Chapter 14

Q. 14.4.1

The decomposition of nitrous oxide at 565 ºC
N_{2}O(g) → N_{2}(g) + ½ O_{2}(g)
is second order in N_{2}O with a rate constant of 1.10 × 10^{-3} M^{-1}s^{-1}. If an experiment is performed where the initial concentration of N_{2}O is 0.108 M, what is the N_{2}O concentration after 1250 seconds?


Verified Solution

You are asked to calculate the concentration of a reactant after a given period of time.
You are given the balanced equation for the reaction, the order of the reaction with respect to the reactant, the rate constant for the reaction, the initial concentration of the reactant, and the period of time.
Because the reaction is second order, use the second-order integrated rate law.

\frac{1}{\left[N_{2}O\right]_{t} }= \frac{1}{\left[N_{2}O\right]_{0} }+kt            t = 1250 seconds

\left[N_{2}O\right]_{0} = 0.108 M           k = 1.10 × 10^{−3}M^{−1}s^{−1}

\frac{1}{\left[N_{2}O\right]_{t} } = \frac{1}{0.108 \text{ M}}+(1.10 \times 10^{-3} M^{-1}s^{-1})(1250 s)

\frac{1}{\left[N_{2}O\right]_{t} }=10.6 \text{ M}^{-1}

\left[N_{2}O\right]_{t} = 0.0940 M