Question

Consider the following elementary reaction: NO2(g) + F2(g) --> NO2F(g) + F (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 F2 in terms of k1, k -1, and the equilibrium concentrations of NO2, NO2F, and F.

Answer 1

To express the equilibrium concentration of F2, the equation [F2] = (k -1[NO2F][F]) / (k1[NO2]) is used, based on the principle that the forward reaction rate equals the reverse reaction rate at equilibrium.

Step-by-step explanation:

To write an expression for the equilibrium concentration of F2 in terms of the rate constants k1 and k -1, and the equilibrium concentrations of NO2, NO2F, and F, one can use the principle that at equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction. For the given reaction NO2(g) + F2(g) → NO2F(g) + F(g), the forward rate is given by k1[NO2][F2] and the reverse rate by k -1[NO2F][F]. Setting these equal to each other gives us:

k1[NO2][F2] = k -1[NO2F][F]

Solving for the equilibrium concentration of F2, we have:

[F2] = (k -1[NO2F][F]) / (k1[NO2])

This expression relates the equilibrium concentration of F2 to the rate constants and the equilibrium concentrations of NO2, NO2F, and F.