A ALEKS-Rosemary Rivera × eSuppose A 250mL Flask × Secure I https/www-awh.aleks.

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A ALEKS-Rosemary Rivera × eSuppose A 250mL Flask × Secure I https/www-awh.aleks.com/alekscgi/x/islexe/to u-igNslkr7/8P3jH-IQT3v1gTPrkicRI6pS60AVDiQYFU2wpW5CMM9duvob Calculating equlbrium composition from an equlibrlum constant Suppose a 500. mL. flask is filled with 1.5 mol of H, and 0.20 mol of HI. The following reaction becomes possible: H2(2)+1)2HI(g) The equilibrium constant K for this reaction is 8.88 at the temperature of the flask. Calculate the equilibrium molarity of I2. Round your answer to two decimal places .M Check

Explanation / Answer

volume = 500 mL = 0.500 L

initial concentration of H2 = number of mol of H2 / volume
= 1.5 mol / 0.500 L
= 3.0 M

initial concentration of HI = number of mol of HI / volume
= 0.20 mol / 0.500 L
= 0.40 M

ICE Table:

                    [H2]                [I2]                [HI]              

initial             3.0                 0                   0.4               

change              +1x                 +1x                 -2x               

equilibrium         3.0+1x              +1x                 0.4-2x            

Equilibrium constant expression is
Kc = [HI]^2/[H2]*[I2]
8.88 = (0.16-1.6*x + 4*x^2)/((3 + 1*x)(1*x))
8.88 = (0.16-1.6*x + 4*x^2)/(3*x + 1*x^2)
26.64*x + 8.88*x^2 = 0.16-1.6*x + 4*x^2
-0.16 + 28.24*x + 4.88*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 4.88
b = 28.24
c = -0.16

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 = 8.006*10^2

roots are :
x = 5.66*10^-3 and x = -5.793

since x can't be negative, the possible value of x is
x = 5.66*10^-3

At equilibrium:
[I2] = x = 5.66*10^-3 M

Answer: 5.66*10^-3 M

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