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The equilibrium constant written from the balanced equation, calculated from equilibrium concentrations, read for what its size means, and compared with the reaction quotient to predict a direction.
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By the end of this lesson you will be able to write the equilibrium constant for a homogeneous equilibrium from its balanced equation, calculate it from equilibrium concentrations, say what a very large or very small value means, find the constant for the reversed equation, and use the reaction quotient to predict which way a mixture will react.
You know that a reversible reaction in a closed container settles at dynamic equilibrium, with constant and usually unequal concentrations, and that the same equilibrium is reached from either side. What is missing is a way to say where it settles, precisely enough to compare two reactions or to predict what a given mixture will do. That is one number.
The equilibrium constant, $K_c$, is the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of its coefficient in the balanced equation. The subscript $c$ says it is written with concentrations in mol/L. A homogeneous equilibrium has every substance in the same phase, such as all gases. The reaction quotient, $Q$, is the same expression worked out for any mixture, at equilibrium or not.
For a balanced equilibrium
$$a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D}$$
the equilibrium constant is
$$K_c = \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}$$
with every concentration measured at equilibrium.
Three things to notice.
Products on the top. A large $K_c$ means the top is large compared with the bottom: at equilibrium the mixture is mostly products. A small $K_c$ means mostly reactants.
The powers are the coefficients. This is the opposite of the rule for rate equations, and it is not a contradiction. A rate describes the route a reaction takes, and the route has to be found by experiment. An equilibrium constant describes the end state, and the end state does not depend on the route: at equilibrium every step of any mechanism is balanced by its own reverse, and when those balances are multiplied together the intermediate steps cancel and only the overall equation is left. So the balanced equation is exactly what $K_c$ needs.
It is a constant only at one temperature. Any equilibrium mixture of hydrogen, iodine and hydrogen iodide at 450 °C gives the same value of about 50, however much of each was put in at the start. At a different temperature the value is different.
Adding a substance, removing one or squeezing the gas moves the position of an equilibrium and leaves the equilibrium constant exactly where it was. Only a change of temperature changes the constant itself.
Another way: picture
Think of a see-saw with products on one end and reactants on the other. The constant says where it comes to rest. A constant of a billion rests hard down on the product side; a constant of a billionth rests hard down on the reactant side; a constant near one sits nearly level.
Another way: steps
To write and use $K_c$:
Work out the same expression for a mixture that has not yet reached equilibrium and you get the reaction quotient, $Q$. Comparing it with $K_c$ tells you which way the mixture must react.
| Comparison | What it means | What happens |
|---|---|---|
| $Q < K_c$ | too little product for equilibrium | net forward reaction, until $Q$ rises to $K_c$ |
| $Q > K_c$ | too much product for equilibrium | net reverse reaction, until $Q$ falls to $K_c$ |
| $Q = K_c$ | already at equilibrium | no net change |
For example, $\mathrm{2NO_2 \rightleftharpoons N_2O_4}$ has $K_c = 170$ L/mol at 25 °C. A mixture with $[\mathrm{NO_2}] = 0.10$ and $[\mathrm{N_2O_4}] = 0.50$ mol/L has $Q = 0.50 \div 0.10^2 = 50$. That is below 170, so more $\mathrm{NO_2}$ pairs up: the brown colour fades until the quotient reaches 170.
The constant belongs to the equation as written. Reverse the equation and the top and bottom swap, so the new constant is $1/K_c$: for $\mathrm{2HI \rightleftharpoons H_2 + I_2}$ at 450 °C it is $1/50 = 0.02$. Double every coefficient and every power doubles, so the constant is squared. Neither change alters the mixture in the flask; they change only how it is described. That is why a quoted $K_c$ is useless without its equation and its temperature.
Units follow the same bookkeeping. Each concentration brings mol/L. When the top and bottom have equal numbers of concentrations the units cancel, as for hydrogen iodide; otherwise they do not, as for ammonia, where two on top and four below leave L²/mol².
Putting the reactants on the top. The convention is products over reactants, and getting it upside down turns a large constant into a small one and reverses every prediction.
Using coefficients as multipliers. $\mathrm{2HI}$ means $[\mathrm{HI}]^2$, not $2[\mathrm{HI}]$.
Carrying the rate-equation rule across. In a rate equation the orders are measured and may ignore the coefficients; in an equilibrium constant the coefficients are the powers, because the end state does not depend on the route.
Using starting concentrations. $K_c$ needs the concentrations at equilibrium. Starting amounts go into $Q$, which is how you predict the direction the mixture will move.
Thinking a large constant means a fast reaction. The constant says where the reaction ends, not how quickly it gets there.
