# NEET Chemistry Chemical Kinetics Questions Solved

BOARD

(a) Explain the following terms :

(i) Order of a reaction

(ii) Molecularity of a reaction

(b) The rate of a reaction increases four times when the temperature changes from 300 K to 320 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature .(R = 8.314 J K-1 mol-1 )

(a) (i)  Order of a reaction : It is the sum of powers of the molar concentrations of reacting species in the rate equation of the reaction.

(ii) Molecularity of a reaction:

1. It is the total number of reacting species (molecules, atoms or ions) which bring the chemical change.

2. It is always a whole number.

3. It is theoritical concept.

4. It is meaningful only for simple reactions or individual steps of a  complex reaction. It is meaningless for overall complex reaction.

(b) Given : ${T}_{1}$= 300K       ${T}_{2}$= 320 K

${K}_{1}$ =K (Consider)

${K}_{2}$= 4K           R= 8.314    Ea= ?

Substituting these values in the formulae,

log $\frac{{k}_{2}}{{k}_{1}}=\frac{{E}_{a}}{2.303R}\left[\frac{{T}_{2}-{T}_{1}}{{T}_{1}{T}_{2}}\right]$

$\therefore$  log$\frac{4K}{K}$ =$\frac{{E}_{a}}{2.303×8.314}\left[\frac{320-300}{320×300}\right]$

or  log 4 =$\frac{{E}_{a}}{19.1471}×\left[\frac{20}{96000}\right]$

or  0.6020 =$\frac{{E}_{a}}{19.1471}×\left[\frac{20}{96000}\right]$

or  Ea$\frac{0.6020×19.1471×96000}{20}$= 55327.46

$\therefore$ Energy of activation, Ea = 55327.46

=55.3 K J mol-1

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