Answer:
a) [tex]k=\frac{9.3x10^{-5}}{M^2s}[/tex]
b) [tex]r=4.37x10^{-7}\frac{M}{s}[/tex]
Explanation:
Hello,
In this case, given that the reaction is second order in NO and first order in O2, the rate law is written as follows:
[tex]r=k[NO]^2[O_2][/tex]
In such a way, for a rate law of 0.00022022 M/s and the concentrations of NO and O2 0.8M and 3.7M respectively, the rate constant is:
[tex]k=\frac{r}{[NO]^2[O_2]}\\\\k=\frac{0.00022022M/s}{(0.8M)^2(3.7M)} \\\\k=\frac{9.3x10^{-5}}{M^2s}[/tex]
Thus, for the new concentrations of NO and O2 0.1M and 0.47M respectively, the rate is:
[tex]r=\frac{9.3x10^{-5}}{M^2s} (0.1M)^2(0.47M)\\\\r=4.37x10^{-7}\frac{M}{s}[/tex]
Regards.
