2A-->B+C given that the initial concentration of A is .022M and that at 69 s, the concentration of B increases to .0095M, what is the rate constant?

i know I need to find the new concentration of A at 69s, but I am having trouble with findthing this. I thought it might be twice the amount of [ B ] but i was incorrect...

To determine the rate constant for the reaction 2A → B + C, you first need to find the change in concentration of A over time.

Given:
Initial concentration of A (A₀) = 0.022 M
Concentration of B at 69 s (B) = 0.0095 M

Since the stoichiometry of the reaction is 2A --> B + C, for every 2 moles of A that react, 1 mole of B is produced. This means that the change in the concentration of A is half the change in concentration of B.

Change in A = (B - A₀)/2

Substituting the given values:
Change in A = (0.0095 M - 0.022 M)/2 = -0.00625 M

Now that you have the change in concentration of A, you can use the integrated rate law for a second-order reaction to find the rate constant.

Integrated rate law for a second-order reaction:
1/[A]t - 1/[A₀] = kt

Rearranging the equation to solve for k:
k = (1/[A]t - 1/[A₀]) / t

Substituting the values:
k = (1/[-0.00625 M] - 1/0.022 M) / 69 s

Simplifying the equation and performing the calculations will give you the rate constant for the reaction.