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N 2 + 3H 2 --------> 2NH 3 If vol. of reacting container is 4L & K c =1.49 * 10^(-5) L 2 mol -2 , then degree of dissociation of above Equilibrium is ?

N2 + 3H2 --------> 2NH3
If vol. of reacting container is 4L & Kc=1.49 * 10^(-5) L2 mol-2, then degree of dissociation of above Equilibrium is ?

Grade:12th pass

2 Answers

Bharat Makkar
34 Points
9 years ago
          N2 + 3H2----------> 2NH3
at t=0    x       3x                0
at eqm  x – a    3x -3a       2a
                        KC = ( 2a/4)2
                                  ( 1 – a/4) ( 3- 3a /4 )3 
NOW NEGLECT “ a “ IN THE DENOMINATOR AND SOLVE FOR “ a “.( I HAVE TAKEN a AS DEGREE OF DISSOCIATION AND a HAS BEEN NEGLECTED IN THE DENOMINATOR AS KC IS SMALL.
Bharat Makkar
34 Points
9 years ago
ON SOLVING YOU WILL GET a = 0.25%

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