Reaction Rate example - Notes PDF

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Reaction rate example worked through Usually, you see a table like this: Experiment

Concentration of [A]

Concentration of [B]

Initial rate (M/s)

1

0.02

0.01

2

2

0.02

0.02

4

3

0.04

0.01

8

Step 1: Look at two experiments where Concentration of A is constant while concentration of B is changing ● In this case, experiment 1 and 2 has Concentration of A as constant (0.02) while Concentration of B changes (0.01 to 0.02) Step 1b: Take note of how the concentration changed by how many factors/times more proportionally (doubled, tripled, quadrupled...etc) Step 2: Take those experiments and look at the change of the Initial Rate in terms of how many factors more it went up/down proportionally (doubled, tripled, quadrupled… etc) ● So we are using experiment 1 and 2 ○ We see that it goes from 2 to 4 so doubled ● We are using experiment 1 and 3 ○ We see that it goes from 2 to 8 Step 3: Compare how the concentration changed with how the initial rate changed (For example: concentration doubled vs rate doubled, concentration doubled vs rate quadrupled) ● From experiment 1 and 2, we see that the rate doubled (from 2 to 4). We also see that the concentration of B also doubled ● From experiment 1 and 3, we see that the concentration doubled from 0.02 to 0.04 and the initial rate changed from 2 to 8 so it quadrupled Step 3b: Compare how the concentration change is related to the initial rate proportion change. In order to do so mathematically, solve this: [concentration change proportion]x = rate change ● For A: we see that concentration change is doubled and rate change is doubled ○ So (2)x = 2, x = 1. So the exponent is 1 so first order with respect to A ● For B: we see that concentration change is doubled and rate change is quadrupled ○ So (2)x = 4, x = 2. So the exponent is 2 so second order with respect to B So overall rate law is rate = k[A][B]2. To find k: plug in the values for A and B and rate from the table

2 = k[.02][.01], K = 10,000. You can use any experiment’s numbers but I just chose experiment 1

The process is the same for if there are three elements: Experiment

[A]

[B]

[C]

Initial Reaction

1

0.2

0.1

0.05

1x10-3

2

0.4

0.1

0.05

2x10-3

3

0.2

0.2

0.05

4x10-3

4

0.2

0.1

0.1

1x10-3

For A: Step 1: Look for two of the concentrations to stay the same while one concentration changes. We see that for experiment 1 and 2, Concentration of A is changing while B and C are not changing Concentration of A is being doubled Step 2: Look at reaction rate change for those experiments For experiment 1 and 2, the rate doubles (from 1 to 2) Step 3: One doubled and the other doubles (so we know the order is 1) but we can also use the “equation” [Concentration change proportion]x = Initial Reaction change 2x = 2. X = 1 For B: Step 1: reactions 1 and 3 have concentration of B change while A and C stay constant Concentration B is doubled Step 2: Initial reaction rate is quadrupled

Step 3: concentration is doubled while rate is quadrupled so order 2 2x = 4, x = 2 For C: Step 1: reactions 1, 4 concentration of C is changing while A and B are constant C is being doubled Step 2: the initial reaction rate stays the same so no change Step 3: concentration is doubled while the rate has no change/stays the same , so order of 0 2x = 1, x = 0 so zero order

So overall rate law is rate = k[A]1[B]2 (no C since it was zero order) To find k: 1x10-3 = k(0.2)(0.1)2 so k = .5

Reaction rate notes: (same as from worked c/p notes) ●

Reaction rates: ○ Reaction order is the relationship between the concentration of species and the rate of a reaction ○ Rate law/rate constant must be determined experimentally (is not associated with the stoichiometric coefficients of balanced equation) ○ The sum of the exponents is the overall reaction order ○ Rate = k[A]x[B]y for equation aA + Bb Cc + Dd ■ K has units ○ Zero-order: increasing the concentration of the reacting species will not speed up the rate of the reaction. Typically found when a material that is required for the reaction to proceed, such as a surface or a catalyst, is saturated by the reactants ■ Graph: [A] vs t, slope is -k is linear and downwards ■ K units: M/s ○ First order: depends on the concentration of only one reactant; the other reactant(s) do not affect the rate ■ Graph ln[A] vs t, slope is -k is linear and downwards ■ K units: s-1 (1/sec) ○ Second order: either A or B is second order or first order in both A and B. ■ Graph: 1/[A] vs time, slope is k ■ K units: M-1s-1 (1/MS)

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Integrated rate law (how it applies to graphs) ○ The math behind this is calculus based so you don’t need to know it, but this may be helpful when looking at the rate graphs ○ Each rate law order can be graphed and put into straight-line form (where you get this k is the slope) ○ Keep in mind the general linear graph has a form of y = mx + b ○ Zero order: [A]t = -kt + [A]0 ○ 1st order: ln [A]t = -kt + ln [A]0 ○ 2nd order: 1/[A]t = kt + 1/[A]0 ○ So you can see where k takes on m, and t is x...