Question

The decomposition of nitrogen dioxide represents a second order process.
2 nitrogen dioxide → dioxygen + 2 nitrogen monoxide
Answer the following questions about the decomposition of nitrogen dioxide. Report all answers to three significant figures.

1. The initial concentration of nitrogen dioxide is 4.317×10^-2 M, however after 2.10 h the concentration decreases to 1.252×10^-2 M. Calculate the rate constant (in /M hr) for this process.

2. If the initial concentration of nitrogen dioxide is 4.317×10^-2 M, calculate the concentration (in M) of nitrogen dioxide after 1.18 hrs.

Answer 1

Answer – We are given, reaction

2 NO­₂ ---> O₂ + 2NO

This is the second order reaction –

1)Initial concentration of [NO₂]o = 0.04317 M , final concentration

[NO₂]t = 0.01252 M , time, t = 2.10 hr

We know,

Second order equation

1/[NO₂] = 1/[NO₂]o + kt

1/ 0.01252M = 1/0.04317 M + k * 2.10 hrs

79.87 M^-1 = 23.16 M^-1 + k* 2.10 hrs

79.87 M^-1 - 23.16 M^-1 = k* 2.10 hrs

56.71 M^-1 = k* 2.10 hrs

So, k = 56.71 M^-1 / 2.10 hrs

          = 27.0 M^-1.hr^-1

the rate constant (in /M hr) for this process 27.0 / M.hrs

2) We are given, [NO₂]o = 0.04317 M , time = 1.18 hrs , k = 27.0 M^-1.s^-1

We know formula

1/[NO₂] = 1/[NO₂]o + kt

1/ [NO₂] = 1/0.04317 M + 27.0 M^-1.s^-1 * 1.18 hrs

1/ [NO₂] = 55.029

So, [NO₂] = 0.0182 M

the concentration (in M) of nitrogen dioxide after 1.18 hrs is 0.0182 M