Introduction to reaction rates
The rate of a chemical reaction is defined as the rate of change in concentration of a reactant or product divided by its coefficient from the balanced equation. A negative sign is used with rates of change of reactants and a positive sign with those of products, ensuring that the reaction rate is always a positive quantity. In most cases, concentration is measured in moles per liter and time in seconds, resulting in units of M/s for the reaction rate. Created by Jay.
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- I didnt understan the part when he says that the rate of the reaction is equal to the rate of O2 (time06:37). How do we know that?
- The rate of reaction is equal to the, R = rate of formation of any component of the reaction / change in time. Here in this reaction O2 is being formed, so rate of reaction would be the rate by which O2 is formed.(25 votes)
- why we chose O2 in determining the rate and compared the rates of N2O5 and NO2 with it?(26 votes)
- We could have chosen any of the compounds, but we chose O₂ for convenience.
O₂ has the smallest coefficient.
If we had chosen NO₂, the rate of decomposition of N₂O₅ would have been half as much, and the rate of formation of O₂ would have been one–fourth as much.
Choosing O₂ just avoids having to use fractions.(16 votes)
- So just to clarify, rate of reaction of reactant depletion/usage would be equal to the rate of product formation, is that right?(7 votes)
- Yes, when we are dealing with rate to rate conversion across a reaction, we can treat it like stoichiometry.
This means the difference in rate between different reactants/products is proportional to their coefficients.(3 votes)
- Why do we need to ensure that the rate of reaction for the 3 substances are equal?(8 votes)
- It is the formal definition that is used in chemistry so that you can know any one of the rates and calculate the same overall rate of reaction as long as you know the balanced equation.(5 votes)
- Firstly, should we take the rate of reaction only be the rate of disappearance/appearance of the product/reactant with stoichiometric coeff. as 1? In the video, can we take it as the rate of disappearance of *2*N2O5 or that of appearance of *4*N2O?
Secondly, doesn't that change the rate of reaction by factors of their stoichiometric coeff., which seems quite unreasonable? Does that mean that the rate of reaction varies with how _we_ represent the reaction?(5 votes)
- This is the answer I found on chem.libretexts.org:
Consider now a reaction in which the coefficients are different:
It is clear that [B] decreases three times as rapidly as [A], so in order to avoid ambiguity when expressing the rate in terms of different components, it is customary to divide each change in concentration by the appropriate coefficient:
- Why the rate of O2 produce considered as the rate of reaction ? Is the rate of reaction always express from ONE coefficient reactant / product(2 votes)
- Great question! The reaction rate is always defined as the change in the concentration (with an extra minus sign, if we are looking at reactants) divided by the change in time, with an extra term that is 1 divided by the stoichiometric coefficient. The reason why we correct for the coefficients is because we want to be able to calculate the rate from any of the reactants or products, but the actual rate you measure depends on the stoichiometric coefficient. For example, in this reaction every two moles of the starting material forms four moles of NO2, so the measured rate for making NO2 will always be twice as big as the rate of disappearance of the starting material if we don't also account for the stoichiometric coefficients.
Since the stoichiometric coefficient for oxygen is 1, the rate of reaction = (change in O2 concentration/change in time) x (1/stoichiometric coefficient for O2 in balanced reaction). Since the stoichiometric coefficient is 1, the rate just looks like rate = change in O2 concentration/change in time. But we could also write the rate in terms of, say, NO2, as follows: rate = (change in NO2 concentration/change in time) x (1/4), where the 4 comes from correcting for the stoichiometric coefficients.
Hope that helps!(10 votes)
- Am I always supposed to make the Rate of the reaction equal to the Rate of Appearance/Disappearance of the Compound with coefficient (1) ?
