Question

For a 10 degree Celsius rise in the temperature, the rate constant doubles for a reaction no. 1, and triples for a reaction no. 2. If the two reaction have comparable pre-exponential factors what is the ratio of their activation energy?

Answer 1

Each reaction rate coefficient $k$ has a temperature dependency, which is usually given by the Arrhenius equation:

$$k = A e ^ {- \frac {E _ {a}}{R T}}$$

$E_{a}$ is the activation energy and $R$ is the gas constant. Since at temperature $T$ the molecules have energies given by a Boltzmann distribution, one can expect the number of collisions with energy greater than $E_{a}$ to be proportional to $e^{-\frac{E_a}{RT}}$. $A$ is the pre-exponential factor or frequency factor.

$$\ln k = \ln A - E _ {a} / R T$$ $$\ln A = \ln k + E _ {a} / R T$$

For the first reaction:

$$\ln k _ {1} = \ln A - E _ {a} / R T \quad \Rightarrow \quad \ln 2 k _ {1} = \ln A - E _ {a} / R (T + 10)$$

For the second reaction

$$\ln k _ {2} = \ln A - E _ {a} / R T \quad \Rightarrow \quad \ln 3 k _ {2} = \ln A - E _ {a} / R (T + 10)$$

From this:

$$\ln 2 k _ {1} = \ln A - E _ {a} / R (T + 10) \Rightarrow \ln A = \ln 2 k _ {1} + E _ {a} / R (T + 10)$$ $$\ln 3 k _ {2} = \ln A - E _ {a} / R (T + 10) \Rightarrow \ln A = \ln 3 k _ {2} + x E _ {a} / R (T + 10)$$ $$\ln 2 k _ {1} + E _ {a} / R (T + 10) = \ln 3 k _ {2} + x E _ {a} / R (T + 10)$$ $$\ln 2 k _ {1} * R (T + 10) + E _ {a} = \ln 3 k _ {2} * R (T + 10) + x E _ {a} \quad (R (T + 10) = \text{const})$$ $$\ln 2 k _ {1} + E _ {a} = \ln 3 k _ {2} + x E _ {a}$$ $$\ln 2 k _ {1} - \ln 3 k _ {2} = x E _ {a} - E _ {a}$$ $$\ln \left(2 k _ {1} / 3 k _ {2}\right) = x E _ {a} - E _ {a}$$

In that moment when $k$ is the same for both reactions:

$$\ln(2/3) = xE_a - E_a$$ $$-0.405 = xE_a - E_a$$ $$x = 0,595$$

The ratio of $E_a$ is 1: 0.595