## Arrhenius equation-

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## Arrhenius equation-

*According to Arrhenius,*

*i) All molecules of a system can not take part in a chemical equation .*

*ii) Only certain number of molecules will react . These reacting molecules are known as active molecules.*

*iii) The molecules which do not take part in chemical reaction are known as passive molecules.*

*iv) An equilibrium exists between active and passive molecules.*

*M (active) ⇌ M (passive)*

*v) When temperature is raised , the above equilibrium shifts to the left then number of active molecules increases. Hence rate of reaction increases with temperature.*

*So basic concept of Arrhenius theory is that passive molecules becomes active due to absorption of energy. Arrhenius proposed a simple equation known as Arrhenius equation.This equation gives a relationship between rate constant (K) and temperature of the system.*

*Arrhenius equation is ,*

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

^{-Ea/RT}

*A is a constant known as frequency factor. It gives frequency of binary collisions of reactant molecules per second per litre.*

*Ea = Energy of activation , R = gas constant , T = temperature in Kelvin*

*Another form of Arrhenius equation ,*

*Taking log of both sides of Arrhenius equation ,*

*ln K = ln A- Ea/RT*

*Now convert ln into log,*

*2.303 log K = 2.303 log A- Ea/RT*

*log K = (- Ea/ 2.303 RT) + log A*

*or*

*log K = log A – (Ea/ 2.303 RT)*

*As the value of Ea increases, the value of K decreases therefore rate of reaction decreases. As value of T increases, the value of K increases. This equation is used to calculate the value of activation energy .*

## Calculation of activation energy-

### i) Graphical method-

*log K = (- Ea/ 2.303 RT) + log A ————eq 1*

*Eq (1) is a type of ‘ y = mx + c ‘ equation and represents a straight line . If values of ‘log K’ are plotted against ‘1/T’ , a straight line should be obtained . Hence ,*

*slope = – Ea/ 2.303 R*

*In order to determine energy of activation of a reaction , its rate constant at different temperatures are measured . Now the values of ‘ log K ‘against ‘1 / T ‘are plotted. The curve obtained is a straight line and slope of line is measured . Slope is related to Ea , as*

*Slope = -Ea / 2.303 R*

*By knowing the ‘value of slope’ and ‘R’ , equation.the activation energy can be calculated with the help of above*

*source : ch302.cm.utexas.edu*

source : Chemistry- TutorCircle

### ii) Rate constant method-

*If K1 and K2 are the rate constants measured at temperatures T1 and T2 respectively , then according to equation,*

*log K = log A – (Ea/ 2.303 RT) ———–eq 1*

*log K _{1} = log A – (Ea/ 2.303 RT1) ————- eq 2*

*log K _{2} = log A – (Ea/ 2.303 RT2) ————- eq 3*

*Subtracting eq( 2) from eq (3)*

*log K _{2} – log K_{1} = (-Ea/ 2.303 RT_{2}) – (- Ea/ 2.303 RT_{1})*

*log [K*_{2} / k_{1]} = [Ea/ 2.303 R][(1 / T_{1}) – (1 / T_{2})] ———- eq 4

_{2}/ k

_{1]}= [Ea/ 2.303 R][(1 / T

_{1}) – (1 / T

_{2})] ———- eq 4

*By knowing the values of K _{1} and K_{2} of a reaction measured at two different temperatures T_{1} and T_{2} respectively , the energy of activation (Ea) can be calculated by eq (4).*