Question

The gas phase decomposition of nitrogen dioxide at 383 °C NO2(g) --> NO(g) + ½ O2(g) is second order in NO2 with a rate constant of 0.540 M-1 s-1. If the initial concentration of NO2 is 0.355 M, the concentration of NO2 will be ______ M after 15.9 seconds have passed.

Answer 1

Answer:

cccfffffff

Explanation:

Integrated rate law for second order kinetics is:

$\frac{1}{[A_t]} = \frac{1}{[A]_0}+kt$

Where, $[A_t]$ is the final initial concentration

$[A_0]$ is the initial concentration = 0.355 M

k is the rate constant = 0.540 (Ms)⁻¹

t is the time = 15.9 seconds

Applying the values in the above expression as:-

$\frac{1}{[A_t]} = \frac{1}{0.355}+0.540\times 15.9\ M=11.4\ M$

11.4 M of $NO_2$ will be left after 15.9 seconds have passed.