Question

8. Calculate the equilibrium constant of the reaction when 8.1 mL of H2 and 9.3 mL of
12 react to form 13.5 mL of HI at 440°C.
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Answer 1

Answer:

25 ml of H2 and 18 ml of it paperwork heated in a sealed glass tube at 465 degree and at aquarium 30. 8 ml of HIV was formed calculate the percentage degree of dissociation of HIV at 465 degree c

Answer 2

Given,

initial volume of H2 = 8.1 ml

initial volume of I2 = 9.3 ml

volume of HI = 13.5 ml

To find :

 equilibrium constant .

Solution :

Let x be the dissociation .

From gas law ,  V is proportional to number of moles (const. T=440 C & P)

               H2  +  I2  -> 2HI

t=0          8.1      9.3      0  

t=eq.     8.1-x    9.3-x    13.5

=>   2x = 13.5

=>     x = 6.75

At equilibrium , volume of H2 = 8.1 - 6.75 = 1.35 ml

and , volume of I2 = 9.3 - 6.75 = 2.55 ml

Kc = [HI]^2/[H2] [I2]

Kc = 52.941

The equilibrium constant is 52.941