IIT JEE Revision - Arrhenius equation

Reaction Rate Depends on Temperature

Temperature has influence on reaction rates. In general, an increase in temperature increases the rate of almost all reactions.

A general approximate rule is that the rate of a reaction becomes almost double for every 10° rise in temperature.

Activation Energy

For many reactions some extra energy is to be supplied to the reactants to initiate the reaction. This excess energy is required to bring the energy of reactants to the energy that is required to start the reaction. The energy of the reactants at which the reaction starts is called threshold energy.

Activation energy is the extra energy supplied to initiate the reaction. Thus activation energy is equal to the difference between the threshold energy and the average kinetic energy of the reacting molecules at the the given temperature (Note as activation energy is being given the temperature of the reactants increases)

Arrhenius Equation

Arrhenius proposed a quantitative relationship between rate constant and temperature

k = Ae^(–E_a/RT)

^{where k = rate constant}

^{A is a constant known as frequency factor. In a JEE problem it was termed as preexponential factor}

–E_a is the activation energy

Both A and –E_a0 are characteristic of the equation

T is the absolute temperature and R is the gas constant

In log form the equation becomes

log k = log A - (E_a)/2.303 RT

As the activation energy –E_a0 increases, the value of k decreases and therefore, the reaction rate decreases.

Find the value of –E_a0

If log k is plotted against 1/T (both found through experiments), the intercept of the line will be equal to - (E_a)/2.303 R. Hence from the slope found from the graph - (E_a) can be found out as -2.303 R multiplied by slope.

Second Method

Measure rate constant at two temperatures k1 and k2 at T1 and T2

log (k2/k1) = (E_a/2.303 R)[ (1/T1) - (1/T2)]

JEE 2009 problem

For a first order reaction A→P, the temperature (T) dependent rate constant(k) was found to follow the equation logk = – (2000)(1/T) + 6.0
The pre-exponential factor A and the activation energy E_a, respectively, are -

Answer:
1.0 × 10⁶6 s^-1 and 38.3 kJ mol^-1