The law of mass action
Consider a general reversible reaction $\ce{A + B <=> C + D}$, where A and B are reactants and C and D are products in the balanced equation. On the basis of experimental studies of many reversible reactions, the Norwegian chemists Cato Guldberg and Peter Waage proposed in 1864 that the concentrations in an equilibrium mixture are tied together by a fixed ratio. In the early days of chemistry concentration was called the "active mass", which is why the relation is also called the law of mass action.
The classic evidence is the gaseous reaction between $\ce{H2}$ and $\ce{I2}$ in a sealed vessel at 731 K, $\ce{H2(g) + I2(g) <=> 2HI(g)}$. NCERT records six experiments — four started from $\ce{H2}$ and $\ce{I2}$ alone, two from $\ce{HI}$ alone, so equilibrium was approached from both directions. A naive ratio such as $\dfrac{[\ce{HI}]}{[\ce{H2}][\ce{I2}]}$ is far from constant across the six runs.
The ratio that does hold constant in every run squares the product term: $\dfrac{[\ce{HI}]^2}{[\ce{H2}][\ce{I2}]}$. The power on each species turns out to be its stoichiometric coefficient in the balanced equation. This is the heart of the law. Two features deserve emphasis. First, the constancy holds whether equilibrium is approached from the reactant side or the product side, confirming that the equilibrium state is independent of the path taken to reach it. Second, the constant depends only on temperature: change the temperature and the value of the constant changes, but at a fixed temperature it is the same for every starting mixture.
Law of chemical equilibrium. At a given temperature, the product of the concentrations of the reaction products (each raised to its stoichiometric coefficient), divided by the product of the concentrations of the reactants (each raised to its coefficient), has a constant value — the equilibrium constant $K_c$.
Individual concentrations change until equilibrium, then stop changing. Beyond that point the law-of-mass-action ratio holds at the single value $K_c$, regardless of the starting amounts.
Writing the Kc expression
For the general reaction $\ce{aA + bB <=> cC + dD}$ the equilibrium constant in terms of molar concentration is
$$K_c = \frac{[\ce{C}]^{c}\,[\ce{D}]^{d}}{[\ce{A}]^{a}\,[\ce{B}]^{b}}$$
Every square-bracketed term is an equilibrium concentration in $\text{mol L}^{-1}$; the subscript "eq" is dropped by convention. The subscript "c" on $K_c$ records that concentrations are used. Two rules of writing follow from NCERT directly.
| Reaction | Kc expression |
|---|---|
| $\ce{H2(g) + I2(g) <=> 2HI(g)}$ | $\dfrac{[\ce{HI}]^2}{[\ce{H2}][\ce{I2}]}$ |
| $\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$ | $\dfrac{[\ce{NH3}]^2}{[\ce{N2}][\ce{H2}]^3}$ |
| $\ce{4NH3(g) + 5O2(g) <=> 4NO(g) + 6H2O(g)}$ | $\dfrac{[\ce{NO}]^4[\ce{H2O}]^6}{[\ce{NH3}]^4[\ce{O2}]^5}$ |
When the system is heterogeneous — more than one phase — the molar concentration of any pure solid or pure liquid is constant and independent of how much is present, so it is folded into the constant and dropped. For the thermal decomposition $\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$, both solids vanish from the expression, leaving $K_c = [\ce{CO2}]$, or equivalently $K_p = p_{\ce{CO2}}$. The pure solids must still be present at equilibrium, but they do not enter $K$.
Kp for gaseous equilibria
For reactions involving gases it is usually more convenient to use partial pressures than concentrations, giving the equilibrium constant the symbol $K_p$. The bridge between pressure and concentration is the ideal-gas equation $pV = nRT$, which rearranges to $p = \left(\dfrac{n}{V}\right)RT$. Since $n/V$ is a molar concentration, for any gas
$$p = [\text{gas}]\,RT \qquad (R = 0.0831 \ \text{bar L mol}^{-1}\text{K}^{-1})$$
At constant temperature the partial pressure of a gas is therefore proportional to its concentration, $p \propto [\text{gas}]$. For $\ce{H2(g) + I2(g) <=> 2HI(g)}$ we may write the constant either as a concentration ratio or as a pressure ratio:
$$K_c = \frac{[\ce{HI}]^2}{[\ce{H2}][\ce{I2}]} \qquad\qquad K_p = \frac{p_{\ce{HI}}^{\,2}}{p_{\ce{H2}}\,p_{\ce{I2}}}$$
The Kp = Kc(RT)^Δn relation
Substituting $p_i = [\,i\,]RT$ into the $K_p$ expression and collecting the $RT$ factors converts pressures into concentrations. For the $\ce{HI}$ equilibrium the $RT$ terms cancel completely and $K_p = K_c$. For $\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$ they do not cancel — the result carries a factor $(RT)^{-2}$, so the two constants differ. For the general gaseous reaction $\ce{aA + bB <=> cC + dD}$ the substitution gives
$$K_p = K_c\,(RT)^{\Delta n}$$
where $\Delta n = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$ in the balanced equation. Because the standard state for pressure is 1 bar, partial pressures must be expressed in bar when computing $K_p$ (recall $1\ \text{bar} = 10^5\ \text{Pa}$). The sign of $\Delta n$ decides everything, as summarised below.
