EXAMPLE#1---- CH 13 # 50 in “Zummie”
PROBLEM OBJECTIVES
 Construct ICE chart structure.
 Apply RULE OF SUCCESSIVE APPROXIMATIONS.
 Apply 5.% rule.
2 SO2(g) + O2(g)   2 SO3(g)
Kp =
0.50 atm
-2x
0.50 atm
-x
0.0
+2x
0.50-2x
-x
+2x
(P SO3)2
(P SO2)2 (P O2)
Kp =
Kp = 0.25
(2x)2
(0.50-2x)2 (0.50-x)
Now, this is some nasty math, a difficult cubic equation, arrgh! It will take to long
to solve – you will never finish an exam. Apply the rule of successive approximations as
follows.
1) Compare the value of x to K, in this problem K < 1 therefore x is small – but not
very small.
2) GUESS – that’s right guess a value for x. X is usually not more than 5-10% of
the starting molarity with this moderate K, and should be about 0.050 atm in
this problem. This guess is made considering the K value and the starting
molarity.
3) Substitute the guessed value of x in all terms except one of them, we will solve
for the x in the (2x)2 in the O2 term. In equilibrium problems, the
denominator is usually where you substitute your guess, and solve for the
numerator value of x.
4) The guessed value will give you a value for x when substituted. Plug this back
into the equation until the solved for value of x is the same in successive
equations.
Kp =
(P SO3)2
(P SO2)2 (P O2)
= 0.25 =
4x2
[0.50 – 2(0.050)]2 [(0.50 – (0.050)]
x=0.067
Kp =
(P SO3)2
(P SO2)2 (P O2)
= 0.25 =
4x2
[0.50 – 2(0.067)]2 [(0.50 – (0.067)]
x=0.060
Kp =
(P SO3)2
(P SO2)2 (P O2)
= 0.25 =
4x2
[0.50 – 2(0.060)]2 [(0.50 – (0.060)]
x=0.063
DOES NOT
EQUAL THE
GUESS,
CONTINUE TO
SUBSTITUTE
IT.
Kp =
(P SO3)2
(P SO2)2 (P O2)
= 0.25 =
4x2
[0.50 – 2(0.063)]2 [(0.50 – (0.063)]
x=0.062
THIS VALUE RETURNS
ITSELF WHEN IT IS
SUBSTITUTED, SO IT IS
THE VALUE WE GO
WITH.