{-1}}{k_1}\cdot[CFCl_2(g)]\cdot [Cl(g)]$

Explanation:

The equilibrium constant is equal to the ratio of the rate constant of the forward reaction to the rate constant of the reverse reaction:

          $K_c=\dfrac{k_{forward}}{k_{reverse}}$

Then, using k₁ and k₋₁ for the rate constants of the forward and the reverse reactions, respectively:

           $k_c=\dfrac{k_1}{k_{-1}}$

The equilibrium equation is:

          CFCl₃(g)   ⇄   CFCl₂(g)   +   Cl(g)

For which the equilibrium constant is:

          $k_c=\dfrac{[CFCl_2(g)]\cdot [Cl(g)]}{[CFCl_3(g)]}=\dfrac{k_1}{k_{-1}}$

Now you can write the equilibrium concentraion of CFCl₃(g) in terms of k₁, k₋₁, [CFCl₂(g)], and [Cl(g)]:

       $[CFCl_3(g)]=\dfrac{k_{-1}}{k_1}\cdot[CFCl_2(g)]\cdot [Cl(g)]$