Chemical Equilibrium
When forward and reverse reactions run at equal pace — the balance point, the constant that describes it, and how to shift it
- What dynamic equilibrium means and why it is reached.
- How to write the equilibrium constant \(K_c\) from the law of mass action.
- The pressure constant \(K_p\) and the relation \(K_p=K_c(RT)^{\Delta n_g}\).
- How the reaction quotient \(Q\) predicts the direction of change.
- The link between \(K\) and Gibbs free energy, and the meaning of degree of dissociation.
- How Le Chatelier's principle predicts the response to changes in concentration, pressure and temperature.
Reversible Reactions & Dynamic Equilibrium
Many reactions are reversible — products re-form reactants as fast as reactants form products. In a closed system, the forward rate falls as reactants deplete while the reverse rate climbs, until the two are equal. At that point concentrations stop changing: the system is at equilibrium.
Equilibrium is dynamic, not static. Both reactions continue at the molecular level; only the net change is zero. Macroscopic properties — colour, pressure, concentration — become constant.
The Law of Mass Action & Kc
The law of mass action states that the rate of a reaction is proportional to the product of the active masses (molar concentrations) of the reactants, each raised to its stoichiometric power. Applying it to forward and reverse rates at equilibrium gives a constant ratio.
Concentrations are those at equilibrium. At a fixed temperature, \(K_c\) is constant no matter the starting amounts — the defining feature of equilibrium.
Kp and Its Relation to Kc
For gaseous equilibria it is often easier to use partial pressures. The constant \(K_p\) is written exactly like \(K_c\) but with partial pressures in place of concentrations. The two are related through the ideal gas law.
Features of the Equilibrium Constant
A few properties of \(K\) are worth fixing firmly, because they govern how it is manipulated.
Writing the reaction backward inverts the constant: \(K_{\text{reverse}}=\dfrac{1}{K_{\text{forward}}}\).
Multiplying coefficients by \(n\) raises the constant to the \(n\)th power: \(K'=K^{n}\).
Adding reactions multiplies their constants: \(K=K_1\times K_2\).
Large \(K\) means products dominate; small \(K\) means reactants dominate; \(K\approx1\) means comparable amounts.
The Reaction Quotient & Direction
The reaction quotient \(Q\) has the same form as \(K\) but uses the concentrations at any moment, not just at equilibrium. Comparing \(Q\) with \(K\) tells you which way the reaction must move.
K and Gibbs Free Energy
Equilibrium is the state of minimum free energy. The connection between the thermodynamics of Chapter 5 and the equilibrium constant is exact and central.
At equilibrium \(\Delta G=0\) and \(Q=K\). A negative \(\Delta G^{\circ}\) gives \(K>1\) (products favoured); a positive \(\Delta G^{\circ}\) gives \(K<1\) (reactants favoured).
Degree of Dissociation
For dissociation equilibria, the degree of dissociation \(\alpha\) is the fraction of the original substance that has dissociated. Building an ICE table (Initial, Change, Equilibrium) in terms of \(\alpha\) turns most equilibrium problems into algebra.
Homogeneous & Heterogeneous Equilibria
A homogeneous equilibrium has all species in one phase; a heterogeneous one spans phases. The key rule for the latter: pure solids and pure liquids have constant "concentration" (their density is fixed), so they are omitted from the equilibrium expression.
For \(\ce{CaCO3(s) <=> CaO(s) + CO2(g)}\), both solids are dropped, leaving simply \(K_p=p_{\ce{CO2}}\). The pressure of carbon dioxide alone defines the equilibrium.
Le Chatelier's Principle
If a system at equilibrium is disturbed, it shifts in the direction that partly undoes the disturbance. This single principle predicts the effect of every common change.
| Change applied | Equilibrium shifts… |
|---|---|
| Add reactant (or remove product) | Forward, toward products |
| Add product (or remove reactant) | Backward, toward reactants |
| Increase pressure (decrease volume) | Toward the side with fewer gas moles |
| Increase temperature | In the endothermic direction |
| Add a catalyst | No shift — equilibrium reached faster |
| Add inert gas at constant volume | No shift |
Concentration, pressure and volume changes shift the position of equilibrium but leave \(K\) unchanged. Only a change in temperature alters the value of the equilibrium constant itself.
Equilibrium in Industry
Industrial chemistry is applied Le Chatelier. The Haber process for ammonia, \(\ce{N2 + 3H2 <=> 2NH3}\) (exothermic, fewer gas moles on the right), is run at high pressure (favours \(\ce{NH3}\)) and a moderate temperature — low enough to favour product, high enough for a workable rate — with an iron catalyst to reach equilibrium quickly.
