Equilibrium – Solving Equilibrium Problems
Date______________________
Knowing the equilibrium constant allows us to predict and calculate what the equilibrium concentration of each
substance will be at equilibrium. This works for all three types of equilibrium (phase, solubility, and chemical).
These questions can involve some tricky algebra, but there are some short cuts we will learn to help us solve
them.
We will look at three different types of equilibrium problems. However, in all cases, the question is solved
exactly the same way, by following these steps:
1.
2.
3.
Problem 1 – Perfect Squares
Consider the reaction of carbon monoxide and water to form carbon dioxide and hydrogen.
CO(g) + H2O(g) ↔ CO2(g) + H2(g)
Suppose you start with 1.00 mol each of CO and H2O in a 50.0 L vessel. How many moles of each substance are
in the equilibrium mixture if the Kc is 0.58?
Homework
Page 472 #5,6
Problem Sheet
Equilibrium Constant Problems
1. What is the equilibrium composition of a reaction mixture if you start with 0.055 mol each of H2 and I2 in a 1.0
L vessel? The reaction is a synthesis. The Kc for this reaction is Kc = 49.7 at 458oC
[Answer: [H2] = [I2] = 1.2 × 10-2 M, [HI] = 8.6 × 10-2 M]
2. Iodine and bromine react to give iodine monobromide, IBr. What is the equilibrium composition of a mixture
at 150oC that initially contained 0.0015 mol each of iodine and bromine in a 5.0 L vessel? The equilibrium
constant Kc for this reaction at 150oC is 1.2  102.
[Answer: [I2] = [Br2] = 4.6  10-5 M, [IBr] = 5.1  10-4 M]
3. Initially a mixture contains 0.850 mol each of N2 and O2 in an 8.00 L vessel. Find the composition of the
mixture when equilibrium is reached at 3900oC. N2 and O2 react to form nitrogen monoxide. Kc at 3900oC is
0.0123.
[Answer: [N2] = [O2] = 0.101 M, [NO] = 0.0112 M]
Equilibrium – Equilibrium problems and the Hundred Rule
Nitrosyl chloride, NOCl(g), decomposes to form nitrogen monoxide, NO(g), and chlorine gas, Cl2(g), according to the
equation:
2NOCl(g) ↔ 2NO(g) + Cl2(g)
At 35 C, Kc = 1.60  10 . Calculate the concentration of all entities at equilibrium if 0.80 mol NOCl(g) is placed in
a 2.00 L container.
o
-5
The steps to solving this problem are the same as previously learned:
1. Set up an ICE table
2. Substitute the expressions for concentrations in the equilibrium constant expression
3. Solve for x.
Step 1 – Set up the ICE table:
Units = mol/L
2NOCl(g)
Initial
0.80/2.00 = 0.40 M
Change
-2x
Equilibrium
0.40 – 2x
2NO(g)
0
+2x
2x
Cl2(g)
0
+x
x
Step 2 – Write the equilibrium equation with the equilibrium concentration expressions:
[NO]2 [Cl2 ]
[NOCl]2
(1)
𝐾𝑐 =
(2)
(2x)2 (x)
𝐾𝑐 =
(0.40 - 2x)2
(3)
𝐾𝑐 =
In equation 3, we can see that we have a problem. There is a cubed variable
in this equation, which we cannot solve (without very complicated math).
However there is a strategy we can employ to solve this problem.
4x 3
(0.40 - 2x)2
Strategy to Solve this Problem
The equilibrium expression that results from this particular equilibrium cannot be solved by taking the square
root of both sides to eliminate the x2 variable. However, there is a small ‘trick’ to solving this problem.
Notice that Kc is very small. What this means is that this equilibrium strongly favours the reactants. In other
words, very little reactant will be converted into product.
We can make a rule of thumb here:
Hundred Rule
If the initial concentration of reactant divided by the equilibrium constant gives a value of greater than 100,
then we can approximate any change ‘x’ to be equal to zero wherever it is subtracted/added.
𝑖𝑓
[𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡]
> 100 𝑡ℎ𝑒𝑛 𝑥 𝑖𝑠 𝑛𝑒𝑔𝑙𝑖𝑔𝑖𝑏𝑙𝑒 𝑤ℎ𝑒𝑛 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑒𝑑 𝑜𝑟 𝑎𝑑𝑑𝑒𝑑.
𝐾𝑐
Step 3 – Solve for x.
(3)
4x 3
𝐾𝑐 =
(0.40 - 2x)2
We can perform a check of the hundred rule on this equation:
[𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡]
0.40 𝑀
=
= 25,000 > 100
𝐾𝑐
1.60 × 10−5
(4)
Therefore, since the hundred rule is confirmed we can say that:
(0.40 − 2𝑥)2 ≅ (0.40)2
Since the value of x is so small
4x 3
𝐾𝑐 =
(0.40)2
4x 3
=
0.16
(5)
1.60 × 10
(6)
2.56 × 10−6 = 4x 3
Cross multiplying the denominator of the right side
(7)
6.4 × 10−7 = x 3
Dividing by the coefficient on the variable
(8)
8.62 × 10−3 = x
Taking the cube root of both sides
−5
Substituting the value of Kc in and calculating the squared value
Now we can substitute in the value of x = 8.62 x 10-3 into our equilibrium expressions:
[NOCl] = 0.40 − 2𝑥 = 0.40 − 2(8.62 × 10−3 ) = 0.38276 ≅ 0.38 M
[NO] = 2𝑥 = 2(8.62 × 10−3 ) = 0.01724 ≅ 1.7 × 10−2 M
[Cl2 ] = 𝑥 = 8.62 × 10−3 ≅ 8.6 × 10−3 M
Using 2 significant digits for our finals answers based on the given value for the initial amount of NOCl of 0.80
mol which has only 2 significant digits.
