The formation of alumina, Al 2 O 3 , can be illustrated by the reaction below: 4 Al (s) + 3 O 2 (g) → 2 Al 2 O 3 (s) At 750 K it takes 267 seconds for the initial concentration of Al 2 O 3 to increase from 6.18 x 10 -5 M to 5.11 x 10 -4 M. What is the average rate of Al?

the formation of aluminum, which is a little too Oh, three can be illustrated by the reaction below where four moles of aluminum solid react with three moles of oxygen gas to produce two moles of aluminum solid. Here it says that 750 Kelvin, it takes 267 seconds for the initial concentration of alumina to increase from 6.18 times 10 to the negative five Mueller to 5.11 times 10 to the negative. For Mueller. What is the average rate of aluminum? All right, so here they're giving us the components to help us find the rate of aluminum. And what we need to find, though, is not the rate of aluminum at the end, but the rate of aluminum solid. So how do we approach a problem like this? Well, what we have to do first is we're actually going to find the rate of our reaction. And if you know the rate of the reaction, you can then use that to find the rate of aluminum. And then once you know the rate of alumina, we can do a rate to rate comparison to find the rate of aluminum solid. So the way we do it is remember, rate of a reaction is just changing concentration over change in time. So the way we've approached this is we do final minus initial in terms of concentration. So 5, 11 times, 10 to the negative for Mueller -6.18 times 10 to the -5 moller Divided by 267 seconds. Here we're not dealing with positive or negative signs or coefficients because this is the rate of the reaction and not in terms of any particular reactant or product here. When I do all that, I get 1.68 times 10 to the negative six Mohler per second. So I just found out the rate of my reaction. Now, if I wanted to, I could find the rate of alumina by dividing it by the coefficient of aluminum. But we don't want to stop there. We want to go, um, final destination rate of just aluminum solid. So to do that, I look at the coefficient in the balanced equation. So there's two alumina for every four aluminum. Okay, so this is malaria is per second of aluminum And four modalities per second of aluminum. All that counsels out and then we'll have our final answer, which comes out to 3.36 times 10 to the negative six molar it ease per second. So this would be my final answer for the average rate of aluminum solid.

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23. Chemistry of the Nonmetals2h 39m
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15. Chemical Kinetics

Stoichiometric Rate Calculations

General Chemistry

15. Chemical Kinetics

Stoichiometric Rate Calculations

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Multiple Choice

The formation of alumina, Al₂O₃, can be illustrated by the reaction below:
4 Al (s) + 3 O₂ (g) → 2 Al₂O₃ (s)
At 750 K it takes 267 seconds for the initial concentration of Al₂O₃ to increase from 6.18 x 10^-5 M to 5.11 x 10^-4 M. What is the average rate of Al?

8.41 x 10^-7 M/s

1.68 x 10^-6 M/s

3.36 x 10^-6 M/s

4.21 x 10^-7 M/s

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Verified step by step guidance

Identify the reaction: 4 Al (s) + 3 O2 (g) → 2 Al2O3 (s). This is a balanced chemical equation showing the formation of alumina (Al2O3) from aluminum (Al) and oxygen (O2).

Determine the change in concentration of Al2O3 over the given time period. The concentration changes from 6.18 x 10^-5 M to 5.11 x 10^-4 M over 267 seconds.

Calculate the change in concentration of Al2O3: Δ[Al2O3] = final concentration - initial concentration.

Use the stoichiometry of the reaction to relate the rate of formation of Al2O3 to the rate of consumption of Al. According to the balanced equation, 4 moles of Al produce 2 moles of Al2O3, so the rate of Al consumption is twice the rate of Al2O3 formation.

Calculate the average rate of Al consumption using the formula: rate = -Δ[Al]/Δt, where Δ[Al] is the change in concentration of Al and Δt is the time interval. Use the stoichiometric relationship to find Δ[Al] from Δ[Al2O3].