Question

The decomposition of nitramide in aqueous solution at 25 °C NH2NO2(aq)N2O(g) + H2O(l) is first order in NH2NO2 with a rate constant of 4.70×10-5 s-1. If an experiment is performed in which the initial concentration of NH2NO2 is 0.384 M, what is the concentration of NH2NO2 after 31642.0 s have passed? M

Answer 1

Answer:

$[NH_2NO_2]=0.0868M$

Explanation:

Hello,

In this case, for the given chemical reaction, the first-order rate law is:

$r=\frac{d[NH_2NO_2]}{dt} =-k[NH_2NO_2]$

Which integrated is:

$[NH_2NO_2]=[NH_2NO_2]_0exp(-kt)$

Thus, the concentration after 31642.0 s for a 0.384-M solution is:

$[NH_2NO_2]=0.384M*exp(-4.70x10^{-5}s^{-1}*31642.0s)$

$[NH_2NO_2]=0.0868M$

Best regards.

Answer 2

Answer:

[A] = 0.0868 M

Explanation:

Rate constant = 4.70×10-5 s-1

First order reaction

Initial concentration, [A]o = 0.384 M

Final concentration, [A] = ?

Time, t = 31642.0 s

All these variables are related by the following equation;

[A] = [A]o e^(-kt)

[A] = 0.384  e^(-4.70×10-5 x  31642.0)

[A] = 0.384 e^(-1.4872)

[A] = 0.384 * 0.2260

[A] = 0.0868 M