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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?
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Solution
Given: Rate = k[NOBr]2 = 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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