# Question Video: Activation Energy from Arrhenius Plot Chemistry • 10th Grade

The rate constant for a reaction, 𝑘, was measured over a range of reaction temperatures, 𝑇. The data were plotted on a graph of ln 𝑘 against 1/𝑇. What is the activation energy for the reaction in terms of the gradient of the plot 𝑚? [A] −𝑅𝑚 [B] 𝑚/𝑅 [C] 𝑚 [D] 𝑅𝑚 [E] −𝑚/𝑅

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### Video Transcript

The rate constant for a reaction, 𝑘, was measured over a range of reaction temperatures, 𝑇. The data were plotted on a graph of log 𝑘 against one divided by 𝑇. What is the activation energy for the reaction in terms of the gradient of the plot 𝑚? (A) Minus 𝑅 multiplied by 𝑚, (B) 𝑚 divided by 𝑅, (C) 𝑚, (D) 𝑅 multiplied by 𝑚, (E) minus 𝑚 divided by 𝑅.

To answer this question, we need to understand the relationship between the rate constant for a reaction 𝑘 and the temperature 𝑇. This is explained using the Arrhenius equation. The Arrhenius equation links the rate constant, which is measured experimentally, to the thermodynamic temperature, which is measured in kelvin. The equation can be seen to have a preexponential part, which is known as the preexponential factor — this is related to collision frequency and orientation — and an exponential part, where we find the activation energy.

The dependence of 𝑘 on 𝑇 is more clearly seen if we linearize the equation by taking the natural logarithm on both sides. This is the log to the base of 𝐸. By rearranging the linearized equation a little further, we can see that it takes the form 𝑦 equals 𝑚𝑥 plus 𝑐, where 𝑚, which is the gradient, is equal to minus 𝐸𝑎 divided by 𝑅. When log 𝑘 is plotted against one over 𝑇, log 𝑘 will always appear on the 𝑦-axis of a graph. So, on our graph of log 𝑘 against one over 𝑇, we would expect a straight line with the 𝑦-intercept equal to log of 𝑎.

The gradient of the straight line can be found by picking two convenient points and dividing the difference in the 𝑦-axis, which is log 𝑘, by the difference in the 𝑥-axis, which is one over 𝑇. Now we know that the gradient referred to as 𝑚 in the question is equal to minus 𝐸𝑎 divided by 𝑅, a little rearrangement of this expression leads us to 𝐸𝑎, which is the activation energy, is equal to minus 𝑅 times 𝑚, which is answer (A).