Consider the reaction NO2 + CO --> NO + CO2 at 225 C where the rate law is Rate

ID: 827322 • Letter: C

Question

Consider the reaction NO2 + CO --> NO + CO2 at 225 C where the rate law is Rate = k[NO2]^2

An experiment is carried out with the initial concentrations: [NO2]0 = [CO]0 = 0.500M

After 3.00 x 10^3 seconds, the molar concentration of NO2 is found to be 0.381 M

a. What are the concentrations of CO and NO, respectrively after 3.00 x 10^3 seconds?

b. Calculate the rate constant k for the reaction. Provide units.

c. How many hours does it take for [NO2] to decrease from 0.500 M to 0.250 M (1 hr = 3600 s)?

cocn. of NO2 is decreased from 0.500 to 0.381 in given time t .. so there was decrese of (5-0.381) = 0.119 M

so concentration of CO and NO would be 0.119 M after the given time t.

rate = k[NO2]^2 THAT IS

-d[NO2]/dt = k[NO2]^2

d[NO2]/[NO2]^2 = -kdt

integrating on both sides

-1/[NO2] = -kt + c

At t=0

-1/0.500 = c = -2

1/[NO2] = kt + 2

now at t = 3 * 10^3 sec [NO2] =0.381

1/0.381 = k( 3 * 10^3) + 2

k = 0.208 * 10^-3

Unit of k is s^(-1) * mol^(-1) * Litre

[NO2] =0.250M

1/0.250 = kt + 2

kt = 2

t = 2/k

=9615.38 Sec

=2.670 Hours

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