Equilibrium Constant

Mix hydrogen and iodine vapour in a sealed flask and the reaction does not run to completion — it slows, and eventually the amounts of $\text{H}_2$ , $\text{I}_2$ and $\text{HI}$ stop changing. At that point the ratio of product to reactant concentrations, each raised to its stoichiometric power, locks onto one fixed number for that temperature: the equilibrium constant, $K$ . Whether you start with pure reactants, pure products, or some mixture in between, the system always relaxes to the same $K$ — only the route differs, never the destination. This single number is one of the most useful in chemistry. It tells you, before doing any experiment, whether a reaction will favour products or reactants, and by comparing $K$ with the reaction quotient $Q$ at any non-equilibrium instant, it tells you which way a mixture will shift to get there. This article builds $K_c$ and $K_p$ from the law of mass action, connects them through the ideal gas law, and uses $Q$ versus $K$ to predict reaction direction — the essential toolkit for everything that follows in chemical and ionic equilibrium.

The core insight

At a fixed temperature, the ratio $[\text{products}]^{\text{powers}}/[\text{reactants}]^{\text{powers}}$ settles at one constant value $K$ regardless of starting composition, and comparing the same ratio $Q$ at any instant with $K$ tells you which direction the reaction will shift.

The core idea

Setting up $K_c$ . For a general reversible reaction at equilibrium,

a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}

the law of mass action gives the equilibrium constant in terms of concentration:

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

where each bracketed term is the equilibrium molar concentration (mol L $^{-1}$ ), and $a, b, c, d$ are the balanced stoichiometric coefficients. $K_c$ is fixed at a given temperature no matter what concentrations you start the reaction with.

Setting up $K_p$ . For gas-phase reactions, partial pressures are often more convenient than concentrations. The equilibrium constant in terms of partial pressure is

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

where each $p$ is the equilibrium partial pressure of that gas (commonly in atm or bar).

Linking $K_p$ and $K_c$ . Because $pV = nRT$ gives $p = (n/V)RT$ = (concentration) $\times RT$ for each gas, substituting into the $K_p$ expression and collecting powers of $RT$ yields

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

where $\Delta n_g = (c+d) - (a+b)$ counts only the gaseous species' coefficients, $R$ is the gas constant ( $8.314\ \text{J K}^{-1}\text{mol}^{-1}$ with SI units, or $0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}$ when pressure is in atm and volume in L), and $T$ is the absolute temperature in kelvin. If $\Delta n_g = 0$ — equal moles of gas on both sides — then $K_p = K_c$ exactly, with no unit conversion needed.

The reaction quotient, $Q$ . $Q$ has exactly the same algebraic form as $K$ , but it can be evaluated at any instant — not only at equilibrium:

Q_c = \dfrac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}\quad(\text{concentrations at a general moment})

Comparing $Q$ against the known $K$ predicts which way the system will move to reach equilibrium:

$Q < K$ : not enough product yet — the reaction runs forward.
$Q > K$ : too much product already — the reaction runs backward.
$Q = K$ : the system is already at equilibrium.

Other properties worth memorising. $K$ depends on temperature alone — never on starting concentrations, total pressure, volume, or the presence of a catalyst (a catalyst speeds up how fast equilibrium is reached, not where it lies). Reversing a reaction inverts its constant, $K'_c = 1/K_c$ ; scaling a balanced equation by a factor $n$ raises the constant to that power, $K_c(\text{new}) = (K_c)^n$ . Pure solids and pure liquids never appear in the $K_c$ or $K_p$ expression, because their "concentration" — density over molar mass — is itself a constant that gets folded into $K$ . A large $K$ ( $\gg 1$ ) signals a product-favoured equilibrium; a small $K$ ( $\ll 1$ ) signals a reactant-favoured one.

Figure

Number line showing Q less than K driving the reaction forward, Q equal to K at rest, and Q greater than K driving it backward

Derivation ALinking Kp and Kc through the ideal gas law

Derive the exact algebraic link between $K_p$ and $K_c$ starting only from the ideal gas law and the definitions of the two constants, and confirm that a positive $\Delta n_g$ makes $K_p$ larger than $K_c$ once $RT$ exceeds one, so that converting between them is never a guess but a direct substitution.

Step 1. Write the ideal gas law for one gaseous species X in the equilibrium mixture: $p_X V = n_X RT$ , so $p_X = \left(\dfrac{n_X}{V}\right)RT = [\text{X}]\,RT$ , since $n_X/V$ is exactly the molar concentration $[\text{X}]$ .

Step 2. Apply this substitution to every gaseous species in $K_p = \dfrac{p_C^{\,c}\,p_D^{\,d}}{p_A^{\,a}\,p_B^{\,b}}$ , replacing each partial pressure with (concentration) $\times RT$ :

K_p = \dfrac{([\text{C}]RT)^c([\text{D}]RT)^d}{([\text{A}]RT)^a([\text{B}]RT)^b}

Step 3. Separate the concentration factors from the $RT$ factors. The concentration part regroups exactly into $K_c$ , and the $RT$ factors combine by the exponent rule $x^m/x^n = x^{m-n}$ :

K_p = \dfrac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}\times(RT)^{(c+d)-(a+b)} = K_c\,(RT)^{\Delta n_g}

Step 4. Read the sign of $\Delta n_g$ . When gas moles increase on going to products ( $\Delta n_g > 0$ ) and $RT > 1$ (true for essentially all reaction temperatures with $R$ in L·atm units), $K_p > K_c$ . When gas moles decrease ( $\Delta n_g < 0$ ), $K_p < K_c$ . When $\Delta n_g = 0$ , the $RT$ factor becomes $(RT)^0 = 1$ and $K_p = K_c$ identically — no conversion needed.

