Topic 11 · page 15 of 21

Activation energy from a graph

The Arrhenius equation links k to temperature. Taking logarithms turns it into a straight line, so a graph of ln k against 1/T has a gradient of minus the activation energy divided by the gas constant.

14 min read11.10

Read first: The rate constant and its units, Activation energy and catalysts

The Arrhenius equation 11.10

Rate constants rise steeply with temperature. The Arrhenius equation describes how:

k=A e−Ea/RT k = A\,e^{-E_\text{a}/RT}
Symbol Meaning Unit
k rate constant depends on the order
A a constant for the reaction (the pre-exponential factor) same as k
EaE_\text{a} activation energy J mol⁻¹
R gas constant, 8.31 J mol⁻¹ K⁻¹
T temperature K

The specification says the equation will be given if it is needed, so you do not have to learn it. You do have to use it. A large EaE_\text{a} makes the exponent more negative and k smaller, and a higher T makes the exponent less negative and k larger.

A straight line from logarithms 11.10

Take natural logarithms of both sides:

ln⁡k=−EaR⋅1T+ln⁡A \ln k = -\frac{E_\text{a}}{R}\cdot\frac{1}{T} + \ln A

This has the form y = mx + c:

y = m x + c ln⁡k\ln k = (−EaR)\left(-\dfrac{E_\text{a}}{R}\right) × 1T\dfrac{1}{T} + ln⁡A\ln A
y axis ln k
x axis 1/T, with T in kelvin
gradient −Ea/R-E_\text{a}/R
intercept ln A

So you plot ln k against 1/T, and the gradient is −Ea/R-E_\text{a}/R. The gradient is negative, because k rises as 1/T falls.

MethodFinding the activation energy from data

  1. Convert each temperature to kelvin (°C + 273) and work out 1/T.
  2. Work out ln k for each rate constant.
  3. Plot ln k (y) against 1/T (x) and draw a line of best fit.
  4. Draw a large gradient triangle on the line, using points on the line, not data points. Gradient = Δy ÷ Δx.
  5. EaE_\text{a} = −gradient × R. This is in J mol⁻¹. Divide by 1000 for kJ mol⁻¹.

A worked graph 11.10

Rate constants for a reaction at five temperatures (illustrative data):

T / K k / s⁻¹ 1/T / 10⁻³ K⁻¹ ln k
290 0.0245 3.448 −3.709
300 0.0505 3.333 −2.986
310 0.0975 3.226 −2.328
320 0.187 3.125 −1.677
330 0.331 3.030 −1.106
A graph of ln k against 1 over T in units of 10 to the minus 3 per kelvin. Five points lie close to a straight line that falls from left to right, from about minus 1.1 at 3.03 to minus 3.7 at 3.45. A dashed gradient triangle on the line runs from 3.00 and minus 0.91 to 3.45 and minus 3.72, so the vertical side is minus 2.81 and the horizontal side is 0.45.
ln k against 1/T. The gradient triangle uses points on the line of best fit.

Worked exampleActivation energy from the gradient

Use the triangle on the graph, with points (3.00, −0.91) and (3.45, −3.72) on the line.

  1. Δy = −3.72 − (−0.91) = −2.81
  2. Δx = (3.45 − 3.00) × 10⁻³ = 0.45 × 10⁻³ K⁻¹
  3. gradient = −2.81 ÷ 0.45 × 10⁻³ = −6.24 × 10³ K
  4. EaE_\text{a} = −gradient × R = 6.24 × 10³ × 8.31 = 5.19 × 10⁴ J mol⁻¹
  5. EaE_\text{a} = 51.9 kJ mol⁻¹

Using the least-squares line through all five points gives the same result, 51.9 kJ mol⁻¹.

Clock reactions 11.10

For a clock reaction the rate is proportional to 1/t (page 6), and the rate is proportional to k when the concentrations are the same in every run. So 1/t is proportional to k. Then ln(1/t) differs from ln k by a constant, which only moves the line up or down. The gradient is the same, so a plot of ln(1/t) against 1/T also has a gradient of −EaE_\text{a}/R. Page 21 does this with the iodine clock.

Check yourself

  1. State what you plot, and what the gradient equals, to find the activation energy from rate constants measured at different temperatures.
  2. A graph of ln k against 1/T has a gradient of −7500 K. Calculate the activation energy, in kJ mol⁻¹. (R = 8.31 J mol⁻¹ K⁻¹)
  3. Calculate 1/T, in units of 10⁻³ K⁻¹ to 3 decimal places, at 25 °C and at 35 °C.
  4. A line on a graph of ln k against 1/T passes through (3.20 × 10⁻³ K⁻¹, −1.50) and (3.40 × 10⁻³ K⁻¹, −2.74). Calculate the gradient and the activation energy in kJ mol⁻¹.
  5. A student plots ln k against the temperature in °C and says the gradient is −EaE_\text{a}/R. Give two things that are wrong.
  6. A clock reaction takes 118 s at 288 K and 61 s at 298 K. Explain why a graph of ln(1/t) against 1/T has the same gradient as a graph of ln k against 1/T.

Answers cover these until you have tried the questions

  1. Plot ln k against 1/T (with T in kelvin). The gradient is −Ea/R-E_\text{a}/R.
  2. EaE_\text{a} = 7500 × 8.31 = 62 325 J mol⁻¹ = 62.3 kJ mol⁻¹.
  3. 25 °C = 298 K, so 1/T = 1/298 = 3.356 × 10⁻³ K⁻¹. 35 °C = 308 K, so 1/T = 1/308 = 3.247 × 10⁻³ K⁻¹.
  4. Δy = −2.74 − (−1.50) = −1.24. Δx = 0.20 × 10⁻³ K⁻¹. Gradient = −1.24 ÷ 0.20 × 10⁻³ = −6.2 × 10³ K. EaE_\text{a} = 6.2 × 10³ × 8.31 = 5.15 × 10⁴ J mol⁻¹ = 51.5 kJ mol⁻¹.
  5. The temperature must be in kelvin, and the x axis must be 1/T, not T. A graph of ln k against T is a curve and does not have the gradient −EaE_\text{a}/R.
  6. For a clock reaction 1/t is proportional to the rate, and so to k (the concentrations are the same in each run). So 1/t = c × k, where c is a constant. Then ln(1/t) = ln k + ln c, which differs from ln k only by a constant. That moves the line up or down but does not change its gradient.