Question

Given the following reaction: 2N2O5=2N2O4+O2, if the rate of oxygen production is 0.15M/min, determine:

Answer 1

Answer:

a. $r_{N_2O_5}=-0.075M/min$

b. $r_{N_2O_4}=0.075M/min$

Explanation:

Hello.

In this case, according to the balanced chemical reaction, we can write the law of rate proportions:

$\frac{r_{N_2O_5}}{-2} =\frac{r_{N_2O_4}}{2} =\frac{r_{O_2}}{1}$

Thus, we proceed as follows:

a. Since the rate of oxygen production is 0.15 M/min, we can make the following setup:

$\frac{r_{N_2O_5}}{-2} =\frac{r_{O_2}}{1}\\\\r_{N_2O_5}=\frac{r_{O_2}}{-2} =\frac{0.15M/min}{-2}\\\\ r_{N_2O_5}=-0.075M/min$

b. Since the rate of oxygen production is 0.15 M/min, we can make the following setup:

$\frac{r_{N_2O_4}}{2} =\frac{r_{O_2}}{1}\\\\r_{N_2O_4}=\frac{r_{O_2}}{2} =\frac{0.15M/min}{2}\\\\ r_{N_2O_4}=0.075M/min$

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