Equilibrium Calculations
Strategy for Equilibrium Calculations
 Reinterpret the problem.
 Make required assumptions.
 Equate quantities according to chemical relationship.
 Solve the equations, and make approximations if necessary.
 Check for validity of the answers.
Equilibrium Calculations
Equilibrium calculations evaluate the quantities in a system that may or may not be at equilibrium. When a set of conditions is given, usually one or more quantities are to be evaluated. From a given set of conditions, we need to reinterpret the problem to prepare equations. Understanding the problem is a major step in problem solving. When the conditions for a problem are incomplete, assumptions can be made to supplement the missing conditions.
You have learned the various expressions of the equilibrium constant in the previous modules. The expression gives the quantitative relationship of various compounds in a system. An unknown quantity can be represented by a symbol. The relationship is then an equation. The unknown quantities can thus be evaluated. We will illustrate how these are done by examples.
By the way, you may recall the modules on stoichiometry and on chemical kinetics. Both topics also involve chemical calculations. The chemical relationships among various quantities of material are complicated. Stoichiometric calculations evaluate composition, quantities of reactants required and quantities of products formed. Kinetic calculations evaluate rates of reactions and dependence of rates on temperature, pressures and concentrations of reactants and products.
Example 1
For the gas phase reaction
\(\ce{H_{2\large{(g)}} + I_{2\large{(g)}} \rightleftharpoons 2 HI_{\large{(g)}}}\)
K_{c} = 50.3 at 731 K. Equal amounts (0.100 M each) are introduced to a container, and then the temperature is raised to 731 K.
Is the system at equilibrium? If not, what direction is the reaction?
SOLUTION
When equal amounts are present,
\(Q_{\large c} = \ce{\dfrac{[HI]^2}{[H2] [I2]}} = \dfrac{(0.1)^2}{(0.1)^2} = 1 < K_{\large c}\: (50.3)\)
There is a tendency to increase \(\ce{[HI]}\) to reach equilibrium at 731 K.
Example 2
For the gas phase reaction
\(\ce{H_{2\large{(g)}} + I_{2\large{(g)}} \rightleftharpoons 2 HI_{\large{(g)}}}\)
K_{c} = 50.3 at 731 K. Equal amounts (0.100 M each) are introduced to a container, and then the temperature is raised to 731 K.
Calculate the concentration of each when the system is at equilibrium.
SOLUTIONFor convenience, we write down the reaction equation and put the concentration of the species below the formula. Since the \(\ce{[HI]}\) concentration will increase, we assume x M of \(\ce{H2}\) or \(\ce{I2}\) to react to give 2x M of \(\ce{HI}\).
\(\begin{array}{cccccl}
\ce{H_{2\large{(g)}} &+ &I_{2\large{(g)}} &\rightleftharpoons &2 HI_{\large{(g)}} &\: }\\
0.100 & &0.100 & &0.100 &\textrm{initial concentration}\\
x & &x & & 2x &\textrm{amounts change}\\
0.1x & &0.1x & &0.1+2x &\ce{equilibrium\:\, concentration}
\end{array}\)
To simplify notation, we ignore the number of significant figures in the formulations. By definition of equilibrium constant, we have
\(K_{\large c} = \ce{\dfrac{[HI]^2}{[H2] [I2]}} = \dfrac{(0.1+2x)^2}{(0.1x)(0.1x)} = 50.3\)
Expanding the above equation gives,
\(46.3 x^2  10.43x + 0.493 = 0\)
Please expand the equation to see if you get the same result. Many students have difficulty in deriving or simplifying this equation.
Solution of the quadratic equation gives x = 0.067 and x' = 0.158. When x' = 0.158, 0.1  0.158 = 0.058. Since concentration should not be a negative value, this solution does not agree with reality. The only acceptable solution is x = 0.067, and thus we have:
\(\mathrm{[H_2] = [I_2] = 0.100  0.067 = 0.033\: M}\)
\(\ce{[HI]} = 2x = 0.100 + 0.134 = \mathrm{0.234\: M}\)
It is always a good idea to check the validity of the answer. We use the result to calculate the equilibrium constant.
\(\dfrac{0.234^2}{0.033^2} = 50.3 = K_{\large c}\).
It is helpful to review the beginning of this page on the problem solving strategy.
Example 3
At 1100 K, K_{c} = 4.20e6 for the gas phase reaction,
\(\ce{2 H2S_{\large{(g)}} \rightleftharpoons 2 H_{2\large{(g)}} + S_{2\large{(g)}}}\)
What concentration of \(\ce{S2}\) can be expected when 0.200 mole of \(\ce{H2S}\) comes to equilibrium at 1100 K in an otherwise empty 1.00L vessel?
