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

`[CFCl_3(g)]=(k_(-1))/(k_1)\cdot[CFCl_2(g)]\cdot [Cl(g)]`

Step-by-step 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=(k_(forward))/(k_(reverse))`

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

`k_c=(k_1)/(k_(-1))`

The equilibrium equation is:

CFCl₃(g) ⇄ CFCl₂(g) + Cl(g)

For which the equilibrium constant is:

`k_c=([CFCl_2(g)]\cdot [Cl(g)])/([CFCl_3(g)])=(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)]=(k_(-1))/(k_1)\cdot[CFCl_2(g)]\cdot [Cl(g)]`