Changing concentration changes the rate but not k. Changing temperature changes k itself, and the Arrhenius equation tells you by how much. Plot ln k against 1/T and the gradient hands you the activation energy.
The rate constantk increases with temperature because a larger fraction of collisions have energy ≥ Ea. The Arrhenius equation is k = Ae−Ea/RT, where A is the pre-exponential factor (linked to collision frequency and orientation), T is in kelvin and R = 8.31 J K−1 mol−1. Taking logs gives ln k = ln A − Ea/RT, a straight line when ln k is plotted against 1/T: the gradient = −Ea/R and the intercept is ln A. A larger Ea means a steeper line, so the rate is more sensitive to temperature.
Fig. 1: an Arrhenius plot; the gradient of ln k against 1/T equals −Ea/R
Section 01
Why temperature changes k.
Raise the temperature in the Boltzmann distribution simulator and watch the shaded area beyond the activation energy grow.
Fraction with enough energy: e−Ea/RT. For Ea = 50.0 kJ mol−1 at 298 K this is 1.70 × 10−9: tiny, but it rises quickly with T.
Rule of thumb: for Ea = 50.0 kJ mol−1, a 10 K rise from 300 K to 310 K multiplies k by 1.91, close to the often-quoted doubling.
Worked example:A = 1.00 × 1010, Ea = 50.0 kJ mol−1, T = 298 K: k = 1.00 × 1010 × e−50000/(8.31 × 298) = 17.0 (same units as A).
Catalysts also change k, by lowering Ea: dropping it from 80 to 60 kJ mol−1 at 298 K multiplies k by about 3.22 × 103 (if A is unchanged).
Stage 1 · Learn
Check what the sim just showed you
Four quick checks. Each correct answer earns XP and records this skill on your profile.
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Which change increases the rate constant k of a reaction?
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What is the gradient of a graph of ln k against 1/T?
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Why does a rise in temperature increase the rate constant?
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An Arrhenius plot has gradient −6000 K. What is Ea?
Section 02
Finding Ea from a graph.
Read off the rate constants at several temperatures in the simulator, then convert each into ln k and 1/T.
Step
What to do
Example
1
convert °C to K, then work out 1/T
300 K → 3.33 × 10−3 K−1
2
work out ln k for each rate constant
k = 2.32 × 10−3 → ln k = −6.07
3
plot ln k (y) against 1/T (x); draw the best straight line
use a large triangle for the gradient
4
Ea = −gradient × R
gradient −6000 K → Ea = 6000 × 8.31 = 49.9 kJ mol−1
Two points only: ln(k2/k1) = (Ea/R)(1/T1 − 1/T2). With k = 2.32 × 10−3 at 300 K and 1.72 × 10−2 at 320 K: Ea = 8.31 × ln(7.41) / (2.08 × 10−4) = 79.9 kJ mol−1.
Rate doubles from 298 K to 308 K:Ea = 8.31 × ln 2 / (1/298 − 1/308) = 52.9 kJ mol−1.
Examiner trap
Always use T in kelvin and R in J K−1 mol−1, so Ea comes out in J mol−1: divide by 1000 for kJ. The gradient is negative; Ea = −gradient × R is positive. Plot ln k, not k, and 1/T, not T. Increasing concentration does not change k; only temperature and catalysts do.
Stage 2 · Exam
Exam-style questions
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The rate constant is 2.32 × 10−3 at 300 K and 1.72 × 10−2 at 320 K. What is Ea?
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The rate of a reaction doubles when the temperature rises from 298 K to 308 K. What is Ea?
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Two reactions have Ea values of 40 and 80 kJ mol−1. Which statement is correct?
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A reaction has Ea = 60.0 kJ mol−1 and k = 0.0100 s−1 at 300 K. What is k at 350 K?
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