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.

Answer 2

The average rate of the reaction will be "$2.57\times 10^{-3} \ M/s$".

According to the question,

• Volume = 0.440 L
• Time = 15.0 s
• Mol of O₂ = 1.7×10⁻²

The reaction will be:

• $2 N_2 O (g) \rightarrow 2 N_2 (g) +O_2 (g)$

Now,

The final concentration of O₂ will be:

→ $[O_2] = \frac{0.017}{0.440}$

          $= 0.0386 \ M$

hence,

The rate of reaction will be:

= $\frac{0.0386-0}{15.0-0}$

= $2.57\times 10^{-3} \ M/s$

Thus the above approach is right.

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