# Effect of Temperature on Rate of Reaction

The rate of reaction increases with rise of temperature. It is expressed as specific rate constant. In general it is approximately 2 to 3 times for rise of temperature from $25^0 C \text{to} 35^0C$.

Hence the ratio of specific rate constant of reaction at rise of $10^0C (25^0 \text{to} 35^0)$ is known as the temperature coefficient.

Temperature coefficient

$= k \hspace{2mm} \text{at} \hspace{2mm} ( t + 10^0 ) / k \hspace{2mm} \text{at} \hspace{2mm} t^0 c = k_{35^0 c} / k_{25^0 c}$

For example, the temperature coefficient for the dissociation of HI,

$2HI \leftrightharpoons H_2 + I_2$

Is 1.8 while that for the reaction of methyl iodide with sodium ethoxide is 2.9 :

$CH_3I + C_2H_5ONa \leftrightharpoons CH_3OC_2H_5 + NaI$

This is due to the fact that only few molecules out of the total number of molecules possess energy equal to or in excess to a minimum energy (threshold energy) which can perform the reaction. Number of such molecules increases rapidly even with a small rise of temperature.

Therefore the molecules must get sufficient amount of energy to carry out the reaction. The excess of energy that the reactant molecules must acquire in order to react to yield products is known as activation energy. This statement may be presented as:

Activation energy =Threshold energy – Average energy actually possessed by molecules.

Effect of temperature on rate of reaction

According to this concept we can say that, non-active molecule (the molecules which possess less than threshold energy) can be activated by absorption of extra energy to perform the reaction.

In other words, there is an energy barrier placed between reactants and products which has to be crossed before reactants can yield products. Actually this is the magnitude of threshold energy.

Magnitude of energy

According to Arrhenius,

$K = Ae^{-Ea/RT}$

where K = rate constant, R = gas constant, T = absolute temperature, A and Ea are two constants known as frequency factor and activation energy respectively. On solving we get, Ea

$\log K = \log A- \dfrac{Ea}{2.303RT}$

At two temperatures $T_1 \text{and}T_2$, the above equation may be written as:

$\log \dfrac{K_2}{D_1} = \dfrac{Ea}{2.303 R} \times \dfrac{T_2 - T_1}{T_1 T_2}$

Another form of Arrhenius equation can be written as:

$\dfrac{dlnK}{dT} = \dfrac{E}{RT^2}$

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