Chemistry 12
Equilibrium Concentrations From Initial Conditions
(using ICEbox Method)
Name:
Keq is a more general term for the equilibrium constant (instead of Kc). In problems dealing with
equilibrium “concentrations”, we will use Keq for our equilibrium constant. We are often required to
calculate the change in concentration during the course of a reaction. These problems will involve
measured starting concentrations, which will have to be changed to equilibrium concentrations before
they can be used in an equilibrium expression. Calculating equilibrium concentrations from a set of initial
conditions takes more calculation steps. In this type of problem the Keq value is often given.
Note: [H2]o means the initial concentration.
Sample problem #1: Given the following equation:
Fe3+(aq) + SCN- (aq)  Fe(SCN)2+(aq)
The initial concentrations of Fe3+ is 0.0064M and SCN- is 0.0010M, when the reaction reaches
equilibrium, the concentration of Fe(SCN)2+ is 4.5 x 10-4-. Calculate the Keq for this reaction.
Solution: The solution technique involves the use of an ICEbox.
[Fe3+]
[SCN-]
Initial
Change
Equilibrium
a) Now, let’s fill in what we know from the question:
[Fe3+]
[SCN-]
Initial
0.0064
0.0010
Change
Equilibrium
[Fe(SCN)2-]
[Fe(SCN)2-]
0
4.5 x10-4
b) Now, let’s fill in the “change” row:
Initial
Change
Equilibrium
[Fe3+]
0.0064
-4.5 x 10-4
[SCN-]
0.0010
-4.5 x 10-4
[Fe(SCN)2-]
0
4.5 x 10-4
4.5 x10-4
This particular reaction has a mole to mole ration of 1:1:1. So we can assume, if 4.5 x10-4 moles of
Fe(SCN)2- were produced at equilibrium then 4.5 x10-4 moles of Fe3+- and 4.5 x10-4 moles of SCNreacted to produce that amount of product.
c) Now, let’s do the math to complete the equilibrium row:
[Fe3+] = 0.0064 - 4.5x 10-4-= 0.0060M
[SCN-] = 0.0010 - 4.5x 10-4-= 5.5x 10-4M
[Fe3+]
[SCN-]
Initial
0.0064
0.0010
-4
Change
-4.5 x 10
-4.5 x 10-4
Equilibrium
0.0060
5.5 x 10-4
d) Use equilibrium concentrations to solve for Keq:
Keq =
[Fe(SCN)2-] = 4.5 x 10-4_______ = 136
[Fe3+][SCN-] (0.0060)(5.5x10-4)
[Fe(SCN)2-]
0
4.5 x 10-4
4.5 x10-4
Sample problem #2: Given the following equation:
H2(g) + I2 (g)  2 HI (g)
Calculate all three equilibrium concentrations when [H2]o and [I2]o = 0.200 M and [HI] o = 0 and Keq= 64.0.
Solution: The solution technique involves the use of an ICEbox.
a) Now, let’s fill in the initial row:
[H2]
0.200
Initial
Change
Equilibrium
[I2]
0.200
[HI]
0
b) Now, for the “change” row.
Initial
Change
Equilibrium
[H2]
0.200
-x
[I2]
0.200
-x
[HI]
0
+2x
The minus sign comes from the fact that the H2 and I2 amounts are going to go down as the reaction
proceeds. “x” signifies that some H2 and I2 will get used up, but we don’t know how much. But, we do
know that equal amounts will get used up because the mole to mole ration is 1:1. The “+” signifies that
more HI will be made as the reaction proceeds on its way to equilibrium. The “2” is important because HI
is being made twice as fast as H2 and I2 are being used up (molar coefficient).
c) Now, let’s do the math to complete the equilibrium row:
[H2 ] = 0.200 – x
[I2 ] = 0.200 – x
[HI ] = 0 + 2x = 2x
[H2 ]
[I2 ]
Initial
0.200
0.200
Change
-x
-x
Equilibrium 0.200 - x
0.200 - x
[HI]
0
+ 2x
2x
d) Now, we are ready to put the values into the equilibrium expression:
Keq =
[HI]2
[H2 ][I2]
64 =
(2x)2
(0.200 –x)(0.200 –x)
√64 = √ (2x)2________
√(0.200 – x)2
8 = 2X______
0.200 – x
8(0.200 – x) = 2x
1.6 – 8x = 2x
1.6 = 10x
0.160 M = x
Or 64 =
(2x)2
(0.200 –x)2
**Note: Here is where your algebra
skills are important!!!
Note: You can do this because you have a perfect square!
e) This is not the end, since the question asked for the equilibrium concentrations, so go back to the
ICEbox and look at the equilibrium row and then solve:
[H2 ] = 0.200 – x = 0.200 – 0.160 = 0.040 M
[I2 ] = 0.200 – x = 0.200 – 0.160 = 0.040 M
[HI] = 2x = 2(0.160) = 0.320 M
Reminder: Multiplying Binomials with F.O.I.L ------
Here's how to solve (4x + 6)(x + 2):
First - multiply the first term in each set of parenthesis:
 4x2
Outside - multiply the two terms on the outside:
 8x
(4x + 6)(x+2)
= 4x2 + 14x + 12
Inside - multiply both of the inside terms:
 6x
Last - multiply the last term in each set of parenthesis:
12
Sample problem #3: Given the equation, COCl2(g)  CO(g) + Cl2(g) Calculate all three
equilibrium concentrations when Keq = 0.680 with [CO]o= 0.500 and [Cl2]o = 1.00 M
Solution: The solution technique involves the use of an ICEbox.
