Question

2.7 ml of H2, 3.1 ml of I2 vapour react at 444°C to form 4.5 ml of HI . Calculate the equilibrium constant at this temperature

Answer 1

let x be the dissociation and from gas laws V proportional. to number of moles(const. T=444 C & P)      

 H2  +  I2  -> 2HI

t=0       2.7      3.1      0  

t=eq.    2.7-x    3.1-x    4.5(2x)

           0.45    0.85     4.5  

2x=4.5

x=2.25

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

Kc = 52.94.

 

Answer 2

Answer:

here is your ans mate

Explanation:

let x be the dissociation

 H2  +  I2  -> 2HI

t=0       2.7      3.1      0  

t=eq.    2.7-x    3.1-x    4.5(2x)

          0.45    0.85     4.5  

2x=4.5

x=2.25

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

Kc = 52.94.