Worked example

Problem. At a certain temperature, 1 mol $\text{H}_2$ and 1 mol $\text{I}_2$ are placed in a 1 L closed flask. At equilibrium, 1.6 mol of $\text{HI}$ is present. For $\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\text{HI(g)}$ , find $K_c$ , find $K_p$ at this temperature (given $T = 500\ \text{K}$ , use $R = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}$ ), and state the direction the reaction would shift if, at some other instant, $Q_c = 40$ .

Step 1. Build the ICE table in mol L $^{-1}$ (volume is 1 L, so moles equal molarity): initial $[\text{H}_2]=[\text{I}_2]=1,\ [\text{HI}]=0$ ; change $-x,-x,+2x$ ; equilibrium $1-x,\ 1-x,\ 2x$ .

Step 2. Use the given equilibrium value: $2x = 1.6 \Rightarrow x = 0.8$ , so $[\text{H}_2]=[\text{I}_2]=1-0.8=0.2\ \text{mol L}^{-1}$ and $[\text{HI}]=1.6\ \text{mol L}^{-1}$ .

Step 3. Evaluate $K_c$ :

K_c = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \dfrac{(1.6)^2}{(0.2)(0.2)} = \dfrac{2.56}{0.04} = 64

Step 4. Find $\Delta n_g$ for $K_p$ : gaseous moles are $2$ on the product side and $1+1=2$ on the reactant side, so $\Delta n_g = 2 - 2 = 0$ . Therefore $K_p = K_c(RT)^0 = K_c = 64$ — no unit conversion is needed because the mole count of gas does not change.

Step 5. Compare $Q_c = 40$ with $K_c = 64$ : since $Q_c < K_c$ , the mixture has too little HI relative to equilibrium, so the reaction shifts forward, converting more $\text{H}_2$ and $\text{I}_2$ into HI until $Q_c$ rises to meet $K_c = 64$ .

Answer: $K_c = 64$ , $K_p = 64$ (no units, since $\Delta n_g = 0$ ), and a mixture with $Q_c = 40$ proceeds forward toward more HI.

Competitive stretch

For $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$ , $K_c = 0.50$ at $T = 700\ \text{K}$ . Here $\Delta n_g = 2-(1+3) = -2$ , so $K_p = K_c(RT)^{-2} = 0.50\times(0.0821\times700)^{-2} = 0.50\times(57.47)^{-2} = 0.50\times3.028\times10^{-4} = 1.5\times10^{-4}$ . The large drop from $K_c$ to $K_p$ is exactly the sign-of- $\Delta n_g$ rule from Step 4 above: negative $\Delta n_g$ combined with $RT>1$ shrinks $K_p$ well below $K_c$ .

Common traps

Writing pure solids or liquids into the $K_c$ / $K_p$ expression — only gases and dissolved (aqueous) species appear; their concentration is baked into the constant, not written out.
Confusing $Q$ and $K$ : $Q$ can be found from concentrations at any moment, while $K$ is that same ratio only at equilibrium, and only $K$ has a fixed value at a given temperature.
Counting non-gaseous species when finding $\Delta n_g$ — only gaseous coefficients enter $\Delta n_g$ for the $K_p$ – $K_c$ conversion.
Believing $K$ changes when you add more reactant or change the container volume — those actions shift the equilibrium position (the amounts present), never the value of $K$ , which tracks temperature alone.

Ask the tutor: Want to practise setting up an ICE table for a trickier stoichiometry, or work through a few more Q-versus-K direction predictions?

Why doesn't the amount of pure solid or liquid affect K?

Because a pure solid or liquid has a fixed density and molar mass, so its "concentration" (moles per litre of that substance itself) never changes no matter how much of it is present — doubling the amount of a solid does not double its concentration, since both moles and volume of the solid double together. This constant value is absorbed into $K$ itself rather than written as a separate factor, which is why you never see $[\text{CaCO}_3(\text{s})]$ sitting inside a $K_c$ expression.

If I start with only products instead of only reactants, do I get a different K?

No — you will reach a different set of equilibrium concentrations, but plugging those equilibrium values into the same $K_c$ expression gives the identical number, provided the temperature is the same. $K$ is a property of the reaction at that temperature, not of how you chose to start it; the system can approach equilibrium from either side and still land on the same $K$ .

How is Q different from K if the formula looks exactly the same?

The formula is identical, but the timing differs. $K$ is that ratio evaluated using concentrations measured only once the system has stopped changing — true equilibrium values at a fixed temperature. $Q$ uses whatever concentrations exist at any chosen instant, equilibrium or not, which is precisely why comparing $Q$ to $K$ tells you whether — and which way — the system still needs to move.

Equilibrium Constant

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads