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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?

Sagot :

Answer:

k = 1.625 mol⁻¹ L s⁻¹

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:

[tex]rate=k\cdot\lbrack NOBr]^2,[/tex]

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

[tex]\begin{gathered} k=\frac{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}[/tex]

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

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