Video: Estimating the Activation Energy of a Reaction from Temperature and Reaction Rate

A reaction was performed repeatedly with the same reactant concentrations but different reaction temperatures 𝑇. The initial rate 𝑟 was measured for each reaction. Which of the following graphs could be used to estimate the activation energy of the reaction? [A] ln 𝑟 against ln 𝑇 [B] 𝑟 against 1/𝑇 [C] 1/𝑟 against 𝑇 [D] 1/𝑟 against ln 𝑇 [E] ln 𝑟 against 1/𝑇

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Video Transcript

A reaction was performed repeatedly with the same reactant concentrations but different reaction temperatures 𝑇. The initial rate 𝑟 was measured for each reaction. Which of the following graphs could be used to estimate the activation energy of the reaction? (A) Natural log 𝑟 against natural log 𝑇. (B) 𝑟 against one divided by 𝑇. (C) One divided by 𝑟 against 𝑇. (D) One divided by 𝑟 against natural log 𝑇. (E) Natural log 𝑟 against one over 𝑇.

In this question, the initial rate 𝑟 is being measured at different temperatures 𝑇. Notice that the same reactant concentrations are used throughout the experiments. So we do not need to concern ourselves with how concentration affects initial rate here. The initial rate of reaction is usually determined by measuring the change in a reactant concentration or a product concentration with respect to time in the early stages of a reaction where the rate is almost linear. The graph shown would be typical for a product concentration against time. The initial rate is determined by finding the slope of this line for an almost linear portion at the start.

The initial rate is directly proportional to the rate constant 𝑘 for a given reaction. We can see how the rate constant 𝑘 varies with temperature by using the Arrhenius equation. The Arrhenius equation links the rate constant, which is proportional to the initial rate, to the thermodynamic temperature, which is measured in kelvin. The dependence of 𝑘 upon 𝑇 is more clearly seen if we linearize the equation by taking the natural log of both sides. That is the log to base of 𝑒, which is written ln.

If we rewrite the equation, we can see that there are two quantities multiplied on the right-hand side. This is a quotient. The natural log of a quotient is the same as taking the natural logs of the quantities in the quotient and adding them together. So by taking natural logs, we have natural log 𝑘 equals natural log 𝐴 plus natural log of 𝑒 to the power of minus 𝐸 𝑎 divided by 𝑅𝑇. Now, a natural log of 𝑒 to the power of minus 𝐸 𝑎 over 𝑅𝑇 is simply the power to which 𝑒 is raised to. This is in fact minus 𝐸 𝑎 divided by 𝑅𝑇. So the equation becomes natural log 𝑘 equals natural log 𝐴 minus 𝐸 𝑎 divided by 𝑅𝑇.

With little rearrangement, we get natural log 𝑘 equals minus 𝐸 𝑎 divided by 𝑅𝑇 plus natural log 𝐴. By separating one over 𝑇 from minus 𝐸 𝑎 divided by 𝑅, we get the linearized version, which takes the form 𝑦 equals 𝑚𝑥 plus 𝑐, where 𝑚 is the slope, which is equal to minus 𝐸 𝑎 divided by 𝑟. From our linearized form of the Arrhenius equation, we can see that if we were to plot a graph of natural log 𝑘 against one over 𝑇, where the temperature is measured in kelvin, we would expect a straight line. Most importantly for this question, the slope of this line will be equal to minus 𝐸 𝑎 divided by 𝑅. The 𝑦-intercept of this particular graph will be equal to the natural log of 𝐴.

Now if we plot a very similar graph, but instead use natural log 𝑅 on the 𝑦-axis, where 𝑅 is the initial rate referred to in the question, we would also expect a straight line. This works because the rate constant 𝑘 is directly proportional to the initial rate 𝑅 for a given temperature. Because 𝑅 is the ideal gas constant and we know the value for the slope, we can easily establish a value for the activation energy 𝐸 𝑎. So 𝐸 𝑎 is equal to the negative one times the slope times 𝑅, the ideal gas constant.

So to answer the question, the graph required in order to estimate the activation energy would be natural log 𝑅 against one over 𝑇. This is answer (E).