Example: Use of reaction quotient to predict direction of a reaction

Which way will each mixture react? Assume K_c = 96.2 M^-1 for the reaction A + B <-> C.

Case 1: [A]₀ = 0.0933 M, [B]₀ = 0.123 M, [C]₀ = 0.389 M

Case 2: [A]₀ = 0.0233 M, [B]₀ = 0.0123 M, [C]₀ = 0.999 M

Case 3: [A]₀ = 0.0333 M, [B]₀ = 0.0333 M, [C]₀ = 0.1067 M

Solution: Calculate Q_c for each case and compare to K_c = 96.2 M^-1 . For case 1,

Since Q_c < K_c , (not enough product or too much reactants) this reaction will move forward. For case 2,

Since Q_c > K_c , (not enough reactants or too much product) this reaction will move backward. For case 3,

Since Q_c = K_c , this system is at equilibrium and will stay as it is.

Our fourth type of equilibrium problem, temperature changes, requires discussion of thermodynamics. By referring to the standard enthalpy of formation tables in any freshman chemistry book, we can find the change in enthalpy for the chemical reaction, and use this value to predict how the reaction equilibrium changes with temperature.

As a review, let’s calculate the standard enthalpy (at 298K = 25^oC) for the reaction we have been considering so far:

The standard enthalpies of formation for each component are:

Using the standard (products – reactants), we get:

This number will be used below.

The equilibrium constant is a function of temperature, which is expressed in the Van’t Hoff equation:

Example: Finding a new equilibrium at a different temperature

Suppose the above reaction, abbreviated again to

If we take this system and change the temperature to 298 K, what will be the new equilibrium concentrations?

When we change the temperature, the old K_c becomes the new Q_c because the reaction is no longer at equilibrium. Our new K_c is much larger after the temperature change, and so we expect the equilibrium to shift to the right (forward reaction predominates).

We can now set up a new start-change-finish diagram:

So,

And we have

Plugging these values into the quadratic equation gives

The first solution gives a negative (physically impossible) value for [A] and [B] so we use the second solution

x = 0.0324 M

[A] = [B] = a - x = 0.0333 – 0.0324 M = 0.0009M

[C] = b + x = 0.1067 + 0.0324 M = 0.1391 M

K_c = 1.72x10⁵. This should not be too surprising, given what we have seen before and the fact that when we calculated [A] and [B], we took the difference between two small numbers that were close together.

This type of calculation is a good candidate for trial and error. Let’s summarize what happens if we do that. The algebra above yields

Starting with x = 0, we got the successive values 0.03301, 0.03297, 0.03297 and stopped there. Now,

x = 0.03297 M

[A] = [B] = a - x = 0.03330 – 0.03297 M = 0.0003266M

[C] = b + x = 0.1067 + 0.03297 M = 0.1397 M

The last theoretical point we need to cover is that of pressure changes. Since by Le Chatelier's rule, the system tries to undo what we do to it, we can state the pressure rule simply: increases in pressure favor the side of the reaction with the fewest moles in the gas phase. In our example reaction, the product side is favored when pressure is increased. It makes the most sense to do these problems in terms of partial pressures, although they can be done in concentrations as well.

Example: Calculation of K_P from K_c

If the value of K_c at 400 K is 96.2 M^-1, what is the value of K_P ?

From the definitions,

Example: Calculation of partial pressures from concentrations

Suppose as before

[A]₀ = [B]₀ = 0.0333 M

[C]₀ = 0.1067 M

What are the corresponding partial pressures and the total pressure? The temperature is again 400 K.

For each component, P_X = [X]RT, so

Example: Effect of pressure change on equilibrium

Suppose the above system suddenly experienced a pressure drop to 4.00 atm at a constant temperature of 400 K. What would the new equilibrium partial pressures become?

Solution:

This problem is interesting because if the total pressure suddenly changes, so do all the partial pressures.

What does not change (until the system goes to equilibrium, which it has not done immediately after the pressure change) is the relative amounts of each component, in other words, the mole fractions (see definitions).

After the pressure change,

Equilibrium will cause a shift to the left, in order to make more gas and so more pressure, since we reduced the pressure and the system tries to undo what we did to it.   Setting up a start-change-finish diagram,

This gives

Plugging in the numbers for each coefficient yields

And so the quadratic solutions become

Rejecting the obviously impossible negative solution. Therefore, the pressures are:

P_A = P_B = a + x = 0.768 + 0.125 atm = 0.893 atm

P_C = b - x = 2.46 - 0.125 atm = 2.34 atm

P = P_A + P_B + P_C = 0.893 + 0.893 + 2.34 atm = 4.13 atm

Plugging these values into as a check gives

So we can be confident our answer is correct.

All other chemical equilibrium problems are variations of the techniques shown in these notes, or combinations of these techniques. By practicing with these problems until new problems can be done easily and without errors (in other words, the check you do at the end agrees with your results), you will be able to perform well on your next chemistry exam.