posted by Swaastikaa on .
The following system is at equilibrium,at 699K in a 5L container.
H2(g)+I2(g)--->2HI(g) Kc = 54.9
Initially,the system had 2.50 moles of HI.What is the moles of H2 at equilibrium?
Can Dr.Bob check if my steps are correct?
Initially,the concentrations of H2 & I2 are 0,whereas concentration of H1 is 2.50mol/5L = 0.50 mol/L .At equilibrium,the concentrations of H2 & I2 are assumed to be x,whereas concentration of HI is 0.50-2x
Using equlibrium constant equation with respect to concentration,
Kc = [HI]^2/[H2][I2]
54.9 = (0.50-2x)^2/x.x
Since 2x is assumed to be negligible,therefore the equation above further simplified to :
54.9 = (0.50)^2/x.x
x^2 = (0.50)^2/54.9
x = 0.06748 mol/L = [H2] = 
Therefore moles of H2 at equilibrium is
0.06748 mol/L X 5L = 0.337 mol
Is my steps and answer correct? Thank you in advance