Question

The reaction          3 BrO- (aq) --> BrO3- (aq) + 2 Br - (aq) in basic solution is second order in BrO-(aq) with a rate constant equal to 0.056 M-1s-1 at 80 degrees Celcius. If [BrO- ]0 = 0.212 M, then what will [BrO- ]  be 1.00 min later?

Answer 1

Answer:

0.124 M

Explanation:

The reaction obeys second-order kinetics:

$r = k[BrO^-]^2$

According to the integrated second-order rate law, we may rewrite the rate law in terms of:

$\dfrac{1}{[BrO^-]_t} = kt + \dfrac{1}{[BrO^-]_o}$

Here:

$k$ is a rate constant,

$[BrO^-]_t$ is the molarity of the reactant at time t,

$[BrO^-]_o$ is the initial molarity of the reactant.

Converting the time into seconds (since the rate constant has seconds in its units), we obtain:

$t = 1.00 min = 60.0 s$

Rearranging the integrated equation for the amount at time t:

$[BrO^-]_t = \dfrac{1}{kt + \dfrac{1}{[BrO^-]_o}}$

We may now substitute the data:

$[BrO^-]_t = \dfrac{1}{0.056 M^{-1}s^{-1}\cdot 60.0 s + \dfrac{1}{0.212 M}} = 0.124 M$