Equilibrium constant

Every reversible reaction eventually stops changing — not because it has finished, but because it has reached a balance point where the forward and reverse reactions proceed at equal rates. The equilibrium constant is the single number that pins down exactly where that balance point sits, for a given reaction at a given temperature. It converts the vague phrase "the reaction reaches equilibrium" into a precise, testable ratio of concentrations or pressures, and it lets you answer two very different-sounding questions with the same formula: how far will this reaction go, and which way is it currently heading if it has not yet arrived? This section builds the equilibrium constant from the law of mass action, develops $K_c$ and $K_p$ and the relation between them, introduces the reaction quotient $Q$ as the tool for predicting direction, and lists the characteristics every equilibrium constant obeys. Every number and result matches the syllabus scope exactly.

The core insight

At a fixed temperature, the ratio of product concentrations to reactant concentrations (each raised to its stoichiometric power) settles at a fixed value $K_c$ for a reversible reaction at equilibrium, and comparing the same ratio $Q$ at any other instant tells you which way the reaction must shift to get there.

The core idea

The law of mass action. For a general reversible reaction at a fixed temperature,

aA + bB \rightleftharpoons cC + dD

the law of mass action states that at equilibrium the ratio of the product of equilibrium concentrations of the products (each raised to its stoichiometric coefficient) to the analogous product for the reactants is a constant, the equilibrium constant $K_c$ :

K_c = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}

where: square brackets denote equilibrium molar concentrations ( $\text{mol L}^{-1}$ ), and $a,b,c,d$ are the coefficients from the balanced equation. $K_c$ is fixed for a given reaction at a given temperature — it does not care what concentrations you started with.

$K_p$ for gaseous systems. When every species is a gas it is often more convenient to work in partial pressures instead of concentrations:

K_p = \dfrac{p_C^{\,c}\,p_D^{\,d}}{p_A^{\,a}\,p_B^{\,b}}

where: $p_i$ is the equilibrium partial pressure of species $i$ . $K_p$ and $K_c$ describe the same equilibrium and are linked by the relation derived below, $K_p = K_c(RT)^{\Delta n_g}$ , where $\Delta n_g$ is the change in moles of gas across the reaction.

The reaction quotient $Q$ . At any stage of a reaction mixture — not necessarily equilibrium — $Q_c$ (or $Q_p$ ) is computed with exactly the same expression as $K_c$ (or $K_p$ ), but using whatever concentrations (or pressures) are present at that instant. Comparing $Q$ to $K$ predicts the net direction of change:

$Q < K$ : the product term is too small relative to equilibrium, so the reaction runs forward.
$Q > K$ : the product term is too large relative to equilibrium, so the reaction runs backward.
$Q = K$ : the system is already at equilibrium and shows no net change.

Characteristics every $K$ obeys. $K$ depends only on temperature for a given reaction — it is unaffected by the initial concentrations chosen, by dilution, or by a catalyst (a catalyst speeds up how fast equilibrium is reached, not where it sits). Pure solids and pure liquids never appear in the $K_c$ / $K_p$ expression because their activity is effectively constant and gets absorbed into $K$ . For the reverse reaction, $K'_c = 1/K_c$ . Scaling a balanced equation by a factor $n$ raises the constant to the power $n$ ( $K^n$ ); adding two equations multiplies their constants ( $K = K_1 \times K_2$ ). The units of $K_c$ depend on $\Delta n_g$ , since each concentration carries units of $\text{mol L}^{-1}$ .

Figure

A number line comparing the reaction quotient Q to the fixed equilibrium constant K, with an arrow pointing forward when Q is less than K, no arrow when Q equals K, and an arrow pointing backward when Q is greater than K

Derivation ADeriving Kp = Kc(RT)^Δn from the ideal gas law

The goal is to connect the two forms of the equilibrium constant, $K_p$ (built from partial pressures) and $K_c$ (built from molar concentrations), into a single formula $K_p = K_c(RT)^{\Delta n_g}$ that lets you convert between them once you know how many moles of gas the reaction gains or loses, and to show exactly why the two constants coincide only when that mole change is zero.

Step 1. Start from the ideal gas equation for any gaseous species $i$ in the mixture, $p_iV = n_iRT$ . Dividing both sides by $V$ gives $p_i = \dfrac{n_i}{V}RT$ , and since $n_i/V$ is just the molar concentration $[i]$ , this is $p_i = [i]RT$ — every partial pressure is its concentration multiplied by $RT$ .

Step 2. Apply this substitution to each species in the general equilibrium $aA + bB \rightleftharpoons cC + dD$ : $p_A = [A]RT$ , $p_B = [B]RT$ , $p_C = [C]RT$ , $p_D = [D]RT$ .

Step 3. Substitute these into the $K_p$ expression: $K_p = \dfrac{p_C^{\,c}p_D^{\,d}}{p_A^{\,a}p_B^{\,b}} = \dfrac{([C]RT)^c([D]RT)^d}{([A]RT)^a([B]RT)^b} = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}\times\dfrac{(RT)^{c+d}}{(RT)^{a+b}}$ .

