Rate of Reaction: Doubling X's Concentration Effect

In summary, the conversation discusses finding the effect of doubling the concentration of substance X on the reaction rate using two data points with varying concentrations of substance X and Y. Three possible methods for calculating the reaction rate are presented, with the first method being the most likely correct one. The importance of clearly defining terms and accurately citing definitions is emphasized.
  • #1
mrxtothaz
14
0

Homework Statement



I have been given a set of data and been asked to find the effect of doubling the concentration of substance X on the reaction rate.

I have chosen two data points to this end:

Mixture 1:
Substance X Concentration: 0.10 mol/L
Substance Y Concentration: 0.05 mol/L
Reaction Time: 20 s

Mixture 2:
Substance X Concentration: 0.05 mol/L
Substance Y Concentration: 0.05 mol/L
Reaction Time: 41 s

So I have chosen two data points where the concentration of substance X has doubled, while the concentration of Y is constant.

Homework Equations



The chemical equation states that for every 2 moles of substance X, one mole of substance Y reacts.

The Attempt at a Solution



I'm not certain how to calculate the reaction rate. Which of these is correct?

1. As the concentration of substance X doubles, the time for reaction is halved. Therefore, the reaction rate is twice as fast.

2. In Mixture 1, 0.1 mol/L of substance X reacts with 0.05 mol/L of substance Y. On this basis, would 0.15 mol/L divided by the time of 20s for the reaction to occur, represent the reaction rate?

Similarly, in Mixture 2, 0.05 mol/L of substance X would react with 0.025 mol/L of substance Y. Therefore, 0.075 mol/L divided by 41s would represent the second reaction rate. Dividing the reaction rate for Mixture 1 by Mixture 2 = 4.1.

By this method, the reaction rate increases by a factor of 4.

3. I recall that the definition for reaction rate is the change in concentration of one of the reactants. By this, I would ignore the concentration of substance Y and only use the amount of substance X that reacts. As such,

Mixture 1: 0.10/20 = 0.005
Mixture 2: 0.05/41 = 0.00122

Again, dividing Mixture 1 by Mixture 2 = 4.1. Therefore, the reaction rate increases by a factor of 4.

As you can see, methods 2 and 3 yield the same result. If one of these is correct, which would be more acceptable to use as a solution?
 
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  • #2
As it turns out, the correct answer is almost certainly #1. The term "reaction time" is not properly defined in the question, and leaves things ambiguous. You are to assume that it is the time taken for some fixed changed in concentration of one specific reactant or product.

Note: The definition you cite in #3 is correct. And that rules out the method you used in #2, which is based on an incorrect definition.

Note 2: Your effort at clearly explaining your thought precess is laudable. In addition, it is good practice to write down the question given to you in the exact same form that you received it. Often, a paraphrase of the question can leave out or obfuscate some relevant piece of information.
 
Last edited:

Related to Rate of Reaction: Doubling X's Concentration Effect

What is the rate of reaction?

The rate of reaction is a measure of how fast a chemical reaction takes place. It is usually expressed as the change in concentration of reactants or products per unit of time.

How does doubling X's concentration affect the rate of reaction?

Doubling X's concentration will result in an increase in the rate of reaction. This is because a higher concentration of reactants means that there are more particles colliding, leading to more successful collisions and therefore a faster reaction.

Is the effect of doubling X's concentration on the rate of reaction always the same?

No, the effect of doubling X's concentration on the rate of reaction can vary depending on the specific reaction and conditions. In some cases, doubling X's concentration may have a larger impact on the rate of reaction, while in others it may have a smaller effect.

Can doubling X's concentration have a negative effect on the rate of reaction?

In some cases, doubling X's concentration can have a negative effect on the rate of reaction. This can occur when the reaction is limited by another factor, such as temperature or the presence of a catalyst. In these cases, doubling X's concentration may not result in a significant increase in the rate of reaction.

What other factors can affect the rate of reaction besides the concentration of X?

The rate of reaction can also be affected by temperature, the presence of a catalyst, and the size and surface area of the reactants. Additionally, the nature of the reactants and the type of reaction can also impact the rate of reaction.

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