The rate of a reaction decreased by 3.555 times when the temperature was changed from $$40\,^\circ\text{C}$$ to $$30\,^\circ\text{C}$$. The activation energy (in $$\text{kJ mol}^{-1}$$) of the reaction is______.

We know that for any chemical reaction the temperature-dependence of the rate constant is expressed by the Arrhenius equation

$$k = A \, e^{-E_a/RT}$$

where $$k$$ is the rate constant, $$A$$ is the pre-exponential factor, $$E_a$$ is the activation energy, $$R$$ is the universal gas constant and $$T$$ is the absolute temperature in kelvin.

Because the order of the reaction does not change with temperature, the rate is directly proportional to the rate constant. Thus the ratio of the two measured rates is the same as the ratio of the two corresponding rate constants. We are told that the rate decreases by a factor of $$3.555$$ when the temperature is lowered from $$40^{\circ}\text C$$ to $$30^{\circ}\text C$$, so we can write

$$\dfrac{k_{30}}{k_{40}}=\dfrac{1}{3.555}$$

Using the logarithmic form of the Arrhenius equation for two temperatures,

$$\ln\!\left(\dfrac{k_{30}}{k_{40}}\right) = -\dfrac{E_a}{R}\left(\dfrac{1}{T_{30}}-\dfrac{1}{T_{40}}\right)$$

First we convert the two Celsius temperatures into kelvin:

$$T_{40}=40^{\circ}\text C + 273 = 313\ \text K$$

$$T_{30}=30^{\circ}\text C + 273 = 303\ \text K$$

Now we substitute the known quantities into the logarithmic equation. On the left-hand side we have

$$\ln\!\left(\dfrac{k_{30}}{k_{40}}\right)=\ln\!\left(\dfrac{1}{3.555}\right)=\ln(1)-\ln(3.555)= -\ln(3.555)$$

Evaluating the natural logarithm,

$$\ln(3.555)\approx 1.269$$

so that

$$\ln\!\left(\dfrac{k_{30}}{k_{40}}\right)\approx -1.269$$

On the right-hand side we have the temperature factor,

$$\dfrac{1}{T_{30}}-\dfrac{1}{T_{40}}=\dfrac{1}{303}-\dfrac{1}{313}$$

Finding a common denominator,

$$\dfrac{1}{303}-\dfrac{1}{313}=\dfrac{313-303}{303\times313}=\dfrac{10}{94839}\ \text{K}^{-1}$$

Numerically,

$$\dfrac{10}{94839}\approx 1.0546\times10^{-4}\ \text{K}^{-1}$$

Now we insert these values into the logarithmic Arrhenius expression:

$$-1.269 = -\dfrac{E_a}{R}\,(1.0546\times10^{-4})$$

Because both sides carry a negative sign, they cancel, giving

$$1.269 = \dfrac{E_a}{R}\,(1.0546\times10^{-4})$$

We isolate $$E_a$$ by multiplying both sides by $$R$$ and dividing by $$1.0546\times10^{-4}$$:

$$E_a = \dfrac{1.269\,R}{1.0546\times10^{-4}}$$

Taking $$R = 8.314\ \text{J mol}^{-1}\text K^{-1}$$, we have

$$E_a = \dfrac{1.269 \times 8.314}{1.0546\times10^{-4}}\ \text{J mol}^{-1}$$

Multiplying in the numerator,

$$1.269 \times 8.314 \approx 10.55$$

and then dividing,

$$E_a = \dfrac{10.55}{1.0546\times10^{-4}}\ \text{J mol}^{-1} \approx 1.000 \times 10^{5}\ \text{J mol}^{-1}$$

Converting joules to kilojoules (by dividing by $$1000$$),

$$E_a \approx 100\ \text{kJ mol}^{-1}$$

So, the answer is $$100$$.