The relation is worth reading in plain language. When the forward reaction produces more gas molecules than it consumes, $\Delta n$ is positive and the $(RT)^{\Delta n}$ factor inflates $K_p$ relative to $K_c$; when the reaction consumes gas, $\Delta n$ is negative and $K_p$ is smaller than $K_c$. Only when the gaseous moles balance exactly do the two constants coincide. This is why the value of a constant is meaningless unless you also state whether it is a $K_p$ or a $K_c$ and which balanced equation it refers to.
| Δn | Relation | Example |
|---|---|---|
| $\Delta n = 0$ | $K_p = K_c$ | $\ce{H2(g) + I2(g) <=> 2HI(g)}$ (2 − 2 = 0) |
| $\Delta n > 0$ | $K_p > K_c$ (if $RT>1$) | $\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$ (2 − 1 = +1) |
| $\Delta n < 0$ | $K_p < K_c$ (if $RT>1$) | $\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$ (2 − 4 = −2) |
Δn counts gaseous moles only — and only the balanced equation's coefficients
When evaluating $\Delta n$ for $K_p = K_c(RT)^{\Delta n}$, ignore pure solids and pure liquids entirely — only gaseous species are counted. A common slip is to include the solid in $\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$; the correct $\Delta n$ is $1 - 0 = +1$, counting only $\ce{CO2}$. The 2024 NEET item asking where "$K_p$ and $K_c$ are NOT equal" is exactly this test: the only option with $\Delta n \neq 0$ is the dissociation of $\ce{PCl5}$.
$K_p = K_c \iff \Delta n = 0$. For every $\Delta n \neq 0$ option, $K_p \neq K_c$.
Worked examples
For $\ce{2NOCl(g) <=> 2NO(g) + Cl2(g)}$, $K_c = 3.75 \times 10^{-6}$ at 1069 K. Calculate $K_p$. (NCERT Problem 6.5)
Count gaseous moles: $\Delta n = (2+1) - 2 = +1$.
$$K_p = K_c(RT)^{\Delta n} = 3.75\times10^{-6}\,(0.0831 \times 1069)^{1}$$
$RT = 0.0831 \times 1069 \approx 88.8$, so $K_p = 3.75\times10^{-6} \times 88.8 \approx 3.3\times10^{-4}$.
$K_p \approx 0.033 \times 10^{-1} = 3.3\times10^{-4}$ (NCERT states $K_p = 0.033$ with $RT$ rounded). The point: a single positive $\Delta n$ multiplies $K_c$ by one power of $RT$.
For $\ce{3O2(g) <=> 2O3(g)}$ at 298 K, $K_c = 3.0 \times 10^{-59}$. If $[\ce{O2}] = 0.040\ \text{M}$ at equilibrium, find $[\ce{O3}]$. (NEET 2022)
$$K_c = \frac{[\ce{O3}]^2}{[\ce{O2}]^3} \implies [\ce{O3}]^2 = K_c\,[\ce{O2}]^3$$
$[\ce{O3}]^2 = 3.0\times10^{-59} \times (0.040)^3 = 3.0\times10^{-59}\times 6.4\times10^{-5} = 1.9\times10^{-63}$.
$[\ce{O3}] = \sqrt{1.9\times10^{-63}} = 4.38\times10^{-32}\ \text{M}$. Note how the coefficient "2" on $\ce{O3}$ and "3" on $\ce{O2}$ enter as the powers — the law of mass action in action.
$\ce{PCl3}$, $\ce{Cl2}$ and $\ce{PCl5}$ are at equilibrium at 500 K with $[\ce{PCl3}] = 1.59\ \text{M}$, $[\ce{Cl2}] = 1.59\ \text{M}$, $[\ce{PCl5}] = 1.41\ \text{M}$. Find $K_c$ for $\ce{PCl5 <=> PCl3 + Cl2}$. (NCERT Problem 6.3)
$$K_c = \frac{[\ce{PCl3}][\ce{Cl2}]}{[\ce{PCl5}]} = \frac{1.59 \times 1.59}{1.41}$$
$K_c = \dfrac{2.528}{1.41} \approx 1.79$. Here $\Delta n = 2-1 = +1$, so $K_p = K_c(RT)$ would differ — a built-in reminder that this reaction sits in the "$K_p \neq K_c$" column.
Knowing the value of $K$ is only useful if you can read what it predicts. See Applications of the Equilibrium Constant for extent of reaction, the reaction quotient $Q$ and predicting direction.
