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?
When equal amounts are present,
[HI]2 (0.1)2 Qc = ---------- = -------- = 1 < Kc (50.3) [H2] [I2] (0.1)2There is a tendency to increase [HI] to reach equilibrium at 731 K.
Calculate the concentration of each when the system is at equilibrium.
H2(g) + I2(g) = 2 HI(g) 0.100 0.100 0.100 initial concentration -x -x 2x amounts change 0.1-x 0.1-x 0.1+2x equilibrium concentrationTo simplify notation, we ignore the number of significant figure in the formulations. By definition of equilibrium constant, we have
[HI]2 (0.1+2x)2 Kc = ---------- = -------------- = 50.3 [H2] [I2] (0.1-x)(0.1-x)Expanding the above equation gives,
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:
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 H2S(g) = 2 H2(g) + S2(g) 0.200 0 0 initial condition -2x 2x x assume x mole of S2 is formed 0.200-2x 2x x equilibrium concentration
The definition of Kc leads to the equation,
[H2]2 [S2] (2x)2 (x) Kc = ----------- = ------------ = 4.20e-6 [H2S]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 - 2x is almost 0.200. With this approximation, we have a simpler equation to solve:
Is the approximation justified? We can calculate the equilibrium constant from the equilibrium concentrations.
The error in Kc = (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 [H2S].
In this example, you have learned the approximation technique.
From the equation, [CO] = [Cl2].
What is the equilibrium concentration of phosgene?
Kc = (0.028)2/(1-0.028) = ?
From the same condition, we can calculate several quantities.
Hint 0.78 M
The initial concentration of H2 and I2 is 0.50 M. [HI] = 2*(0.50-0.11) M
Can the equilibrium constant be calculated?
Hint 50.3 M
At equilibrium, the concentrations are [H2] = [I2] = 0.11, and [HI] = 2*(0.50-0.11) = ? Use these values to calculate Kc.
Hint 0.22 M
H2(g) + I2(g) = 2 HI(g). 1.0-x 1.0-x 2xSolve the equation,
What is [H2] if the container has a volume of 2.0 L?