Putting It to Work
Problem. For \(\ce{H2 + I2 <=> 2HI}\), equilibrium concentrations are \([\ce{H2}]=[\ce{I2}]=0.5\) and \([\ce{HI}]=2.0\ \text{M}\). Find \(K_c\).
Solution. Apply the mass-action expression:
Problem. For \(\ce{N2 + 3H2 <=> 2NH3}\) at \(500\ \text{K}\), \(K_c=0.061\). Find \(K_p\).
Solution. Gas moles change \(4\to2\), so \(\Delta n_g=-2\); use \(R=0.0821\):
Problem. A reaction has \(K_c=50\). At a moment \(Q_c=10\). Which way does it proceed?
Solution. Compare \(Q\) with \(K\):
There is too little product, so net change is toward the right.
Problem. \(\ce{PCl5}\) is \(50\%\) dissociated at total pressure \(1\ \text{atm}\). Find \(K_p\).
Solution. With \(\alpha=0.5,\ P=1\) in \(K_p=\dfrac{\alpha^2}{1-\alpha^2}P\):
Problem. For \(\ce{2SO2 + O2 <=> 2SO3}\) (exothermic), predict the effect of (a) raising pressure, (b) raising temperature.
Solution. (a) The right side has fewer gas moles \((3\to2)\), so higher pressure shifts the equilibrium forward, raising \(\ce{SO3}\). (b) The forward reaction is exothermic, so higher temperature shifts it backward, lowering \(\ce{SO3}\) and decreasing \(K\).
Problem. A reaction has \(\Delta G^{\circ}=-5.7\ \text{kJ mol}^{-1}\) at \(298\ \text{K}\). Find \(K\).
Solution. From \(\Delta G^{\circ}=-RT\ln K\) with \(R=8.314\):
Chapter Summary
Equilibrium is equal forward and reverse rates, with constant macroscopic properties.
\(K_c\) from concentrations, \(K_p\) from pressures, with \(K_p=K_c(RT)^{\Delta n_g}\).
\(Q
\(\Delta G^{\circ}=-RT\ln K\) ties equilibrium to thermodynamics.
Omit pure solids and liquids from \(K\).
Systems oppose disturbances; only temperature changes \(K\).
Problems
Write the balanced equation, build an ICE table where amounts change, and decide whether \(K_c\) or \(K_p\) is asked. Difficulty rises down the list.
- Write the \(K_c\) expressions for \(\ce{2NO2 <=> N2O4}\) and \(\ce{CaCO3(s) <=> CaO(s) + CO2(g)}\).
- At equilibrium \([\ce{N2}]=0.2,\ [\ce{H2}]=0.4,\ [\ce{NH3}]=0.6\ \text{M}\). Find \(K_c\) for \(\ce{N2 + 3H2 <=> 2NH3}\).
- For a reaction \(K_c=4\). Find \(K_p\) at \(300\ \text{K}\) if \(\Delta n_g=+1\).
- If \(K\) for \(\ce{A <=> B}\) is \(8\), find \(K\) for \(\ce{2B <=> 2A}\).
- A reaction has \(K_c=25\) and a current \(Q_c=25\). State what this means.
- \(1\ \text{mol}\) each of \(\ce{H2}\) and \(\ce{I2}\) in a \(1\ \text{L}\) flask reach equilibrium with \(K_c=49\) for \(\ce{H2 + I2 <=> 2HI}\). Find the equilibrium \([\ce{HI}]\).
- \(\ce{N2O4}\) is \(20\%\) dissociated at \(1\ \text{atm}\). Find \(K_p\) for \(\ce{N2O4 <=> 2NO2}\).
- State and justify the effect of increasing pressure on \(\ce{N2 + O2 <=> 2NO}\).
- For an exothermic reaction, sketch how \(K\) changes as temperature rises, and explain using Le Chatelier.
- A reaction at \(298\ \text{K}\) has \(K=1\times10^{-3}\). Find \(\Delta G^{\circ}\).
- \(\ce{PCl5}\) dissociates such that the equilibrium total pressure is \(2\ \text{atm}\) with \(\alpha=0.4\). Find \(K_p\).
- For \(\ce{2SO2 + O2 <=> 2SO3}\), explain how adding more \(\ce{O2}\) affects the yield of \(\ce{SO3}\) and the value of \(K\).