# Chemistry

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!

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1. 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

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posted by Autumn
2. 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.

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posted by DrBob222
3. I got: 2nd order with respect to NO
Thank you for the explanation!

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posted by Autumn
4. So, are you saying that the answer is Rate=k[O2][NO]^2???????

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5. 2nd order with NO is right.
Yes, x is 2, y is 1 so
rate = k(NO)^2(O2)

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posted by DrBob222
6. 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)

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]

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posted by Doc48

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