Equilibrium • Topic 1 of 3

Physical & Chemical Equilibrium

Equilibrium is the state in a reversible process where the rates of the forward and backward changes become equal, so the measurable properties of the system stop changing with time. It is not a dead stop — it is dynamic: both directions keep going at the same rate, like an escalator on which people walk down while it moves up, leaving the crowd unchanged.

Physical equilibrium

In physical processes no new substance forms; only the state changes. At equilibrium the two phases coexist with constant concentrations.

  • Liquid ↔ Vapour (in a closed vessel): rate of evaporation = rate of condensation, giving a constant saturated vapour pressure at a fixed temperature.
  • Solid ↔ Liquid at the melting point: ice and water coexist at $0^\circ\text{C}$.
  • Dissolution: in a saturated solution the rate of dissolving = rate of crystallising, so $[\text{solute}]$ stays fixed.

Chemical equilibrium & the law of mass action

For a reversible reaction $aA+bB \rightleftharpoons cC+dD$, the rate of any reaction is proportional to the product of active masses (molar concentrations) of the reactants raised to their stoichiometric coefficients. At equilibrium the forward and reverse rates are equal, which leads to a constant ratio — the equilibrium constant:

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

For gaseous reactions we may use partial pressures to define $K_p$. The two constants are linked by

$$K_p=K_c(RT)^{\Delta n}$$

where $\Delta n=(c+d)-(a+b)$ is the change in the number of gaseous moles. When $\Delta n=0$, $K_p=K_c$. A large $K_c$ ($\gg 1$) means products dominate; a small $K_c$ ($\ll 1$) means reactants dominate.

Reaction quotient $Q$

The reaction quotient $Q_c$ has the same form as $K_c$ but uses the current (non-equilibrium) concentrations. Comparing $Q$ with $K$ predicts the direction of net change: if $QK$ the reverse reaction proceeds; if $Q=K$ the system is already at equilibrium.

Le Chatelier's principle

If a system at equilibrium is disturbed, it shifts in the direction that counteracts the disturbance.

  • Concentration: adding a reactant shifts forward; removing a product shifts forward.
  • Pressure (gases): increasing pressure shifts toward the side with fewer gaseous moles.
  • Temperature: raising $T$ favours the endothermic direction. Only temperature changes the value of $K$.
  • Catalyst: speeds both directions equally — reaches equilibrium faster but does not shift it.
Forward and reverse rates becoming equal at equilibriumRateTimeforward rate (falls)reverse rate (rises)equilibrium: rates equal
Solved Examples
1
Worked Example
For $N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)$, write the expressions for $K_c$ and $K_p$ and state the relation between them.
Solution
  1. Write $K_c$ from the law of mass action: $K_c=\dfrac{[NH_3]^2}{[N_2][H_2]^3}$.
  2. Write $K_p$ using partial pressures: $K_p=\dfrac{p_{NH_3}^2}{p_{N_2}\,p_{H_2}^3}$.
  3. Find $\Delta n=2-(1+3)=-2$, so $K_p=K_c(RT)^{-2}$.

Answer: $K_p=K_c(RT)^{-2}$.

2
Worked Example
At equilibrium a $1\ \text{L}$ vessel contains $0.4\ \text{mol}$ $H_2$, $0.4\ \text{mol}$ $I_2$ and $1.6\ \text{mol}$ $HI$ for $H_2+I_2\rightleftharpoons 2HI$. Calculate $K_c$.
Solution
  1. Concentrations (volume $=1\ \text{L}$): $[H_2]=0.4$, $[I_2]=0.4$, $[HI]=1.6\ \text{mol L}^{-1}$.
  2. Write $K_c=\dfrac{[HI]^2}{[H_2][I_2]}=\dfrac{(1.6)^2}{0.4\times 0.4}$.
  3. Compute: $\dfrac{2.56}{0.16}=16$.

Answer: $K_c=16$ (no units, since $\Delta n=0$).

3
Worked Example
For $PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$, $K_c=0.04\ \text{mol L}^{-1}$ at $250^\circ\text{C}$ ($523\ \text{K}$). Find $K_p$. Take $R=0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}$.
Solution
  1. Find $\Delta n=(1+1)-1=+1$.
  2. Use $K_p=K_c(RT)^{\Delta n}=0.04\times(0.0821\times 523)^1$.
  3. Compute $RT=0.0821\times 523=42.9$, so $K_p=0.04\times 42.9$.

Answer: $K_p\approx 1.72\ \text{atm}$.

4
Worked Example
For a reaction $K_c=50$ at a certain temperature. A mixture has $Q_c=10$. Predict the direction of the net reaction.
Solution
  1. Compare $Q_c$ with $K_c$: $Q_c=10$ and $K_c=50$.
  2. Since $Q_c
  3. The system responds by making more products.

Answer: The reaction proceeds in the forward (left-to-right) direction.

5
Worked Example
For $2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$ ($\Delta H<0$), state the effect on equilibrium of (i) increasing pressure and (ii) increasing temperature.
Solution
  1. Count gaseous moles: reactants $=3$, products $=2$; fewer moles on the product side.
  2. Increasing pressure shifts toward fewer moles → forward, making more $SO_3$.
  3. The reaction is exothermic, so raising $T$ favours the endothermic (reverse) direction, reducing $SO_3$ and lowering $K$.

Answer: (i) shifts forward (more $SO_3$); (ii) shifts backward (less $SO_3$).

6
Worked Example
$1\ \text{mol}$ of $PCl_5$ is heated in a $2\ \text{L}$ vessel and $40\%$ dissociates into $PCl_3$ and $Cl_2$. Calculate $K_c$.
Solution
  1. Moles dissociated $=0.40$, so at equilibrium $PCl_5=0.60$, $PCl_3=0.40$, $Cl_2=0.40\ \text{mol}$.
  2. Divide by volume $2\ \text{L}$: $[PCl_5]=0.30$, $[PCl_3]=0.20$, $[Cl_2]=0.20\ \text{mol L}^{-1}$.
  3. $K_c=\dfrac{[PCl_3][Cl_2]}{[PCl_5]}=\dfrac{0.20\times 0.20}{0.30}=\dfrac{0.04}{0.30}$.

Answer: $K_c\approx 0.133\ \text{mol L}^{-1}$.

Key Points

  • Equilibrium is dynamic: the forward and reverse rates are equal, so concentrations stay constant while both reactions continue.
  • Law of mass action gives $K_c=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}$; for gases $K_p=K_c(RT)^{\Delta n}$ with $\Delta n$ = change in gaseous moles.
  • A large $K$ means products are favoured; a small $K$ means reactants are favoured.
  • Reaction quotient $Q$ predicts direction: $QK$ goes reverse, $Q=K$ is equilibrium.
  • Le Chatelier's principle: a disturbed system shifts to oppose the change; only temperature alters the value of $K$, a catalyst only speeds attainment.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.At equilibrium in a closed system:
Explanation: Equilibrium is dynamic — both reactions continue at equal rates, so net concentrations are constant.
Q2.For $N_2+3H_2\rightleftharpoons 2NH_3$, the value of $\Delta n$ used in $K_p=K_c(RT)^{\Delta n}$ is:
Explanation: $\Delta n=2-(1+3)=-2$.
Q3.If $Q_c>K_c$ for a reaction, the net reaction proceeds:
Explanation: $Q>K$ means too many products, so the system shifts reverse to restore equilibrium.
Q4.Adding a catalyst to a system at equilibrium:
Explanation: A catalyst speeds both directions equally; equilibrium is reached faster but its position and $K$ are unchanged.
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