# M14Q3: Combining Reactions and their Equilibrium Constants

Learning Objectives

- Write an equilibrium constant expression, such as K
_{c}, in terms of reactants and products and the stoichiometry of the reaction. - Describe how an equilibrium constant for a balanced chemical equation would change if different stoichiometric coefficients were used in the balanced chemical equation, if the reaction were reversed, or if several equations were added together to give a new net equation, and calculate the new equilibrium constant.

Now that we can write equilibrium reactions and expressions (*K _{eq}*), what happens if we change all the coefficients in the reactions? What about if two reactions are happening to create a third, unknown reaction, and we want to know if that third reaction is reactant or product favored? In this section, we will be looking at how to manipulate the equilibrium constant in these scenarios.

# Equilibrium Constant Expressions and Stoichiometry

Consider the reaction:

Reaction 1: 2 NOCl(g) ⇌ 2 NO(g) + Cl_{2}(g) K_{eq,1} =

This reaction and its resulting equilibrium constant will represent the ratio between the products and reactants at equilibrium. But what if we were more interested in studying one mole of NOCl reacting to form NO and Cl_{2}? If we divide the entire reaction by 2, the equilibrium reaction and equilibrium constant expression would be:

Reaction 2: NOCl(g) ⇌ NO(g) + ½ Cl_{2}(g) K_{eq,2} =

Notice the difference between the two equilibrium constant expressions. The products are still in the numerator and reactants in the denominator, but the exponents that each species is raised to differs because the coefficients are different. And in fact each coefficient in *K _{eq,2}* is half that of

*K*, or each species is raised to the ½ power.

_{eq,1}K_{eq,1} = (K_{eq,2})^{½}

# Equilibrium Constant Expressions and Reversing Reactions

Sometimes we need to flip a reaction in order to properly represent the system we are trying to study. An example would be if we wanted to study:

Reaction 1: 2 NOCl(g) ⇌ 2 NO(g) + Cl_{2}(g) K_{eq,1} =

But we only had information about the opposite reaction:

Reaction 2: 2 NO(g) + Cl_{2}(g) ⇌ 2 NOCl(g) K_{eq,2} =

How do these two equilibrium constant expressions differ?

K_{eq,1} = K_{eq,2} =

When we flip a reaction, the reactants and the products switch sides, so the numerator and the denominator in the equilibrium constant expression also switch and they are inverses of eachother.

K_{eq,1} =

# Equilibrium Constant Expressions for Multiple Reactions

Suppose we wanted to study the reaction:

Reaction 1: 2 NOCl(g) + O_{2}(g) ⇌ N_{2}O_{4}(g) + Cl_{2}(g) K_{eq,1} =

Unfortunately for us, this is a completely unstudied reaction and we have no idea if this reaction is reactant- or product-favored, whether it will occur fast or slow, and whether it is safe for us to even perform the reaction! Luckily, we have two other similar reactions that we can use to give us insight into our desired reaction.

Reaction 2: 2 NOCl(g) ⇌ 2 NO(g) + Cl_{2}(g) K_{eq,2} =

Reaction 3: 2 NO(g) + O_{2}(g) ⇌ N_{2}O_{4}(g) K_{eq,3} =

Notice that if we add Reaction 2 and Reaction 3 together, we are able to get our desired reaction, Reaction 1. But how do we know what the value of K_{eq,1} is? Let’s compare K_{eq,1}, K_{eq,2}, and K_{eq,3} a little more closely. What do we have to do with K_{eq,2} and K_{eq,3} in order to get K_{eq,1}?

K_{eq,1} = K_{eq,2} = K_{eq,3} =

Perhaps you noticed that if we multiply *K _{eq,2}* and

*K*, it will be equal to

_{eq,3}*K*.

_{eq,1}K_{eq,2} × K_{eq,3} = = K_{eq,1}

## Summary of How to Manipulate *K*_{eq}

_{eq}

- If you multiply a reaction by a coefficient, you also raise the equilibrium constant expression to that same coefficient. If you divide a reaction by a coefficient, try to reframe the process into multiplication (i.e. dividing by two is the same as multiplying by ½).
- If you flip a reaction, you take the inverse of the equilibrium constant expression.
- If you add reactions together, you multiply the equilibrium constant expressions.
- If you encounter more than one of these steps in a single problem, steps 1 & 2 can be done in any order, but always perform steps 1 & 2 prior to step 3.

## Key Concepts and Summary

If K_{eq} is known for a specific reaction, K_{eq} can also be calculated if the chemical reaction is modified by changing the coefficients, reversing the reaction, or combining the given reaction with another reaction. Approaching this as you have approached Hess’ Law in the past is a great strategy, except remember that changing the coefficients requires you to raise K_{eq} to that change, reversing a reaction requires you to take the inverse of K_{eq}, and combining reactions requires you to multiply the K_{eq} for each reaction combined.

### Chemistry End of Section Exercises

- Acetic acid is a weak acid. It reacts with water to a small extent to form acetate and hydronium ions. The equilibrium constant for the reaction is 1.8 x 10
^{-5}.

CH_{3}COOH(aq) + H_{2}O(l) ⇌ CH_{3}COO^{–}(aq) + H_{3}O^{+}(aq)

What is the K_{c}for the following reactions?- 2 CH
_{3}COOH(aq) + 2 H_{2}O(l) ⇌ 2 CH_{3}COO^{–}(aq) + 2 H_{3}O^{+}(aq) - CH
_{3}COO^{–}(aq) + H_{3}O^{+}(aq) ⇌ CH_{3}COOH(aq) + H_{2}O(l) - ½ CH
_{3}COOH(aq) + ½ H_{2}O(l) ⇌ ½ CH_{3}COO^{–}(aq) + ½ H_{3}O^{+}(aq) - 4 CH
_{3}COO^{–}(aq) + 4 H_{3}O^{+}(aq) ⇌ 4 CH_{3}COOH(aq) + 4 H_{2}O(l)

- 2 CH
- Given the equations:
2 N
_{2}(g) + O_{2}(g) ⇌ 2 N_{2}O(g) K_{c}= 1.2 x 10^{-35}N_{2}O_{4}(g) ⇌ 2 NO_{2}(g) K_{c}= 4.6 x 10^{-3}½ N_{2}(g) + O_{2}(g) ⇌ NO_{2}(g) K_{c}= 4.1 x 10^{-9}Calculate the value of K

_{c}for 2 N_{2}O_{4}(g) ⇌ 2 N_{2}O(g) + 3 O_{2}(g).

### Answers to Chemistry End of Section Exercises

- (a) 3.2 × 10
^{-10}(b) 5.6 × 10^{4}(c) 4.2 × 10^{-3}(d) 9.5 × 10^{18} - 9.0 × 10
^{-7}

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