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Chemical Equilibrium Examples II

**Reaction Quotient
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**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^{o}C) for
the reaction we have been considering so far:

The standard enthalpies of formation for each component are:

- DH
^{o}_{f}(PCl_{3}) = -306.4 kJ mol^{-1} - DH
^{o}_{f}(Cl_{2}) = 0 kJ mol^{-1} - DH
^{o}_{f}(PCl_{5}) = -398.9 kJ mol^{-1}

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:

Either version may be used, depending on which is easier to solve for the desired
variable. Note that there are two K’s and two T’s, to be understood that K_{2}
goes with T_{2} and K_{1} goes with T_{1} . The enthalpy of
reaction DH^{o}_{rxn}
is considered constant. In reality, this number varies some with temperature, so this
equation is an *approximate* form of a much more complicated equation and should be
used over relatively *small temperature ranges* (no more than +/- 200K about 298K).
Also, please note that exp(x) = e^{x}.

Note that there are five different variables in this equation: DH^{o}_{rxn }, T_{1 },
T_{2 }, K_{1} and K_{2 }. Since the equation can be solved for any
one of them (more on the T’s later), this equation alone can generate at least three
completely different kinds of problems, solving for DH^{o}_{rxn} , K’s and T’s.

**Example:** Finding a new equilibrium at a different temperature

Suppose the above reaction, abbreviated again to

Is at equilibrium at 400 K. As before, the equilibrium constant is 96.2 M^{-1}
at this temperature. The equilibrium concentrations are

- [A] = [B] = 0.0333 M = a
- [C] = 0.1067 M = b

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

Solution:

The first thing to do is find the value of K_{c} at 298 K. To do this, use the
first version of the Van’t Hoff equation and the value of DH^{o}_{rxn} we calculated above

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:

Component |
A |
B |
C |

Start | [A]_{0} = a |
[B]_{0} = a |
[C]_{0} = b |

Change |
- x |
- x |
+ x |

Finish | a - x | a - x | b + x |

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**

If these values are plugged in to check K_{c} we get an
answer which is terribly far from our original value,

K_{c} = 1.72x10^{5}. 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**

Plugging these values into the definition of K_{c} yielded 1.31x10^{6}, as expected. The moral is: watch out for
situations like this, which will ruin your precision! The difference of two small numbers
which are close together is always such a case.

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}
?

Solution:

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.

Solution:

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,

Component |
A |
B |
C |

Start | P_{A0} = a |
P_{B0} = a |
P_{C0} = b |

Change |
+ x |
+ x |
- x |

Finish | a + x | a + x | b - x |

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.

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