Answer:- Rate in change of carbon dioxide is \frac{7.2*10^-^2m}{s}[/tex] .
Solution:- For the given reaction the rate could be written as:
rate = [tex]-d\frac{[C_2H_4]}{dt}=-\frac{1}{3}d\frac{O_2}{dt}=\frac{1}{2}d\frac{CO_2}{dt}=\frac{1}{2}d\frac{H_2O}{dt}[/tex]
Here the negative sign indicates a decrease in concentration of the reactants and the coefficients are inverted. For example the coefficient of [tex]O_2[/tex] is 2, so it is written as [tex]\frac{1}{2}[/tex] .
From above expression, [tex]-d\frac{[C_2H_4]}{dt}=\frac{1}{2}d\frac{CO_2}{dt}[/tex]
Let's plug in the value for the rate of decrease in concentration of ethylene.
[tex]\frac{3.6*10^-^2m}{s}=\frac{1}{2}d\frac{CO_2}{dt}[/tex]
[tex]d\frac{CO_2}{dt}=\frac{2*3.6*10^-^2m}{s}[/tex]
[tex]d\frac{CO_2}{dt}=\frac{7.2*10^-^2m}{s}[/tex]
So, the rate in change(increase) of carbon dioxide is [tex]\frac{7.2*10^-^2m}{s}[/tex] .
