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

[A] = 0.0868 M

Step-by-step 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

Answer 2

`[NH_2NO_2]=0.0868M`

Step-by-step explanation:

Hello,

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

`r=(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.