hii tanuj

Acoording to your question,first of all write the equation-

H2+I2=2HI

initial moles    x + x = 0

at equilibrium 1-x + 1-x= 2x

let me explain now,suppose initial moles are x for both reactant and 0 moles for product.now at equilibrium some moles of reactant must be lost suppose it will become 1-x and product will become x but as there is 2 moles of product so it will become 2x.

as kc is given 0.0156 so now we can write it as

kc=(2x) whole square/(1-x) whole square

0.0156=(2x) whole square/(1-x) whole square

0.12=2x/1-x

x=0.05 so now at equilibrium H2 will be 1-0.05=0.95,I2 will be 1-0.05=0.95 and HI will be 2x=0.1

i hope this information helps you.

thank you