Question

pls no silly messages i am asking this 2nd time plss guys pls help a sis in need...
3 moles of H2 and 1 mole of Iodine were taken in a 1L flask . find the composition of the eqlbm mixture if Kc =2 for the reaction H2 + I2 = 2HI . Also find the % of H2 reacted ?​

Answer 1

Answer:

We're asked to find the equilibrium concentration of  

HI  with an initial concentration of  

1  mol H 2  and  1 mol I 2 .

The equilibrium constant expression for this reaction is

K c = [ HI ] 2 [ H 2 ] [ I 2 ] = 49

The initial concentrations of both  

H 2  and  I 2  are  1 l mol 2 l L = 0.5 M

So our initial concentrations for each species are

Initial:

H 2 :  0.5 M  I 2 :  0.5 M HI :  0

From the coefficients of the chemical equation, we can predict the changes in concentration with the quantity  x :

Change:

H 2 :  − x  I 2 :  − x  HI :  + 2 x

and the final concentrations are the sum of the initial and change:

Final:

H 2 :  0.5 M − x  I 2 :  0.5 M − x HI :  2 x

It's algebra time! Let's plug these into our equilibrium constant expression to start solving for  

x :  K c = ( 2 x ) 2 ( 0.50 − x ) ( 0.50 − x ) = 49  4 x 2

49 (0.25 −x + x 2 )  49 x 2 − 49 x + 12.25

4 x 2  45 x 2 − x + 12.25

0  x = 49 ± √ ( − 49 ) 2 − 4 ( 45 ) ( 12.25 ) 2 ( 45 )

= 0.7  or  0.389

We can neglect the solution that is greater than  0.5 , because then the equilibrium concentrations of hydrogen and iodine would be negative! So the one to use is  0.389 .

Therefore, the final equilibrium concentration of  

HI  is [ HI ] = 2 x = 2 ( 0.389 ) = 0.778 M

Explanation:

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