Solved problem

Q_p =	p(CH₃OH)
	p(CO)p(H₂)²

Consider preparation of CH₃OH from 30 kPa CO and 70 kPa H₂ 225°C.
CO(g) + H₂(g)

CH₃OH(g)
Calculate p(CH₃OH) at equilibrium. K_p at this temperature is 0.049.

Step 1: Enter given initial pressures.
Step 2: Assign changes with x = change in p(CH₃OH)
Step 3: Divide equilibrium pressures by 100 kPa.

p(H₂) = (70 - 2x)/100; p(CO) =(30 - x)/100; p(CH₃OH) = x/100

0.049 =	x/100
	(0.3 - x/100)(0.7 - 2x/100)²

Step 4: Substitution in Q gives

	p(CO)/kPa	p(H₂)/kPa	p(CH₃OH)/kPa
initial	30	70	0
change	-x	-2x	+x
final	30 – x	70 – 2x	x

The equation in Step 4 would contain x³, x² and x terms when multiplied out.
Since K_p is small, the solution was simplified by first making some approximations.

Step 5: Solve for x.
0.0072 = x/100 kPa
x = p(CH₃OH) = 0.72 kPa
x/100 << 0.3 and 2x/100 << 0.7

Step 6: Use approximations in Q

0.049 =	x/100
	(0.3)×(0.7)²

Step 5: Approximations
x/100 much less than 0.3
2x/100 much less than 0.7

If K_p is not small, see other methods in problems 1 and 2 (or solve using a spreadsheet!).
Ultimately, however, the ability to recognize when approximations can be made is very useful!!!