The following table shows how the initial rate of this reaction depends on the concentrations of the two reactants

[NO] [O2] Initial rate
0.0050 0.0050 0.02
0.0050 0.0075 0.03
0.010 0.0075 0.12

Use the data to determine the order of reaction with respect to each of the reagents.

Sorry if the way I formatted the table looks really weird, thanks in advance!

rate 2 = k2(NO)^x(O2)^y is the equation for #2 and

rate 1 = k1(NO)^x(O2)^y is the equation for #1 trial.
I picked 2 and 1 because they have one reactant rate = to the same in another trial. Then divide first equation by second to get the final result of y = 1.

Put in the numbers.and remember that k1 = k2 so we cancel them.
0.03 (0.005)^x(0.0075)^y
------ = -----------------------------
0.02 (0.005)^x(0.005)^y

Note that (0.005)^x/0.005)^x cancels along with k1 and k2.
1.5 = (1.5)^y so y must be 1;'i.e., the rxn is 1st order with respect to O2.
Do the same kind of thing to find x.

You will want to use #3 and #2 so that the (O2)^y/(O2)^y part cancels. Post your work if you get stuck.

I got: 2nd order with respect to NO

Thank you for the explanation!

2nd order with NO is right.

Yes, x is 2, y is 1 so
rate = k(NO)^2(O2)

Orders of Rxn are 'rate trends' ... That is, look at the change in rate from initial concentration to 2x, 3x, etc. the initial concentration for each reactant. The general rule is hold all concentrations constant and vary one component at a time and note change in rates...

If one changes concentration and change in rate = zero => 0 order reaction

If one doubles concentration and change in rate doubles => 1st order reaction
If one triples concentration and change in rate triples => 1st order reaction
( it’s a 1 to 1 linear trend)

If one doubles concentration and change in rate increases 4x => 2nd order
If one triples concentration and change in rate increases 9x => 2nd order
( it’s an exponential trend)

For your problem
Test………..[NO]…………. [O2]…….….. Initial rate
…1………..0.0050………0.0050………….0.02
…2………..0.0050………0.0075………….0.03
…3………..0.0100………0.0075………….0.12

Rxns 1&2 => holding [NO] constant, Δ[O₂] = 1.5x => ΔRate = 1.5 => 1:1 => 1st order in [O₂]
Rxns 2&3 =>holding [O₂] constant, Δ[NO] = 2x => ΔRate = 4x => 1:4 => 2nd order in [NO]

Rate = k[NO]²[O₂]¹ => k = Rate/[NO]²[O₂]¹
Choose any row, k will be the same for all rows; i.e., constant.
Row 1 => k = 0.02/(0.0050)²(0.0050) = 1.6 x 10⁵ L²∙sˉ¹∙molˉ²
Row 2 => k = 0.03/(0.0050)²(0.0075) = 1.6 x 10⁵ L²∙sˉ¹∙molˉ²
Row 3 => k = 0.12/(0.0100)²(0.0075) = 1.6 x 10⁵ L²∙sˉ¹∙molˉ²

Numerical Rate Law => Rate = = 1.6 x 10⁵ L²∙sˉ¹∙molˉ²[NO]²[O₂]

In case the table is hard to read:

[NO] / mol dm^-3
Experiment 1: 0.0050
Experiment 2: 0.0050
Experiment 3: 0.010

[O2] / mol dm^-3
Experiment 1: 0.0050
Experiment 2: 0.0075
Experiment 3: 0.0075

Initial Rate / mol dm^-3 s^-1
Experiment 1: 0.02
Experiment 2: 0.03
Experiment 3: 0.12

So, are you saying that the answer is Rate=k[O2][NO]^2???????

To determine the order of reaction with respect to each of the reagents, we need to examine how the initial rate changes as the concentration of each reactant changes while keeping the other reactant's concentration constant.

Let's start by comparing the initial rates for the first two rows of the table, where the concentration of [NO] remains constant at 0.0050 M, and only the concentration of [O2] changes:

[NO] [O2] Initial rate
0.0050 0.0050 0.02
0.0050 0.0075 0.03

We can see that as the concentration of [O2] increases from 0.0050 M to 0.0075 M, the initial rate increases from 0.02 to 0.03. This suggests that there is a direct relationship between the concentration of [O2] and the initial rate.

To determine the order of reaction with respect to [O2], we can compare the effect of doubling the concentration of [O2]. Let's compare the first and third rows, where [O2] changes from 0.0050 M to 0.0075 M, while [NO] remains constant:

[NO] [O2] Initial rate
0.0050 0.0050 0.02
0.010 0.0075 0.12

Here, we can see that doubling the concentration of [O2] from 0.0050 M to 0.0075 M results in an increase in the initial rate from 0.02 to 0.12. This suggests that the change in initial rate is more significant than the change in [O2] concentration, indicating a second-order relationship between [O2] and the initial rate.

Now, let's analyze the effect of changing the concentration of [NO] while keeping [O2] constant. Comparing the first and third rows, where [NO] changes from 0.0050 M to 0.010 M:

[NO] [O2] Initial rate
0.0050 0.0050 0.02
0.010 0.0075 0.12

We can observe that doubling the concentration of [NO] from 0.0050 M to 0.010 M results in an increase in the initial rate from 0.02 to 0.12. This implies that the change in initial rate is proportional to the change in [NO] concentration, indicating a first-order relationship between [NO] and the initial rate.

Therefore, based on the data provided, the reaction is first order with respect to [NO] and second order with respect to [O2].