Question

The reaction of NOBr(g) to form NO(g) and Br2(g) is second order: 2NOBr(g) → 2NO(g) + Br2(g). The rate constant is 0.556 L mol^(-1) s^(-1) at some temperature. If the initial concentration of NOBr in the container is 0.28 M, how long will it take for the concentration to decrease to 0.049 M? t = _____s.

Name: the reaction of nobrg to form nog and br2g is second order 2nobrg 2nog br2g the rate constant is 0556 l mol 1 s 1 at some temperature if the initial concentration of nobr in the container is 31797
Uploaded: 2022-04-12T18:34:48-08:00
Duration: 59 s
Channel: Ronald Prasad
Description: the reaction of nobrg to form nog and br2g is second order 2nobrg 2nog br2g the rate constant is 0556 l mol 1 s 1 at some temperature if the initial concentration of nobr in the container is 31797

          The reaction of NOBr(g) to form NO(g) and Br2(g) is second order: 2NOBr(g) → 2NO(g) + Br2(g). The rate constant is 0.556 L mol^(-1) s^(-1) at some temperature. If the initial concentration of NOBr in the container is 0.28 M, how long will it take for the concentration to decrease to 0.049 M? t = _____s.

Added by Nuria G.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Ronald Prasad

04/12/2022

Step 1

Step 1: Use the integrated rate law for a second-order reaction: \(\frac{1}{[NOBr]_t} = \frac{1}{[NOBr]_0} + kt\) Show more…

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The reaction of NOBr(g) to form NO(g) and Br2(g) is second order: 2NOBr(g) → 2NO(g) + Br2(g). The rate constant is 0.556 L mol^(-1) s^(-1) at some temperature. If the initial concentration of NOBr in the container is 0.28 M, how long will it take for the concentration to decrease to 0.049 M? t = _____s.