Question

The reaction 2NO(g) + O2(g) 2NO2(g) is second order in NO and first order in O2. When [NO] = 0.8 M and [O2] = 3.7 M, the observed rate of the reaction is 0.00022022 M/s. (a) What is the value of the rate constant? (d) What is the rate of reaction when [NO] = 0.1 M and [O2] = 0.47 M?

Answer 1

a)
`k=(9.3x10^(-5))/(M^2s)`

b)
`r=4.37x10^(-7)(M)/(s)`

Step-by-step 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:

`r=k[NO]^2[O_2]`

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:

`k=(r)/([NO]^2[O_2])\\\\k=(0.00022022M/s)/((0.8M)^2(3.7M)) \\\\k=(9.3x10^(-5))/(M^2s)`

Thus, for the new concentrations of NO and O2 0.1M and 0.47M respectively, the rate is:

`r=(9.3x10^(-5))/(M^2s) (0.1M)^2(0.47M)\\\\r=4.37x10^(-7)(M)/(s)`

Regards.