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 NO_{2} ---> O_{2} + 2NO

This is the second order reaction –

**1)**Initial concentration of [NO_{2}]o =
0.04317 M , final concentration

[NO_{2}]t = 0.01252 M , time, t = 2.10 hr

We know,

Second order equation

1/[NO_{2}] = 1/[NO_{2}]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, [NO_{2}]o = 0.04317 M
, time = 1.18 hrs , k = 27.0 M^{-1}.s^{-1}

We know formula

1/[NO_{2}] = 1/[NO_{2}]o + kt

1/ [NO_{2}] = 1/0.04317 M + 27.0
M^{-1}.s^{-1} * 1.18 hrs

1/ [NO_{2}] = 55.029

So, [NO_{2}] = 0.0182 M

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

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