SOLUTIONWe write the equation and place quantities at the initial condition and at various stages below the formula. We assume x mol of \(\ce{S2}\) is formed, and this leads to the formation of 2x moles of \(\ce{H2}\). Since 2x moles of \(\ce{H2S}\) is required, the equilibrium concentration of \(\ce{H2S}\) is 0.200  2x. Thus, we have
\(\begin{array}{cccccl}
\ce{2 H2S_{\large{(g)}} &\rightleftharpoons &2 H_{2\large{(g)}} &+ &S_{2\large{(g)}} &\:}\\
0.200 & &0 & &0 &\ce{initial\: condition}\\
2x & &2x & &x &\ce{assume\: x\: mole\: of\: S2\: is\: formed}\\
0.2002x & &2x & &x &\ce{equilibrium\:\, concentration}
\end{array}\)
The definition of K_{c} leads to the equation,
\(K_{\large c} = \ce{\dfrac{[H2]^2 [S2]}{[H2S]^2}} = \dfrac{(2x)^2 (x)}{ (0.2002x)^2} = 4.20\textrm{e}6\)
This is a cubic equation, and there is no general method to solve this type of equations. Fortunately, since the equilibrium constant is very small, we expect x to be a small value. Thus, 0.200  2x is almost 0.200. With this approximation, we have a simpler equation to solve:
\(\begin{align}
4 x^3 &= (0.200)^2 \times 4.20\textrm{e}6\\
&= 1.68\textrm{e}7\\
x^3 &= \dfrac{1.68\textrm{e}7}{4} = 4.2\textrm{e}8\\
x &= (4.2\textrm{e}8)^{1/3}\\
&= 3.5\textrm{e}3 = \ce{[S2]}
\end{align}\)
\(\ce{[H2]} = 2 x = 7.0\textrm{e}3\)
DISCUSSION
Is the approximation justified? We can calculate the equilibrium constant from the equilibrium concentrations.
\(\dfrac{4 (0.0035)^3}{(0.2000.007)^2} = 4.6\textrm{e}6\)
The error in K_{c} = (4.64.2)e6/4.2e6 = 4.8%. This is within 5 % criterion, but we are approaching the limit for using approximation. In the approximation, we used 0.200 M instead of 0.193 M as \(\ce{[H2S]}\).
In this example, you have learned the approximation technique.
Questions
 A 1.00L container contains 1.00 M of phosgene, which decomposes according to the reaction,
\(\ce{COCl_{2\large{(g)}} \rightleftharpoons CO_{\large{(g)}} + Cl_{2\large{(g)}}}\)
At equilibrium, the concentration of \(\ce{Cl2}\) is 0.028 M. What is the concentration of \(\ce{CO}\)?
 A 1.00L container contains 1.00 M of phosgene, which decomposes according to the reaction,
\(\ce{COCl_{2\large{(g)}} \rightleftharpoons CO_{\large{(g)}} + Cl_{2\large{(g)}}}\)
At equilibrium, the concentration of \(\ce{Cl2}\) is 0.028 M. What is the equilibrium constant?
 A 2.00L container contains 1.00 mole each of \(\ce{H2}\) and \(\ce{I2}\) gases. When the system reaches equilibrium, the concentration of \(\ce{I2}\) is 0.11. The equilibrium equation is
\(\ce{H_{2\large{(g)}} + I_{2\large{(g)}} \rightleftharpoons 2 HI_{\large{(g)}}}\)
What is the concentration of \(\ce{HI}\)?
 A 2.00L container contains 1.00 mole each of \(\ce{H2}\) and \(\ce{I2}\) gases. When the system reaches equilibrium, the concentration of \(\ce{I2}\) is 0.11. The equilibrium equation is
\(\ce{H_{2\large{(g)}} + I_{2\large{(g)}} \rightleftharpoons 2 HI_{\large{(g)}}}\)
What is the equilibrium constant?
 A 1.00L container contains 1.00 mole each of \(\ce{H2}\) and \(\ce{I2}\) gases. The equilibrium constant K_{c} = 50 for the equilibrium
\(\ce{H_{2\large{(g)}} + I_{2\large{(g)}} \rightleftharpoons 2 HI_{\large{(g)}}}\)
What is the concentration of \(\ce{H2}\)?
Solutions

0.028
From the equation, \(\mathrm{[CO] = [Cl_2]}\).Discussion...
What is the equilibrium concentration of phosgene? 
8.0e4
\(K_{\large c} = \dfrac{(0.028)^2}{10.028} =\: ?\)
From the same condition, we can calculate several quantities. 
0.78 M
The initial concentration of \(\ce{H2}\) and \(\ce{I2}\) is 0.50 M. \(\mathrm{[HI] = 2\times(0.500.11)\: M}\)Discussion...
Can the equilibrium constant be calculated? 
50.3 M
At equilibrium, the concentrations are \(\ce{[H2]} = \ce{[I2]} = 0.11\), and \(\ce{[HI]} = 2\times(0.500.11) =\: ?\) Use these values to calculate K_{c}. 
0.22 M
\(\begin{alignat}{5}
\ce{H_{2\large{(g)}} &+ &I_{2\large{(g)}} &\rightleftharpoons &&\:2 HI_{\large{(g)}}}\\
1.0x &&1.0x & &&\:\:\:\:2x
\end{alignat}\)Solve the equation,
\(\dfrac{(2x)^2}{(1.0x)^2} = 50\)
A easy way to solve the equation is to take square roots on both sides of the equation as a first step.Discussion...
What is \(\ce{[H2]}\) if the container has a volume of 2.0 L?
Contributors
Chung (Peter) Chieh (Chemistry, University of Waterloo)