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## Arrhenius Equation – The Effect of Temperature on Reaction Rate

Generally, the rate of a reaction increase with increasing temperature. However, there are very few exceptions. The magnitude of this increase in rate is different for different reactions. As a rough rule, for many reactions near room temperature, reaction rate tends to double when the temperature is increased by 10°C.

A large number of reactions are known which do not take place at room temperature but occur readily at higher temperatures. Example: Reaction between H_{2} and O_{2} to form H_{2}O takes place only when an electric spark is passed.

Arrhenius suggested that the rates of most reactions vary with temperature in such a way that the rate constant is directly proportional to e^{-(Ea/RT)} and he proposed a relation between the rate constant and temperature.

k = Ae^{-(Ea/RT)} …………. (1)

Where A the frequency factor,

R the gas constant,

E_{a} the activation energy of the reaction and,

T the absolute temperature (in K)

The frequency factor (A) is related to the frequency of collisions (number of collisions per second) between the reactant molecules. The factor A does not vary significantly with temperature and hence it may be taken as a constant.

E_{a} is the activation energy of the reaction, which Arrhenius considered as the minimum energy that a molecule must have to posses to react.

Taking logarithm on both side of the equation (1)

y = c + mx

The above equation is of the form of a straight line y = mx + c.

A plot of ln k Vs 1/T gives a straight line with a negative slope – \(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\) line with a negative slope –\(\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\). If the rate constant for a reaction at two different temperatures is known, we can calculate the activation energy as follows.

At temperature T = T_{1}; the rate constant k = k_{1}

This equation can be used to caluculate E_{a} from rate constants K_{1} and k_{2} at temperatures

T_{1} and T_{2}.

**Example 1**

The rate constant of a reaction at 400 and 200K are 0.04 and 0.02 s^{-1} respectively. Caluculate the value of activation energy.

Solution:

According to Arrhenius equation

E_{a} = log(2) × 2.303 × 8.314 JK^{-1}mol^{-1} × 400K

E_{a} = 2.305 J mol^{-1}

**Example 2**

Rate constant k of a reaction varies with temperature T according to the following Arrhenius equation log k = log A – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\)(\(\frac{1}{T}\)) Where E_{a} is the activation energy. When a graph is plotted for log K Vs \(\frac{1}{T}\) a straight line with a slope of – 4000k is obtained. Caluculate the activation energy.

Solution:

log k = logA – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\)(\(\frac{1}{T}\))

y = c + mx

m = – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\)

E_{a} = – 2.303 R m

E_{a} = – 2.303 × 8.314 JK^{-1} mol^{-1} × (-4000K)

E_{a} = 76, 589 J mol^{-1}

E_{a} = 76.589 kJ mol^{-1}