Answer:
a. [tex]r_{N_2O_5}=-0.075M/min[/tex]
b. [tex]r_{N_2O_4}=0.075M/min[/tex]
Explanation:
Hello.
In this case, according to the balanced chemical reaction, we can write the law of rate proportions:
[tex]\frac{r_{N_2O_5}}{-2} =\frac{r_{N_2O_4}}{2} =\frac{r_{O_2}}{1}[/tex]
Thus, we proceed as follows:
a. Since the rate of oxygen production is 0.15 M/min, we can make the following setup:
[tex]\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[/tex]
b. Since the rate of oxygen production is 0.15 M/min, we can make the following setup:
[tex]\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[/tex]
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