Suppose a 500.mL flask is filled with 0.50 mol of I2 and 0.90 mol. of HI. The fo

Question

Suppose a 500.mL flask is filled with 0.50 mol of I2 and 0.90 mol. of HI. The following reaction becomes possible:

The equilibrium constant K for this reaction is 4.51 at the temperature of the flask. Calculate the equilibrium molarity of H2. Round your answer to one decimal place.

Solution

I2                H2           HI

I    1 M             0             1.8 M

C    -x              -x               +2x

E 1-x                x               1.8+2x

(1.8+2x)(1.8+2x)/1-x (x)=4.51

3.24+3.6x+3.6x+4x^2=4.51x^2+4.51X

3.24+7.2x+4x^2=4.51x^2+4.51X

0.51x^2-2.69x-3.24=0

Explanation / Answer

Solution

I2                H2           HI

I    1 M             0             1.8 M

C    -x              -x               +2x

E 1-x               -x               1.8+2x

(1.8+2x)(1.8+2x)/(1-x) (-x)=4.51

(3.24+3.6x+3.6x+4x^2)/(-x+x^2)=4.51

3.24+7.2x+4x^2= -4.51x + 4.51x^2

0.51*x^2 - 11.71*x - 3.24 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 0.51

b = -11.71

c = -3.24

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 = 143.734

roots are :

x = 23.234 and x = -0.273

Here x can’t be positive as [H2] at equilibrium is -x and it will become negative

so,

x = -0.273 M

[H2] = -x = 0.273 M

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