Question

Slove: \[\ce{2NOBr -> 2NO_{2(g)} + Br_{2(g)}}\]

For the above reaction, the rate law is rate = k[NOBr]². If the rate of reaction is 6.5 × 10⁻⁶ mol L⁻¹ s⁻¹ at 2 × 10⁻³ mol L⁻¹ concentration of NOBr, calculate the rate constant k for the reaction.

Answer 1

For a reaction,

\[\ce{2NOBr -> 2NO_{2(g)} + Br_{2(g)}}\]

Given: Rate = k[NOBr]²

∴ k = `"Rate"/([NOBr]^2)`

= `(6.5 xx 10^6)/([2 xx 10^-3]^2)`

= `6.5/4`

∴ k = 1.625 mol⁻¹ L⁻¹ s⁻¹