[COCl2]
Initial
Change
Equilibrium
[CO]
[Cl2]
Answer:
Initial
Change
Equilibrium
Keq =
[COCl2]
0
+x
x
[CO]
0.500
-x
0.500 – x
[Cl2]
1.00
-x
1.00 – x
[CO][Cl2]
[COCl2]
0.680 = (0.500 – x)(1.00 – x )
x
0.680 x = 0.5 – 0.5x -1x + x2
0.680x = x2 – 1.5 x + 0.5
0 = x2 - 1.5 x – 0.68 x + 0.5
0 = x2 – 2.18 x + 0.5
****Put into form: ax2 + bx +c =0
x = ( -b   b2 – 4 ac )
2a
x = -(-2.18)   (-2.182)-(4 )( 1 )(0.5)
2
Then, using the quadratic formula;
x = (2.18  2.75
*****Gives two solutions!
2
x = (2.18+ 1.66)
or
x = (2.18 - 1.66)
2
2
= +1.92
= +0.26
We need to decide which the correct answer is:
 If we choose x = 1.92 this means:
[COCl2] = x = 1.92,
[CO] = 0.500 – x = 0.500 –1.92 = -1.42 OOPS! Impossible to get a negative (-)
concentration!
 So this leaves us with x = 0.26 this means:
[COCl2] = x = 0.26 M
[CO] = 0.500 – x = 0.500 –0.26 = 0.24M
[Cl2] = 1.00 – x = 1.00 - 0.26 = 0.74 M, which are the correct answers!
Equilibrium Practise:
Part A:
1. A chemist was studying the following reaction: SO2(g) + NO2(g) → NO(g) + SO3(g)
In a 1.0 L container, the chemist added 1.7 × 10−1 mol of SO2(g) to 1.1 × 10−1 mol of NO2(g). At
equilibrium, the concentration of SO3(g) was found to be 0.089 mol/L. What is the value of Kc for the
reaction at this temperature?
2. A 1.0 L reaction vessel contained 0.750 mole of CO(g) and 0.275 mole of H2O(g). After 1 h, equilibrium
was reached according to the equation:
CO(g) +H2O(g) → CO2(g) + H2(g) Analysis showed that 0.25 mol
of CO2 was present at equilibirum. What is Keq for this reaction?
3. A 0.80 M solution of hydrogen sulfide decomposes according to the equation:
2 H2S(g) → 2 H2(g) + S2(g)
At equilibrium the concentration of S2 was found to be 0.22 M. Find Keq for this reaction.
4. Suppose 0.50 mol of NOCl was introduced into an empty 1.00 L flask and allowed to come to
equilibrium according to the equation: 2 NOCl(g) → 2NO(g) + Cl2(g) If 0.10 mol of Cl2 was found at
equilibrium, what is Keq for the reaction?
5. In the following system, 4 HCl (g) + O2 (g)  2 H2O (g) + 2 Cl2 (g)
The initial concentration of water is 0.80 M and the initial concentration of chlorine is 1.2M. At equilibrium
the concentration of oxygen is 0.25 M. Find the equilibrium concentrations of HCl, H2O and Cl2.
6. For the system, N2 (g) + O2 (g)  2NO(g) , 1.0 M of N2 and 0.80 M of O2 are used to start the system. The
equilibrium concentration of O2 is 0.60 M. Determine the following.
a. equilibrium concentration of N2 ____________
b. equilibrium concentration of NO ____________
c. Calculate Keq __________________________
7. When 0.250 moles of CO2 is placed in a 1.00 liter container, some of it decomposes by the reaction
2CO2 (g)  2 CO (g) + O2 (g) . Write the equilibrium expression for the system. The equilibrium
concentration of CO is 0.040 mol/L. Calculate the following:
a. equilibrium concentration of O2 ______________________
b. equilibrium concentration of CO2 _____________________
c. Keq _________________________
Part B:
1. Phosgene, COCl2 (g), is an extremely toxic gas. It was used during World War I. Today it is used to
manufacture pesticides, pharmaceuticals, dyes, and polymers. It is prepared by mixing carbon monoxide
and chlorine gas. CO(g) + Cl2(g) → COCl2 (g) If 0.055 mol of CO(g) and 0.072 mol of Cl2(g) are placed
in a 5.0 L container, at a certain temperature, the equilibrium constant is 20. What are the equilibrium
concentrations of the mixture?
2. At 250C, Keq = 0.145 for the following reaction in carbon tetrachloride: 2BrCl(g) ↔ Br2(g) + Cl2(g). If
the initial concentrations of BrCl in the solution are 0.050 M, what will their equilibrium concentrations be?
3. Hydrogen bromide decomposes at 700 K. 2HBr(g) → H2(g) + Br2 (g) If Kc = 4.2 × 10−9 and 0.090
mol of HBr is placed in a 2.0 L reaction vessel and heated to 700 K. What is the equilibrium concentration
of each gas?
4. Calculate all three equilibrium concentrations when 0.500 mole each of H2 and Br2 are mixed in a
2.00L container and Keq = 36.0. (H2 + Br2  2 HBr)
Answers: A. 1) 4.66 2) 5 3) 0.33 4) 0.044 5) [HCl] = 1.0M, [H2O] = 0.300M, [Cl2] = 0.700M 6) [N2] = 0.800M [NO] = 0.400M Keq= 0.33 7) [O2]
= 0.020M, [CO2] = 0.210M, Keq= 7.6 x 10-4 B. 1) [CO]= 0.009M, [Cl2]= 0.0119M, [COCl2]=0.0025M 2) [BrCl]=0.0284M, [Br2] and [Cl2]=0.0108M 3)
[H2] and [Br2]= 2.89 x 10-6M [HBr]=0.045M 4) [H2] and [Br2]= 0.062M [HBr]=0.376M