Manipulating the equilibrium constant
Because $K_c$ is built from the balanced equation, rewriting that equation changes $K_c$ in a predictable way. NCERT Table 6.4 lists the three operations. Reverse the reaction and the constant inverts; multiply all coefficients by $n$ and the constant is raised to the power $n$; add two reactions and their constants multiply.
| Operation on the equation | New equilibrium constant | Worked illustration ($\ce{H2 + I2 <=> 2HI}$, $K_c = x$) |
|---|---|---|
| Reverse the reaction | $K' = \dfrac{1}{K}$ | $\ce{2HI <=> H2 + I2}$ gives $K' = \dfrac{[\ce{H2}][\ce{I2}]}{[\ce{HI}]^2} = \dfrac{1}{x}$ |
| Halve the coefficients ($n = \tfrac12$) | $K'' = K^{1/2}$ | $\tfrac12\ce{H2} + \tfrac12\ce{I2} <=> \ce{HI}$ gives $K'' = x^{1/2}$ |
| Multiply coefficients by $n$ | $K''' = K^{n}$ | $n\ce{H2} + n\ce{I2} <=> 2n\,\ce{HI}$ gives $K''' = x^{n}$ |
| Add two reactions | $K = K_1 \times K_2$ | Combine sub-steps by multiplying their constants |
Because $K_c$ and its inverse $K'_c$ carry different numerical values, it is essential to quote the exact balanced equation a constant belongs to. The "add and multiply" rule powers the most demanding PYQs: NEET 2017 supplied $K_1$, $K_2$ and $K_3$ for three sub-reactions and asked for $K$ of a target reaction assembled from them — solved by combining $\text{(3)}\times 3 + \text{(2)} - \text{(1)}$, giving $K = \dfrac{K_2\,K_3^{\,3}}{K_1}$. Multiplication of a step by 3 cubes its constant; reversing the third step would invert it.
The three permitted moves on a balanced equation map onto fixed operations on $K$: reversing inverts it, scaling every coefficient by $n$ raises it to the power $n$, and adding reactions multiplies their constants — each following directly from the form of the $K_c$ expression.
Units of the equilibrium constant
$K_c$ is computed by substituting concentrations in $\text{mol L}^{-1}$; $K_p$ by substituting partial pressures in Pa, kPa, bar or atm. Whether the result carries a unit depends on whether the exponents of numerator and denominator match — that is, on $\Delta n$. When they are equal the units cancel and $K$ is dimensionless; otherwise $K$ carries a unit.
| Reaction | Δn | Unit of Kc | Unit of Kp |
|---|---|---|---|
| $\ce{H2(g) + I2(g) <=> 2HI(g)}$ | 0 | none | none |
| $\ce{N2O4(g) <=> 2NO2(g)}$ | +1 | $\text{mol L}^{-1}$ | bar |
Equilibrium constants can also be written as pure numbers by referencing standard states: 1 bar for a pure gas and 1 mol L$^{-1}$ for a solute. A partial pressure of 4 bar then becomes $4\ \text{bar} / 1\ \text{bar} = 4$, a dimensionless quantity. This convention is what allows $K$ to be slotted into thermodynamic relations such as $\Delta_r G^\circ = -RT\ln K$ without unit clashes.
"K has no units" is true only when Δn = 0
A frequent error is to treat all equilibrium constants as dimensionless. They are not, unless standard states are explicitly invoked or $\Delta n = 0$. For $\ce{N2O4 <=> 2NO2}$, $K_c$ genuinely carries $\text{mol L}^{-1}$ and $K_p$ carries bar. Conversely, students sometimes assign a unit to $\ce{H2 + I2 <=> 2HI}$ where there is none.
Unit of $K_c = (\text{mol L}^{-1})^{\Delta n}$; unit of $K_p = (\text{bar})^{\Delta n}$. With $\Delta n = 0$, both are unitless.
The equilibrium constant in one screen
- Law of mass action: $K_c = \dfrac{[\ce{C}]^c[\ce{D}]^d}{[\ce{A}]^a[\ce{B}]^b}$ — products over reactants, each to its coefficient (Guldberg–Waage, 1864).
- Kp–Kc bridge: $K_p = K_c(RT)^{\Delta n}$ with $R = 0.0831\ \text{bar L mol}^{-1}\text{K}^{-1}$ and $\Delta n$ counting gaseous moles only.
- Equal only at $\Delta n = 0$: e.g. $\ce{H2 + I2 <=> 2HI}$ has $K_p = K_c$; $\ce{PCl5 <=> PCl3 + Cl2}$ has $K_p \neq K_c$.
- Manipulation: reverse $\to 1/K$; scale by $n \to K^n$; add reactions $\to$ multiply constants.
- Units: $K_c$ in $(\text{mol L}^{-1})^{\Delta n}$, $K_p$ in $(\text{bar})^{\Delta n}$ — dimensionless when $\Delta n = 0$ or when standard states are specified.
- Heterogeneous: pure solids and liquids do not appear in $K$.