NEET Chemistry Chemical Kinetics Questions Solved


(a) Explain the following terms :

    (i) Rate of a reaction

   (ii) Activation energy of a reaction

(b) The decomposition of phosphine, PH3, proceeds according to the following equation: 

     4PH3 (g) P4(g) + 6 H2(g)

It is found that the reaction follows the following rate equation:

      Rate = K[PH3].

The half - life of PHis 37.9 s at 120 C.

(i) How much time is required for 3/4 th of PHto decompose?

(ii) What fraction of the original sample of PHremains behind after 1 minute?

(a) (i) Rate of a reaction: The change in the concentration of any one of the reactants or products per unit time is called rate of a reaction.

(ii) The minimum extra amount of energy absorbed by the reactant molecules to form the activated complex is called activation energy. 

The activation energy of the reaction decreases by the use of catalyst.

(b) (i) According to the formula :

            K =2.303tlog [A][A]               .......(1)

   and  K =0.693t12                              ........(2)

On comparing both equations

     K =0.693t12 =2.303tlog [A][A]

or  t =2.303×t120.693log [A][A]

Let the initial concentration = a i.e. A

then t =2.303×37.90.693log aa-34a

or  t =2.303×37.90.693log 4

or t =87.28370.693 × 0.6020  = 75.8

 Time required to decompose, t =75.8 seconds

(ii)  K =2.303tlog aa-x

      0.69337.9s = 2.303 60slog aa-x            

      log aa-x=0.69337.9s ×60 2.303 

      log aa-x= 0.4764  

     aa-x = antilog 0.4764 = 2.995

If a=1, then (a- x) i.e. fraction left may be calculated as:

1(a-x)= 2.995

 Fraction of original sample,

(a- x) = 12.995= 0.334

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