Step 4. Recognise the first factor as exactly $K_c$ , and collect the $(RT)$ powers using the exponent rule $(RT)^{c+d}/(RT)^{a+b} = (RT)^{(c+d)-(a+b)}$ . Naming $\Delta n_g = (c+d)-(a+b)$ , the moles of gaseous product minus moles of gaseous reactant, this gives

K_p = K_c(RT)^{\Delta n_g}

where: $R = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}$ (or the SI value, matched to how $K_p$ 's pressure units are expressed), $T$ is the absolute temperature in kelvin, and $\Delta n_g$ counts gaseous species only.

Step 5. Read off the special case: if $\Delta n_g = 0$ (equal moles of gas on both sides), $(RT)^0 = 1$ and $K_p = K_c$ exactly, regardless of temperature. Whenever $\Delta n_g \neq 0$ , the two constants differ numerically and, in general, carry different units — they still describe the identical equilibrium, just measured on two different scales.

Worked example

Using $Q$ to predict reaction direction, then checking the $K_p$ – $K_c$ relation for the same system. At a certain temperature, the equilibrium constant for $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ is $K_c = 54$ . A reaction vessel at this temperature is found to contain $[\text{H}_2] = 0.10\ M$ , $[\text{I}_2] = 0.20\ M$ , and $[\text{HI}] = 0.40\ M$ at some instant.

Step 1. Write the $Q_c$ expression, identical in form to $K_c$ : $Q_c = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$ .

Step 2. Substitute the given (non-equilibrium) concentrations: $Q_c = \dfrac{(0.40)^2}{(0.10)(0.20)} = \dfrac{0.16}{0.02} = 8.0$ .

Step 3. Compare $Q_c$ with $K_c$ : $Q_c = 8.0$ and $K_c = 54$ , so $Q_c < K_c$ . The product term is too small relative to equilibrium, so the reaction must run forward, converting more $\text{H}_2$ and $\text{I}_2$ into $\text{HI}$ until $Q_c$ rises to $54$ .

Step 4. Now find $\Delta n_g$ for this same reaction: moles of gaseous product $= 2$ (from $2\text{HI}$ ), moles of gaseous reactant $= 1 + 1 = 2$ (from $\text{H}_2$ and $\text{I}_2$ ). So $\Delta n_g = 2 - 2 = 0$ .

Step 5. Apply $K_p = K_c(RT)^{\Delta n_g} = K_c(RT)^0 = K_c$ . Because this reaction has equal moles of gas on both sides, $K_p = K_c = 54$ numerically — no unit conversion is ever needed for this particular equilibrium, unlike ammonia synthesis where $\Delta n_g = -2$ and $K_p \neq K_c$ .

Competitive stretch

If instead the reverse reaction $2\text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g)$ were written, its equilibrium constant would be $K'_c = 1/K_c = 1/54 \approx 0.0185$ — the reciprocal, by the reversal rule, still with $\Delta n_g = 0$ so $K'_p = K'_c$ as well.

Common traps

Including pure solids or liquids in the $K_c$ / $K_p$ expression. Only gases and dissolved species with genuinely variable concentrations appear; a pure solid or liquid's constant contribution is folded into $K$ itself.
Confusing $Q$ and $K$ . $K$ is the fixed value strictly at equilibrium; $Q$ uses the identical formula but can be evaluated at any instant — the comparison between the two, not either number alone, tells you the direction of change.
Forgetting $\Delta n_g$ counts gaseous species only. Solids, liquids, and aqueous species never enter the mole-count difference used in $K_p = K_c(RT)^{\Delta n_g}$ , even if they appear in the balanced equation.
Assuming $K$ changes with concentration, pressure, or a catalyst. $K$ depends only on temperature and the reaction as written; changing the amounts you start with moves the equilibrium position, never the value of $K$ .

Ask the tutor: Want to work through more $Q$ -versus- $K$ direction problems, or see how these constants feed into calculating equilibrium concentrations from initial amounts?

Why does Kp not equal Kc whenever Δn_g is not zero, even though they describe the same equilibrium

Because $K_p$ is built from partial pressures and $K_c$ from concentrations, and the two are linked by $p_i = [i]RT$ for every gaseous species. When you substitute this into the $K_p$ expression, a leftover factor of $(RT)^{\Delta n_g}$ survives unless the powers of $RT$ from products and reactants exactly cancel, which only happens when $\Delta n_g = 0$ . So $K_p$ and $K_c$ are two different numerical measures of the identical equilibrium state, related by a temperature-dependent conversion factor whenever the mole count of gas changes.

Does the reaction quotient Q ever become negative or does it use the same units as K

$Q$ can never be negative, because it is built from concentrations or pressures, which are always positive, raised to positive powers — the same reasoning that keeps $K$ positive. $Q$ is computed with exactly the same formula and the same units as the corresponding $K$ (so $Q_c$ compares against $K_c$ , and $Q_p$ against $K_p$ , never mixed). The only difference between $Q$ and $K$ is which set of concentrations or pressures — instantaneous or equilibrium — is plugged in.

If I double every coefficient in a balanced equation, does the equilibrium constant just double too

No — it gets squared, not doubled. Multiplying a balanced equation by a factor $n$ raises every concentration term in the $K$ expression to $n$ times its original power, and since raising a product of powers to a further power multiplies the exponents, the new constant is the old one raised to the power $n$ , i.e. $K^n$ . So doubling the equation ( $n=2$ ) gives $K_{\text{new}} = K^2$ , not $2K$ — a common arithmetic slip worth checking with a small numeric example.

Equilibrium constant

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads