AP Chemistry 7.5-7.6: Magnitude and Properties of K

Equilibrium constant $K$

The equilibrium constant is the value of the reaction quotient when a reversible reaction is at equilibrium: $K=Q_{\text{eq}}$. At a fixed temperature, it is constant for a given balanced reaction.

What does the magnitude of $K$ indicate about an equilibrium mixture?

A large $K$ indicates that products are favored at equilibrium, while a small $K$ indicates that reactants are favored. Values much greater than 1 favor products, and values much less than 1 favor reactants.

What does $K\approx 1$ indicate about an equilibrium mixture?

Neither reactants nor products are strongly favored; appreciable amounts of both are present at equilibrium.

Does a large equilibrium constant mean the reaction is fast?

No. $K$ describes the relative composition at equilibrium, not the rate at which equilibrium is reached. Reaction rate is a kinetic property.

How does the equilibrium constant depend on initial concentrations?

At a fixed temperature, $K$ for a given reaction is independent of the initial amounts of reactants and products. Different starting mixtures may approach the same $K$ from different compositions.

Which condition changes the numerical value of an equilibrium constant?

Temperature changes the numerical value of $K$. Changing initial concentrations, pressures, or adding a catalyst does not change $K$ for a given reaction at a fixed temperature.

What is the difference between changing the reaction mixture and changing $K$?

Changing concentrations or partial pressures changes the reaction quotient and may shift the system to a new equilibrium composition, but it does not change $K$ at constant temperature. Changing temperature can change $K$.

How does reversing a chemical equation affect its equilibrium constant?

The new equilibrium constant is the reciprocal of the original: $K_{\text{reverse}}=\dfrac{1}{K_{\text{forward}}}$.

How does multiplying every coefficient in a balanced equilibrium equation by a factor $x$ affect $K$?

The new constant is raised to that factor: $K_{\text{new}}=(K_{\text{old}})^x$.

What happens to the equilibrium constant when two equilibrium reactions are added?

After canceling species that appear on opposite sides, the overall equilibrium constant is the product of the individual constants: $K_{\text{overall}}=K_1K_2$.

How would you derive the equilibrium constant for a target reaction from several known reactions?

Reverse, multiply, or otherwise manipulate the known equations until they sum to the target equation. Apply reciprocal or exponent rules to their $K$ values, then multiply the modified constants.

Why must intermediate species be canceled when combining equilibrium reactions?

Only species in the net balanced equation belong in the overall equilibrium expression. Canceling intermediates in the equations corresponds to multiplying the individual equilibrium constants.

A reaction has $K=0.50$. What is the equilibrium constant for the reversed reaction, and what is the constant if the reaction is doubled?

For the reversed reaction, $K=1/0.50=2.0$. If all coefficients are doubled without reversing, $K=(0.50)^2=0.25$.

Why can the same $K$ result from starting with different initial mixtures?

At a fixed temperature, each initial mixture changes until it reaches equilibrium. Although the equilibrium compositions may differ, the equilibrium quotient for the same balanced reaction is the same constant.