Question

For the reaction 2HI=H2+I2.one mole of HI is introduced into a vessel at constant temperature.calculate the no of moles of I2 when equilibrium is reached (Kc=0.0156)
Plz help me and give complete solution

Answer 1

2HI>>H2 +I2
letting x be the conc of (I2)
The initial concentration of HI is 1mole/vol

K=x^2/(1-(2x))

.0156-.312x-x^2=0

x=(.312+-sqrt(.312^2+.0624))/(-2)

x= .0438 moles

check my math, it is easy to make an error.

Answer 2

I can't understand .312x how this value come i mean from where?

Answer 3

K=x^2/(1-(2x)) = 0.0156
It comes from 0.0156 * 2

Answer 4

Right