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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 × 10–6 mol L–1 s–1, when the concentration of NOBr is 2 × 10–3 mol L–1 - Chemistry

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Question

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

Numerical
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Solution

Given: Rate = k[NOBr]= 6.5 × 10–6 mol L–1 s–1

[NOBr] is 2 × 10–3 mol L–1

To find: Rate constant (k)

Calculation: rate = k[NOBr]2

k = `"rate"/["NOBr"]^2`

`= (6.5 xx 10^-6  "mol L"^-1"s"^-1)/((2 xx 10^-3 "mol L"^-1)^2)`

= 1.625 mol-1 L s–1 

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Chapter 6: Chemical Kinetics - Short answer questions (Type- I)

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