Rate of Reaction: Doubling X's Concentration Effect

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The discussion focuses on determining the effect of doubling the concentration of substance X on reaction rate using two mixtures with varying concentrations. The first mixture has a higher concentration of X and a shorter reaction time, suggesting a faster reaction rate, while the second mixture has a lower concentration and longer reaction time. Three methods for calculating reaction rates are proposed, with the first method indicating that doubling X's concentration halves the reaction time, thus doubling the rate. The discussion concludes that the first method is the most acceptable, as it aligns with the definition of reaction rate and clarifies the ambiguity in the term "reaction time." Overall, the reaction rate increases by a factor of four when comparing the two mixtures.
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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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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.
 
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