Consider the following reaction:

$2N_{2}O(g)2N_2(g)+O_2(g)$

Part A

Express the rate of the reaction in terms of the change in concentration of each of the reactants and products.

Rate $=\frac{1}{2}\frac{[N_{2}O]}{t}=-\frac{1}{2}\frac{[N_2]}{t}=\frac{[O_2]}{t}$

Rate $=\frac{[N_{2}O]}{t}=\frac{1}{2}\frac{[N_2]}{t}=-\frac{1}{2}\frac{[O_2]}{t}$

Rate $=-\frac{[N_{2}O]}{t}=-\frac{1}{2}\frac{[N_2]}{t}=\frac{1}{2}\frac{[O_2]}{t}$

Rate $=-\frac{1}{2}\frac{[N_{2}O]}{t}=\frac{1}{2}\frac{[N_2]}{t}=\frac{[O_2]}{t}$

Previous Answers

Correct

Part B

In the first $16\mathrm{.}0s$ of the reaction, $4\mathrm{.}4{10}^{-2}$mol of $O_2$ is produced in a reaction vessel with a volume of $0\mathrm{.}300L.$ What is the average rate of the reaction over this time interval?

Express your answer in moles per litre per second to two significant figures.

Rate $=$

$M*s^{-1}$