Verification of the Hundred Rule
At this point, we should check the percentage change in the concentration of our reactants. With the Hundred
Rule, we made an assumption that the amount of reactant converted to product was negligible. As long as the
percentage change in the initial concentration is less than 5%, we are ok to make this assumption.
% 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 =
𝑖𝑛𝑖𝑡𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 − 𝑓𝑖𝑛𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡
× 100%
𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡
% 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 =
0.40 M − 0.38 M
× 100%
0.40 M
% 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 5%
Therefore, since our percentage difference is within 5%, the assumption is valid (if only just barely!).
Equilibrium Problems with the Hundred Rule
1. In a 10.0 L reaction vessel, 5.00 mol of hydrogen gas and 8.00 mol of iodine gas are added at 58 oC. If the Kc at
that temperature is 3.50×10-6, what concentration of hydrogen iodide gas will be produced at equilibrium?
[Ans – [H2] = 0.499 M [I2] = 0.799 [HI] = 1.18 x 10-3 M]
2. In a 25.0 L vessel, 30.0 mol of chlorine gas is added. Then 20.0 mol of hydrogen gas is added to produce
hydrogen chloride gas. How much hydrogen will remain in the vessel at equilibrium if the equilibrium constant
at 250oC is 7.85×10-5?
[Ans – [Cl2] = 1.20 M [H2] = 0.800 [HCl] = 8.68 x 10-3 M]
Textbook – Page 476 #7, 8
Equilibrium – Equilibrium Problems Using Quadratic Formula
A chemical equilibrium exists in a closed container wherein phosphorous trichloride and chlorine gas are in
equilibrium with the product phosphorous pentachloride.
PCl5 ↔ Cl2 + PCl3
If the initial amounts of phosphorous trichloride and chlorine gas were 0.50 M and 0.070 M respectively, and
the equilibrium constant Kc = 96.2 at 400 K, what are the equilibrium concentrations of all 3 chemical species?
This problem is still solved just like before.
1. Set up an ICE table
2. Write the equilibrium expression and substitute in the equilibrium concentration expressions.
3. Solve for x.
Step 1 – Set up the ICE table:
Units = mol/L
PCl5(g)
Initial
0.50
Change
-x
Equilibrium
0.50 – x
Cl2(g)
0
+x
x
PCl5(g)
0
+x
x
Step 2 – Write the equilibrium equation with the equilibrium concentration expressions:
We can perform a check of the hundred rule on this
equation:
𝑥∙𝑥
0.50 − 𝑥
(1)
𝐾𝑐 =
(2)
96.2 =
[𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡] 0.050 𝑀
=
= 5.2 × 10−4 < 100
𝐾𝑐
96.2
𝑥2
0.50 − 𝑥
Therefore, we cannot use the hundred rule here.
To solve this problem, we will need to employ the quadratic formula. Recall that the quadratic equation has the
general formula of:
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0
Which has the solution of:
𝑥=
−𝑏 ± √𝑏 2 − 4𝑎𝑐
2𝑎
Therefore, you must reorganize the equilibrium expression into the form of ax2 + bx + c = 0.
(3)
48.1 − 96.2𝑥 = x 2
Cross multiply the denominator of the right hand side
(4)
0 = x 2 + 96.2𝑥 − 48.1
Put the equation into the form of ax2 + bx + c = 0
(5)
𝑥=
−𝑏 ± √𝑏 2 − 4𝑎𝑐
2𝑎
Solve using the quadratic equation: a = 1, b = 96.2, c = 48.1
(6)
(7)
𝑥=
−96.2 ± √96.22 − 4(1)(−48.1)
2(1)
𝑥=
−96.2 √96.22 − 4(−48.1)
±
2
2
It’s useful to separate the equation into two fractions
separated by the ± sign.
(8)
𝑥 = −48.1 ± 48.59742792
We can see that if we choose the negative sign, we will
get a negative answer, which is meaningless.
(9)
𝑥 = 0.49743 or -96.697
The quadratic equation has two solutions. Only one of
the solutions will result in realistic equilibrium
concentrations.
Now we can substitute in the value of x = 0.49743 into our equilibrium expressions:
[PCl5 ] = 0.50 − 𝑥 = 0.50 − (0.49743) = 2.57208 × 10−3 ≅ 2.6 × 10−3 M
[Cl2 ] = [PCl3 ] = 𝑥 = 0.49743 ≅ 5.0 × 10−1 M
Equilibrium Problems with Quadratic Equation
1. In a sealed container, nitrogen dioxide is in equilibrium with dinitrogen tetroxide.
2NO2(g) ↔ N2O4(g)
Kc = 1.15 at 55oC
Find the equilibrium concentration of nitrogen dioxide and dinitrogen tetroxide if the initial concentration of
nitrogen dioxide is 0.650 M.
[Ans. – [NO2(g)] = 0.357 M [N2O4(g)] = 0.147 M]
2. Hydrogen and iodine gas placed in a sealed container at 490oC will form hydrogen iodide. The equilibrium
constant for this equilibrium mixture is 46.0. If initially 0.40 mol of hydrogen and 0.6 mol of iodine are injected
into a 500. mL electrically heated reaction vessel, what are the concentrations of all the entities at equilibrium?
[Ans. – [H2] = 0.090 M [I2] = 0.49 M [HI] = 1.4 M]
Additional Textbook Problems
Page 480 #9, 10