Question

Suppose a 500.mL flask is filled with 1.5mol of CO , 0.60mol of H2O and 1.8mol of CO2 . The following reaction becomes possible: +COgH2Og +CO2gH2g The equilibrium constant K for this reaction is 3.35 at the temperature of the flask. Calculate the equilibrium molarity of H2 . Round your answer to one decimal place.

Answer 1

Answer – We are given, volume = 500 mL , moles of CO = 1.5 moles

H₂O = 0.60 moles , moles of CO₂ = 1.8 moles , K = 3.35

First we need to calculate the molarity of each-

[CO] = 1.50 moles / 0.500 L = 3.0 M

[H₂O] = 0.60 moles / 0.500 L = 1.2 M

[CO₂] = 1.8 moles / 0.500 L = 3.6 M

Now we need to put ICE chart -

   CO(g) + H₂O(g) <---> CO₂(g) + H₂(g)

I 3.0          1.2               3.6            0

C -x           -x                +x           +x

E 3.0-x    1.2-x             3.6+x      +x

We know,

K = [CO₂(g)] [H₂(g)] / [CO(g)] [H₂O(g)]

3.35 = (3.6+x)(x) / (3.0-x)(1.2-x)

3.35[(3.0-x)(1.2-x)] = 3.6x + x²

3.35(x²-4.2x+3.6) = 3.6x+x²

3.35x²-14.07x +12.06 = 3.6x + x²

3.35x²-x² -14.07x-3.6x + 12.06

2.235x²-17.67x +12.06 = 0

Using the quadratic equation

x = -b+/- √(b²-4ac / 2a

x = 0.754

so, at equilibrium, x = [H₂] = 0.754 M