Question

Consider the following elementary reaction: CFC13(g)-CFC12(9)+Clg) Suppose we let k1 stand for the rate constant of this reaction, and k1 stand for the rate constant of the reverse reaction Write an expression that gives the equilibrium concentration of Cl in terms of k, k_1, and the equilibrium concentrations of CFCI3 and CFCI2 1. K-1 [ci]

Answer 1

Answer:

$[Cl]_{eq}=\frac{k_1[CFCl_3]_{eq}}{k_{-1}[CFCl_2]_{eq}}$

Explanation:

Hello,

In this case, considering and rewriting the given chemical reaction, one has:

$CFCl_{3}(g)\rightleftharpoons CFCl2(g)+Cl(g)$

Thus, the rate equation is written as:

$-r_{CFCl_3}=k_1C_{CFCl_3}-k_{-1}C_{CFCl_2}C_{Cl}$

In such a way, for the equilibrium condition, the rate remains constant, that is equal to zero, hence:

$0=k_1C_{CFCl_3}-k_{-1}C_{CFCl_2}C_{Cl}$

Solving for the concentration of Cl, one finally obtains:

$C_{Cl}=\frac{k_1C_{CFCl_3}}{k_{-1}C_{CFCl_2}}$

Or:

$[Cl]_{eq}=\frac{k_1[CFCl_3]_{eq}}{k_{-1}[CFCl_2]_{eq}}$

Best regards.