Question

For the reaction 2NOBr→2NO2+Br2, the rate law is rate =k[NOBr]^2. If the rate of a reaction is 6.5×10−6molL−1s−1, when the concentration of NOBr is 2×10−3molL−1.What would be the rate constant of the reaction?

Answer 1

k = 1.625 mol⁻¹ L s⁻¹

Step-by-step explanation:

What is given?

rate = 6.5 x 10⁻⁶ mol L⁻¹s⁻¹,

[NOBr] = 2 x 10⁻³ mol L⁻¹.

Step-by-step solution:

We want to find the rate constant, k of the reaction based on the rate law:

`rate=k\cdot\lbrack NOBr]^2,`

so we just have to solve for 'k' and replace the given values:

`\begin{gathered} k=(rate)/(\lbrack NOBr]^2), \\ \\ k=\frac{6.5\cdot10^{-6\text{ }}mol\text{ L}^(-1)s^(-1)}{(2\cdot10^(-3)\text{ mol L}^(-1))^2}, \\ \\ k=1.625\text{ mol}^(-1)L^s^(-1) \end{gathered}`

The answer would be k = 1.625 mol⁻¹ L s⁻¹