Can y\'all help me with this? (a) What is the rate law for this reaction? Rate =

Question

Can y'all help me with this?

(a) What is the rate law for this reaction?

Rate = k [NO2(g)] [O3(g)]Rate = k [NO2(g)]2 [O3(g)]     Rate = k [NO2(g)] [O3(g)]2Rate = k [NO2(g)]2 [O3(g)]2Rate = k [NO2(g)] [O3(g)]3Rate = k [NO2(g)]4 [O3(g)]

(b) What is the value of the rate constant?





(c) What is the reaction rate when the concentration of NO2(g) is 1.58 M and that of O3(g) is 2.73 M if the temperature is the same as that used to obtain the data shown above?

M/s

Experiment [NO2(g)] (M) [O3(g)] (M) Rate (M/s) 1 0.989 0.989 89300 2 0.989 1.98 1.79e+05 3 1.98 0.989 1.79e+05 4 1.98 1.98 3.58e+05

Explanation / Answer

Step 1

Experimental data is used to find the order of each reactant and then we will be able to predict the rate law for this equation.

Rate law:

Rate = k [NO2]m [O3]n

Here m and n are the orders of the species NO2 and O3 respectively.

In order to write rate law, we must need these values.

Prediction of orders based on given data.

Method:

Choose two experimental runs where one of the two reactants concentration is same / constant.

In given table, we use exp run 1 and 2 (concentration of NO2 is constant) and take a ratio of rate law to find order with respect to O3.

Run 1 …. Rate 1

Run 2 …. Rate 2

Let’s take a ratio of rate 2 / rate 1 , k is constant and it’s the same for all the runs so it gets cancelled.

Rate 2 / Rate 1 = ([NO2]m [O3]n)2 / ( [NO2]m [O3]n)1

[NO2] is constant so,

Rate 2 / Rate 1 = ( [O3]n)2 / ( [O3]n)1

Lets plug in the values

1.79E5 / 89300 = (1.98/0.989)n

2 = 2n

We know if n = 1 then above equation is satisfied so n= 1 and order with respect to O3 is 1.

Determination of n ( order with respect to NO2)

Lets look at exp runs 1 and 3.

Rate 3 /Rate 1 = ( [NO2]m )3 / ( [NO2]m )1

Lets plug in the value s

1.79 E5 / 89300 = ( 1.98 / 0.989)m

2 = 2m

m = 1 and the order with respect to NO2 is 1 as well.

So, the rate law becomes,

Rate = k [NO2] [O3]

b)

Calculation of rate constant.

Lets use exp run 1

Rate law

Rate 1 = k [NO2][O3]

89300 = k (0.989)(0.989)

k = 91297.5 per s

( c )

Given: [NO2] = 1.58 M

[O3] = 2.73 M

Let’s use given rate constant.

Rate = 91297.5 per s x (1.58) x (2.73)

Rate = 393802.6 M/s

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