- 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.

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.

Is the system at equilibrium? If not, what direction is the reaction?

*Solution*

When equal amounts are present,

[HI]There is a tendency to increase [HI] to reach equilibrium at 731 K.^{2}(0.1)^{2}Q_{c}= ---------- = -------- = 1 <K(50.3) [H_{c}_{2}] [I_{2}] (0.1)^{2}

**
Calculate the concentration of each when the system is at equilibrium.
**

HTo simplify notation, we ignore the number of significant figure in the formulations. By definition of equilibrium constant, we have_{2}(g) + I_{2}(g) = 2 HI(g) 0.100 0.100 0.100 initial concentration -x-x2xamounts change 0.1-x0.1-x0.1+2xequilibrium concentration

[HI]Expanding the above equation gives,^{2}(0.1+2x)^{2}K= ---------- = -------------- = 50.3 [H_{c}_{2}] [I_{2}] (0.1-x)(0.1-x)

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:

[HI] = 2

It is always a good idea to check the validity of the answer. We use the result to calculate the equilibrium constant.

It is helpful to review beginning of this page on the problem solving strategy.

2 H_{2}S(g) = 2 H_{2}(g) + S_{2}(g) 0.200 0 0 initial condition -2x2xxassumexmole of S_{2}is formed 0.200-2x2xxequilibrium concentration

The definition of *K _{c}* leads to the equation,

[H_{2}]^{2}[S_{2}] (2x)^{2}(x)K= ----------- = ------------ = 4.20e-6 [H_{c}_{2}S]^{2}(0.200-2x)^{2}

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 - 2*x* is almost 0.200. With this approximation, we have
a simpler equation to solve:

= 1.68e-7

= 3.5e-3 = [S

[H

*
Discussion*:

Is the approximation justified? We can calculate the equilibrium
constant from the equilibrium concentrations.

The error in *K _{c}* = (4.6-4.2)e-6/4.2e-6 = 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 [H

In this example, you have learned the approximation technique.

**A 1.00-L container contains 1.00 M of phosgene, which decomposes according to the reaction,**COCl At equilibrium, the concentration of Cl_{2}(g) = CO(g) + Cl_{2}(g)._{2}is 0.028 M. What is the concentration of CO?0.028*Hint*

From the equation, [CO] = [Cl_{2}].*Discussion...*

What is the equilibrium concentration of phosgene?**A 1.00-L container contains 1.00 M of phosgene, which decomposes according to the reaction,**COCl At equilibrium, the concentration of Cl_{2}(g) = CO(g) + Cl_{2}(g)._{2}is 0.028 M. What is the equilibrium constant?8.0e-4*Hint*

*K*= (0.028)_{c}^{2}/(1-0.028) = ?

From the same condition, we can calculate several quantities.**A 2.00-L container contains 1.00 mole each of H**_{2}and I_{2}gases. When the system reached equilibrium, the concentration of I_{2}is 0.11. The equilibrium equation isH What is the concentration of HI?_{2}(g) + I_{2}(g) = 2 HI(g)0.78 M*Hint*

The initial concentration of H_{2}and I_{2}is 0.50 M. [HI] = 2*(0.50-0.11) M**Discussion...**

Can the equilibrium constant be calculated?**A 2.00-L container contains 1.00 mole each of H**_{2}and I_{2}gases. When the system reached equilibrium, the concentration of I_{2}is 0.11. The equilibrium equation isH What is the equilibrium constant?_{2}(g) + I_{2}(g) = 2 HI(g)50.3 M*Hint*

At equilibrium, the concentrations are [H_{2}] = [I_{2}] = 0.11, and [HI] = 2*(0.50-0.11) = ? Use these values to calculate*K*._{c}**A 1.00-L container contains 1.00 mole each of H**_{2}and I_{2}gases. The equilibrium constant*K*= 50 for the equilibrium_{c}H What is the concentration of H_{2}(g) + I_{2}(g) = 2 HI(g)._{2}?0.22 M*Hint*

H

Solve the equation,_{2}(g) + I_{2}(g) = 2 HI(g). 1.0-*x*1.0-*x*2*x*(2 A easy way to solve the equation is to take square roots on both sides of the equation as a first step.*x*)^{2}/(1.0-*x*)^{2}= 50.*Discussion...*

What is [H_{2}] if the container has a volume of 2.0 L?