Question

In the first 15.0 s of the reaction, 1.7×10−2 mol of O2 is produced in a reaction vessel with a volume of 0.440 L . What is the average rate of the reaction over this time interval?

Answer 1

Answer:

$Rate=2.57x10^{-3}\frac{M}{s}$

Explanation:

Hello,

In this case, for the reaction:

$2N_2O(g) \rightarrow 2N_2(g)+O_2(g)$

We can easily compute the average rate by firstly computing the final concentration of oxygen:

$[O_2]=\frac{0.017mol}{0.440L}=0.0386M$

Then, we compute it by using the given interval of time: from 0 seconds to 15.0 seconds and concentration: from 0 M to 0.0386M as oxygen is being formed:

$Rate=\frac{0.0386M-0M}{15.0s-0s}\\ \\Rate=2.57x10^{-3}\frac{M}{s}$

Regards.