Chemical Equilibrium: MDCAT Chemistry notes
Chemical Equilibrium MDCAT notes: dynamic equilibrium, law of mass action, Kc expressions and units, Kp and Kc, magnitude of K and reaction quotient.
Reversible reactions and dynamic equilibrium
A reversible reaction can proceed in both directions. The forward reaction is the one written from left to right in the equation; the reverse (backward) reaction goes from right to left.
At the start the forward rate is highest. As reactants are used up it falls, while the reverse rate rises as products build up. Eventually the two rates become equal. This state is chemical equilibrium:
- It is dynamic: both reactions continue at equal rates.
- There is no net change in concentration. The concentrations become constant, but they are generally not equal to each other.
- It can be approached from either side and is reached only in a closed system.
- A catalyst helps equilibrium to be reached sooner, but does not change the equilibrium position or $K_c$.
The ultimate fate of any reversible reaction in a closed vessel is this state of no net concentration change, not completion.
Law of mass action and Kc
The rate of a reaction is proportional to the product of the active masses (molar concentrations) of the reactants. For $a\mathrm{A} + b\mathrm{B} \rightleftharpoons c\mathrm{C} + d\mathrm{D}$:
$$K_c = \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}$$
$K_c$ is always written as products over reactants, each raised to the power of its coefficient. Working backwards, an expression $\dfrac{[\mathrm{C}]^2}{[\mathrm{A}][\mathrm{B}]}$ comes from $\mathrm{A + B \rightleftharpoons 2C}$. $K_c$ changes only with temperature.
Units of Kc
| Reaction | $\Delta n$ | Unit of $K_c$ |
|---|---|---|
| $\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$ | 0 | No unit |
| $\mathrm{PCl_5 \rightleftharpoons PCl_3 + Cl_2}$ | +1 | $\mathrm{mol\,dm^{-3}}$ |
| $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$ | −2 | $\mathrm{dm^6\,mol^{-2}}$ |
| $\mathrm{2SO_2 + O_2 \rightleftharpoons 2SO_3}$ | −1 | $\mathrm{dm^3\,mol^{-1}}$ |
Here $\Delta n$ = moles of gaseous products − moles of gaseous reactants. The unit is $(\mathrm{mol\,dm^{-3}})^{\Delta n}$.
Kp and Kc
For gas reactions $K_p$ uses partial pressures. The two are related by $K_p = K_c(RT)^{\Delta n}$.
- $\Delta n = 0$: $K_p = K_c$ (equal moles of gaseous reactants and products).
- $\Delta n \gt 0$: $K_p \gt K_c$, e.g. $\mathrm{2NOCl \rightleftharpoons 2NO + Cl_2}$ (2 to 3 moles).
- $\Delta n \lt 0$: $K_p \lt K_c$, e.g. ammonia synthesis.
Meaning of the magnitude of K
- Large K (ratio above 1): equilibrium lies towards products; products are more than reactants.
- Very small K: very little forward reaction; mostly reactants at equilibrium.
- K near 1: appreciable amounts of both.
Reaction quotient Qc
$Q_c$ has the same form as $K_c$ but uses concentrations at any moment.
- $Q_c \lt K_c$: too few products; reaction moves forward.
- $Q_c = K_c$: at equilibrium.
- $Q_c \gt K_c$: too many products; more reactants are needed, so the reaction moves backward.
Key formulas
- $K_c = \dfrac{[\text{products}]^{\text{coeff}}}{[\text{reactants}]^{\text{coeff}}}$
- $K_p = K_c(RT)^{\Delta n}$
- Unit of $K_c$: $(\mathrm{mol\,dm^{-3}})^{\Delta n}$
Common MDCAT traps
- At equilibrium concentrations are constant, not equal and not zero.
- A catalyst makes equilibrium arrive earlier; temperature, pressure and concentration can shift it.
- If $Q_c \gt K_c$, the reaction goes backward, not forward.
- For $\mathrm{PCl_5}$ decomposition the unit is $\mathrm{mol\,dm^{-3}}$, not its reciprocal.
Quick revision
- Equilibrium is dynamic: forward rate equals reverse rate.
- $K_c$ depends only on temperature.
- $K_p = K_c$ when $\Delta n = 0$.
- Small $K_c$ means little forward reaction.
- Coefficients in the equation become powers in $K_c$.