For $\mathrm{PCl_5 \rightleftharpoons PCl_3 + Cl_2}$ at equilibrium, $[\mathrm{PCl_5}] = 0.040$, $[\mathrm{PCl_3}] = 0.020$ and $[\mathrm{Cl_2}] = 0.10$ mol/L.
All three are equilibrium concentrations, so they go straight into $K_c$.
$K_c = \dfrac{[\mathrm{PCl_3}][\mathrm{Cl_2}]}{[\mathrm{PCl_5}]} = \dfrac{0.020 \times 0.10}{0.040}$.
Products on the top, every coefficient 1.
$K_c = 0.002 \div 0.040 = 0.05$ mol/L.
Two concentrations on top and one below leave mol/L.
For $\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$ at 450 °C, $K_c = 50$. A flask holds $[\mathrm{H_2}] = [\mathrm{I_2}] = 0.10$ and $[\mathrm{HI}] = 1.0$ mol/L.
These are not known to be equilibrium values, so this is a quotient.
$Q = 1.0^2 \div (0.10 \times 0.10) = 100$.
The same expression as the constant.
$Q = 100$ is above $K_c = 50$, so the mixture has too much hydrogen iodide: net reverse reaction until the quotient falls to 50.
Above the constant means too much product.
$Q = [\mathrm{B}]^2 \div [\mathrm{A}] = 0.16 \div 0.5 = 0.32$.
B is squared because its coefficient is 2.
Compare: $0.32$ is below $0.8$, so …
…the mixture has too little product and reacts forward until its quotient reaches 0.8.
For the gas-phase equilibrium $\mathrm{2A + 3B \rightleftharpoons 2C}$, write the expression for the equilibrium constant $K_c$, using $x$ for $[\mathrm{A}]$, $y$ for $[\mathrm{B}]$ and $z$ for $[\mathrm{C}]$.
Answer:
Hydrogen and iodine reach equilibrium in a sealed flask: $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$. At equilibrium $[\mathrm{H_2}] = 0.1$ mol/L, $[\mathrm{I_2}] = 0.5$ mol/L and $[\mathrm{HI}] = 0.4$ mol/L. Calculate $K_c$.
Answer:
At a certain temperature the equilibrium $\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}$ has $K_c = 2$. What is $K_c$ at the same temperature for the equilibrium written the other way round, $\mathrm{PCl_3(g) + Cl_2(g) \rightleftharpoons PCl_5(g)}$? A fraction is fine.
Answer:
Butane and methylpropane interconvert over a catalyst: $\mathrm{butane \rightleftharpoons methylpropane}$, with $K_c = 4$ at the working temperature. A mixture is made with $0.1$ mol/L of butane and $0.3$ mol/L of methylpropane. What happens next?
For $\mathrm{2NO_2(g) \rightleftharpoons N_2O_4(g)}$, $K_c = 4$ at a fixed temperature. For each mixture, calculate the reaction quotient and say which way the mixture will react.
| [NO2] in mol/L | [N2O4] in mol/L | reaction quotient | what the mixture does | |
|---|---|---|---|---|
| mixture X | 0.3 | 0.72 | ||
| mixture Y | 0.3 | 0.18 | ||
| mixture Z | 0.2 | 0.16 |
Three equilibria at 25 °C have very different constants. Match each constant to what an equilibrium mixture of that reaction contains.
| almost entirely products | almost entirely reactants | reactants and products in comparable amounts | |
|---|---|---|---|
| Kc = 10 to the power 13 | |||
| Kc = 10 to the power −13 | |||
| Kc = 1.5 |
Gas sampled from an ammonia plant's reactor, once it has reached equilibrium, contains $[\mathrm{N_2}] = 0.1$ mol/L, $[\mathrm{H_2}] = 0.2$ mol/L and $[\mathrm{NH_3}] = 0.03$ mol/L. For $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$, calculate $K_c$ at the reactor temperature, in L²/mol².
Answer: L²/mol²
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For $\mathrm{A(g) + B(g) \rightleftharpoons C(g)}$, $K_c = 9$ L/mol at a fixed temperature. An equilibrium mixture has $[\mathrm{A}] = 0.8$ mol/L and $[\mathrm{B}] = 0.8$ mol/L. What is $[\mathrm{C}]$, in mol/L?
Answer: mol/L
You can write, calculate and interpret an equilibrium constant, and compare it with a reaction quotient. Say out loud why the powers in an equilibrium constant are the coefficients when the powers in a rate equation are not, and what a quotient above the constant means for the mixture. Next: keeping track of starting amounts, changes and equilibrium amounts in one table.
10. Your turn: for $\mathrm{A \rightleftharpoons 2B}$, $K_c = 0.8$ mol/L. A mixture has $[\mathrm{A}] = 0.5$ and $[\mathrm{B}] = 0.4$ mol/L. Which way does it react?, step 3