Topic 13: Chemical Equilibria

Write equilibrium concentrations of products over reactants and use the balanced-equation coefficients as powers. Square brackets mean concentration in mol dm$^{-3}$; coefficients are exponents, not multiplying factors.

For $\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$, $K_c=[\ce{CO2}]$. The two solids are present in the equilibrium but omitted from the expression. For a homogeneous mixture, include every participating species in that phase.

Use only gaseous species. Each $p(X)$ is the equilibrium partial pressure of gas X in atm, and each balanced coefficient becomes its power. Solids, liquids and aqueous species are omitted from $K_p$.

For $\ce{Mg(NO3)2(s) <=> MgO(s) + 2NO2(g) + 1/2O2(g)}$, $K_p=p(\ce{NO2})^2p(\ce{O2})^{1/2}$. The solid terms do not appear.

Keep initial, change and equilibrium amounts separate. If 1.60 mol $\ce{SO3}$ forms from $\ce{2SO2 + O2 <=> 2SO3}$, then 1.60 mol $\ce{SO2}$ and 0.80 mol $\ce{O2}$ are consumed before mole fractions are calculated.

$\Delta n$ is products minus reactants for the species actually present in the chosen constant expression. If $\Delta n=0$, the units cancel. State units only where appropriate and retain enough figures during intermediate steps.

Pressure has no effect on equilibrium composition when both sides contain the same total moles of gas. Count gaseous coefficients only; pure solids and liquids do not enter this comparison.

In a heterogeneous system, changing the amount or surface area of a pure solid does not change the equilibrium composition while some solid remains. It can change how quickly equilibrium is reached.

Industrial conditions balance equilibrium yield with rate and cost. A high temperature may reduce the equilibrium yield of an exothermic product yet be chosen because the low-temperature rate is too slow.

After a concentration or pressure disturbance, the reaction proceeds in the direction that restores the composition quotient to the unchanged equilibrium-constant value. The new equilibrium composition can differ even though K is identical.

A catalyst lowers the activation energy for both directions. It shortens the time taken to reach equilibrium but does not alter the equilibrium composition, yield or constant.

The direction also works backwards as evidence. If lowering temperature makes $K_c$ smaller, the forward reaction is endothermic. If raising temperature makes $K_p$ smaller, the forward reaction is exothermic.

A new K requires new equilibrium partial pressures or concentrations. For an exothermic forward reaction at higher temperature, product terms decrease relative to reactant terms, so the calculated constant becomes smaller.

Use a complete causal chain: identify whether the forward direction is endothermic or exothermic; state how the temperature change alters K; then state which side must become more abundant so the equilibrium expression attains that new K.

Hot pure water can have pH below 7 while remaining neutral because $[\ce{H+}]=[\ce{OH-}]$; the larger $K_w$ increases both concentrations.

For an exothermic forward reaction, $-\Delta H/T$ is positive. Raising T makes this positive surroundings contribution smaller, so $\Delta S_{total}$ and $\ln K$ decrease; K decreases.

For an endothermic forward reaction, $-\Delta H/T$ is negative. Raising T makes it less negative, so $\Delta S_{total}$ and $\ln K$ increase; K increases. This comparison assumes $\Delta S_{system}$ is approximately constant over the stated range.

Describe the equilibrium position as lying toward products or reactants. A large K says the equilibrium is product-favoured; it does not mean that the equilibrium is currently 'shifting' right.

When K is already very large, a pressure increase that favours products may deliver only a small extra equilibrium yield. The small gain may not justify higher compression cost or risk.

A value such as $K_p=1.55 imes10^6$ indicates an equilibrium far toward products. A value such as $K_c=4.85$ indicates more products than reactants for a comparable simple mixture, but appreciable amounts of both remain.