Question

The reaction 2A → A₂ was experimentally determined to be second order with a rate constant, k, equal to 0.0265 M^–1min^–1. If the initial concentration of A was 3.00 M, what was the concentration of A (in M) after 180.0 min?

  M ?

Answer 1

Given, the balanced second order reaction,

2A A₂

Also given,

Rate constant(k) = 0.0265 M^-1 min^-1

The initial concentration of A (N_o) = 3.00 M

Time (t) = 180.0 min

N_t = ? The concentration of A(in M) after time "t".

We know, The second order rate law equation,

1 / [N_t] = kt + 1 / [N_o]

Substituting the known values in the formula,

1 / [N_t] = 0.0265 M^-1 min^-1 x 180.0 min. + 1 / 3.00 M

1 / [N_t] = 4.77 + 0.3333

1 / [N_t] = 5.1033

[N_t] = 1 / 5.1033

[N_t] = 0.196 M