Question

Show proper I.C.E. setup and calculations please! units and correct significant figures also! Thanks so much!

1. Initially, 0.10 mol of HI, 0.50 mol of H₂ , and 0.50 mol of I ₂ are placed into a 2.0L reaction vessel at 200degrees celsius. Calculate the equilibrium concentrations for HI, H_2, and I₂. Place answers below.

2HI (g) <----> H₂ (g) + I₂ (g)   K_c= 1.8

Answer 1

2HI (g) <----> H2 (g) + I2 (g)     Kc= 1.8

ICE table

	[HI]	[H2]	[I2]
initial	0.10 mol	0.50 mol	0.50 mol
change	-2x	+x	+x
equilibrium	0.10-2x	0.50+x	0.50+x

Kc=1.8=[H2] [I2]/[HI]^2=(0.50+x) (0.50+x)/(0.10-2x)

1.8=(0.50+x) ^2/(0.10-2x)

Or.1.8-3.6x=x^2+4.6x+0.07

Or,x^2+4.6x+0.07=0 [quadratic equation of the form ax^2+bx+c=0 has the solution

x=[-b±(b^2-4ac)^1/2]/2a

so solving for x,

x=[-4.6±((4.6)^2-4*1*(0.07))^1/2]/2*1

x=0.015 and -4.58

concentration cannot be negative so x=0.015

equilibrium concentration of [HI]=0.10-2x=0.10-2*0.015=0.07

equilibrium concentration of [H2]=0.50+x=0.50+0.015=0.515

equilibrium concentration of [I2]= 0.50+x=0.50+0.015=0.515