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The temperature dependence of rate of a chemical reaction can be accurately explained by Arrhenius equation. It was first proposed by Dutch chemist   J.H. Vant’s Hoff but Swedish chemist Arrhenius provides its physical justification and interpretation.
Where   K= Rate constant
              A= Arrhenius constant or frequency factor or pre exponential factor
              R= Universal gas constant =25/3 joule per mole per second
              Ea= Activation Energy
  T= temperature
-Ea/RT= Boltzmann factor or fraction of molecule having equal or greater than Activation energy or fraction of molecule that have kinetic energy greater than activation energy.
ILLUSTRATIVE EXAMPLE (1):  The activation energy for the reaction is 209.5 k J per mole at 581 K.
 Calculate the faction of molecule of reactant s having energy equal to or greater than activation energy.
SOLUTION: We known that fraction of molecule that have kinetic energy greater than activation energy is given by
Important cases of Arrhenius equation:
Case (1):  If T approaches to infinite

It means the entire reactant molecule will be active; and crossing over the energy barrier.  It will be possible when either Ea is Zero or temperature (T) is infinite. These are practically not possible.

The maximum value of K is A when temperature tends to infinity.

Graph indicate larger the activation energy, smaller is the value of rate constant (K)
 For free radical combination reaction  (Ea=0 ) thus K=A ,that means  for free radical  reaction rate  constant equal to  Arrhenius factor and becomes independence of temperature .
Case (2):  We know that (Ea) activation energy is always positive, thus K always increasing with increasing temperature whether reaction is exothermic or endothermic

Mathematical prove: since activation energy is always positive it can be never negative this is proven as:
Case (3): However in many complex reactions it is observed that rate constant found to be decreases with increasing temperature.
For example
By equation (1) and (2)

Since the overall reaction is  exothermic and  Kc  decreases  with increasing temperature  as well as the decrease  in Kc  out weight the increase  in K with temperature, thus K obs  show a decrease with increase in temperature.
Case (4): larger the activation energy greater the effect of temperature on rate constant.
Case (5): At lower temperature, increase in temperature causes more changes in value of rate constant
Case (6):
“It means if Ea=RT then rate constant become about 37% of the Arrhenius constant.

ILLUSTRATIVE EXAMPLE (2): Consider the following reaction
The activation energy of backward reaction exceeds that of the forward reaction by 2RT (in J per mole). If the pre-exponential factor of forward reaction is 4 times of the reverse reaction. The absolute value of (G) Gibbs energy at STP (in j per mole) for the reaction at 300 K is
 (Given ln2= 0.7, RT= 2500 J per Mole at 300 K and G is the Gibbs energy) (JEE Advanced 2018)
SOLUTION:   Given condition
ILLUSTRATIVE EXAMPLE (3): Plots showing the variation of rate constant (K) with temperature (T) are given below. The plot that follow Arrhenius equation is (JEE Advanced 2010)

Hence on increasing temperature rate constant (K) increases exponentially.

So option (A) is correct

Calculation Activation Energy:
Take both side natural logarithm and obtained as
(1): The above equation is the straight line with negative slope
(2): The slope of above equation gives Activation energy and intercept gives frequency factor
(3) Dependence of rate constant on temperature for two reactions is given as: 
Slope of reaction (2) is greater than (1) hence reaction (2) has higher activation energy so reaction (2) is more sensitive to temperature.
Calculation Activation Energy at two different temperatures (T1 and T2):
ILLUSTRATIVE EXAMPLE (4): In Arrhenius equation for a certain reaction, the value of A and Ea (activation energy) are 4 × 1013 sec–1 and 98.6 kJ mol–1 respectively. At what temperature, the reaction will have specific rate constant 1.1 × 10–3 sec–1?
SOLUTION: According to Arrhenius equation
ILLUSTRATIVE EXAMPPLE (5): The energy of activation for a reaction is 100 kJ mol–1. Presence of a catalyst lowers the energy of activation by 75%.What will be effect on rate of reaction at 20ºC, other things being equal?
SOLUTION: The Arrhenius equation is

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