- No, in the example given, it just happens to be the case that the rate of reaction given to us is for the compound with mole coefficient 1. However, the method remains the same. Just figuring out the mole ratio between all the compounds is the way to go about questions like these.(2 votes)
- Why can I not just take the absolute value of the rate instead of adding a negative sign?(5 votes)
- I came across the extent of reaction in a reference book what does this mean??(4 votes)
- the extent of reaction is a quantity that measures the extent in which the reaction proceeds. It is usually denoted by the Greek letter ξ. The extent of a reaction has units of amount (moles). It was introduced by the Belgian scientist Théophile de Donder.(3 votes)
- Why is the concentration of A reducing with an increase in time?(3 votes)
- A is a reactant so it will be consumed and turned into B as the reaction progresses. Mathematically, Jay showed this at3:32.(4 votes)
- The rate of a chemical reaction is defined as the change in the concentration of a reactant or a product over the change in time, and concentration is in moles per liter, or molar, and time is in seconds. So we express the rate of a chemical reaction in molar per second. Molar per second sounds a lot like meters per second, and that, if you remember your physics is our unit for velocity. So, average velocity is equal to the change in x over the change in time, and so thinking about average velocity helps you understand the definition for rate of reaction in chemistry. If we look at this applied to a very, very simple reaction. So we have one reactant, A, turning into one product, B. Now, let's say at time is equal to 0 we're starting with an initial concentration of A of 1.00 M, and A hasn't turned into B yet. So at time is equal to 0, the concentration of B is 0.0. Let's say we wait two seconds. So, we wait two seconds, and then we measure the concentration of A. Obviously the concentration of A is going to go down because A is turning into B. Let's say the concentration of A turns out to be .98 M. So we lost .02 M for the concentration of A. So that turns into, since A turns into B after two seconds, the concentration of B is .02 M. Right, because A turned into B. So this is our concentration of B after two seconds. If I want to know the average rate of reaction here, we could plug into our definition for rate of reaction. Change in concentration, let's do a change in concentration of our product, over the change in time. So, the Rate is equal to the change in the concentration of our product, that's final concentration minus initial concentration. So the final concentration is 0.02. So, we write in here 0.02, and from that we subtract the initial concentration of our product, which is 0.0. So, 0.02 - 0.0, that's all over the change in time. That's the final time minus the initial time, so that's 2 - 0. So the rate of reaction, the average rate of reaction, would be equal to 0.02 divided by 2, which is 0.01 molar per second. So that's our average rate of reaction from time is equal to 0 to time is equal to 2 seconds. We could do the same thing for A, right, so we could, instead of defining our rate of reaction as the appearance of B, we could define our rate of reaction as the disappearance of A. So the rate would be equal to, right, the change in the concentration of A, that's the final concentration of A, which is 0.98 minus the initial concentration of A, and the initial concentration of A is 1.00. So 0.98 - 1.00, and this is all over the final time minus the initial time, so this is over 2 - 0. Now this would give us -0.02. - 0.02 here, over 2, and that would give us a negative rate of reaction, but in chemistry, the rate of reaction is defined as a positive quantity. So we need a negative sign. We need to put a negative sign in here because a negative sign gives us a positive value for the rate. So, now we get 0.02 divided by 2, which of course is 0.01 molar per second. So we get a positive value for the rate of reaction. All right, so we calculated the average rate of reaction using the disappearance of A and the formation of B, and we could make this a little bit more general. We could say that our rate is equal to, this would be the change in the concentration of A over the change in time, but we need to make sure to put in our negative sign. We put in our negative sign to give us a positive value for the rate. So the rate is equal to the negative change in the concentration of A over the change of time, and that's equal to, right, the change in the concentration of B over the change in time, and we don't need a negative sign because we already saw in the calculation, right, we get a positive value for the rate. So, here's two different ways to express the rate of our reaction. So here, I just wrote it in a little bit more general terms. Let's look at a more complicated reaction. Here, we have the balanced equation for the decomposition of dinitrogen pentoxide into nitrogen dioxide and oxygen. And let's say that oxygen forms at a rate of 9 x 10 to the -6 M/s. So what is the rate of formation of nitrogen dioxide? Well, if you look at the balanced equation, for every one mole of oxygen that forms four moles of nitrogen dioxide form. So we just need to multiply the rate of formation of oxygen by four, and so that gives us, that gives us 3.6 x 10 to the -5 Molar per second. So, NO2 forms at four times the rate of O2. What about dinitrogen pentoxide? So, N2O5. Look at your mole ratios. For every one mole of oxygen that forms we're losing two moles of dinitrogen pentoxide. So if we're starting with the rate of formation of oxygen, because our mole ratio is one to two here, we need to multiply this by 2, and since we're losing dinitrogen pentoxide, we put a negative sign here. So this gives us - 1.8 x 10 to the -5 molar per second. So, dinitrogen pentoxide disappears at twice the rate that oxygen appears. All right, let's think about the rate of our reaction. So the rate of our reaction is equal to, well, we could just say it's equal to the appearance of oxygen, right. We could say it's equal to 9.0 x 10 to the -6 molar per second, so we could write that down here. The rate is equal to the change in the concentration of oxygen over the change in time. All right, what about if we wanted to express this in terms of the formation of nitrogen dioxide. Well, the formation of nitrogen dioxide was 3.6 x 10 to the -5. All right, so that's 3.6 x 10 to the -5. So you need to think to yourself, what do I need to multiply this number by in order to get this number? Since this number is four times the number on the left, I need to multiply by one fourth. Right, so down here, down here if we're talking about the change in the concentration of nitrogen dioxide over the change in time, to get the rate to be the same, we'd have to multiply this by one fourth. All right, finally, let's think about, let's think about dinitrogen pentoxide. So, we said that that was disappearing at -1.8 x 10 to the -5. So once again, what do I need to multiply this number by in order to get 9.0 x 10 to the -6? Well, this number, right, in terms of magnitude was twice this number so I need to multiply it by one half. I need to get rid of the negative sign because rates of reaction are defined as a positive quantity. So I need a negative here. So that would give me, right, that gives me 9.0 x 10 to the -6. So for, I could express my rate, if I want to express my rate in terms of the disappearance of dinitrogen pentoxide, I'd write the change in N2, this would be the change in N2O5 over the change in time, and I need to put a negative one half here as well. All right, so now that we figured out how to express our rate, we can look at our balanced equation. So, over here we had a 2 for dinitrogen pentoxide, and notice where the 2 goes here for expressing our rate. For nitrogen dioxide, right, we had a 4 for our coefficient. So, the 4 goes in here, and for oxygen, for oxygen over here, let's use green, we had a 1. So I could've written 1 over 1, just to show you the pattern of how to express your rate.