ivera Suppose A 250mL Flask re httpswwwawhalekscomalekscgi
Solution
volume = 500 mL = 0.500 L
initial concentration of H2 = number of mol of H2 / volume
 = 0.40 mol / 0.500 L
 = 0.80 M
initial concentration of HCl = number of mol of HCl / volume
 = 1.4 mol / 0.500 L
 = 2.8 M
ICE Table:
 
                     [H2]                [Cl2]               [HCl]             
 initial             0.8                 0                   2.8               
change +1x +1x -2x
equilibrium 0.8+1x +1x 2.8-2x
Equilibrium constant expression is
 Kc = [HCl]^2/[H2]*[Cl2]
 0.473 = (2.8-2x)^2 / (0.8+x)*x
 0.473 = (7.84-11.2*x + 4*x^2)/((0.8 + 1*x)(1*x))
 0.473 = (7.84-11.2*x + 4*x^2)/(0.8*x + 1*x^2)
 0.3784*x + 0.473*x^2 = 7.84-11.2*x + 4*x^2
 -7.84 + 11.5784*x-3.527*x^2 = 0
 This is quadratic equation (ax^2+bx+c=0)
 a = -3.527
 b = 11.58
 c = -7.84
 
 Roots can be found by
 x = {-b + sqrt(b^2-4*a*c)}/2a
 x = {-b - sqrt(b^2-4*a*c)}/2a
 
 b^2-4*a*c = 23.45
 
 roots are :
 x = 0.9549 and x = 2.328
 
 x can\'t be 2.328 as this will make the concentration negative.so,
 x = 0.9549
 
 At equilibrium:
 [H2] = 0.8+1x = 0.8+1*0.95486 = 1.75486 M
 Answer: 1.75 M

