# Rate of Reaction

## 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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