Temperature Needed to Double a Reaction Rate

Given

• Activation energy Ea = 50.0 kJ/mol

• k2/k1 = 2.00 (the rate doubles)

• T1 = 300 K

• Ideal gas constant R = 8.314 J/(mol K)

Find

The absolute temperature T2 at which the rate constant is twice its value at 300 K.

Principle

The Arrhenius equation relates the rate constant to temperature through the activation energy: k = A e^(-Ea/RT). Taking the ratio at two temperatures cancels the pre-exponential factor A, giving a two-point form that connects k2/k1 directly to the temperatures. Because temperature sits inside an exponential, even small increases can multiply the rate.

Formula

Solution

1. Convert Ea to joules to match R: Ea = 50.0 kJ/mol = 50,000 J/mol.

2. Compute the left side: ln(k2/k1) = ln(2) = 0.693.

3. Compute Ea/R = 50,000 / 8.314 = 6014 K.

4. Rearrange: (1/T1 - 1/T2) = 0.693 / 6014 = 1.153 x 10^-4 K^-1.

5. Evaluate 1/T1 = 1/300 = 0.0033333 K^-1, so 1/T2 = 0.0033333 - 0.0001153 = 0.0032180 K^-1.

6. Invert: T2 = 1 / 0.0032180 = 310.7 K, which rounds to 311 K.

7. Sanity check: T2 > T1 as expected for a faster rate, and the increase is small, which is typical for Arrhenius behavior.

Unit check

Ea/R has units (J/mol) / (J/(mol K)) = K, and multiplying by (1/T1 - 1/T2) in K^-1 gives a dimensionless number, matching ln(k2/k1), which is dimensionless. Temperatures are in kelvin throughout, so the final T2 = 311 K is an absolute temperature.

What this means physically

A modest 11 K rise (about 3.7% in absolute temperature) doubles the rate for this reaction, showing how steeply rates depend on temperature. This is why chemists heat reactions to speed them up — and why refrigeration slows spoilage reactions. Larger activation energies make the rate even more temperature-sensitive.