5.1.2(f)-(h) - Equilibrium constants and position of equilibrium

K as a Target Ratio

A numerical comparison such as K much greater than 1 must use the stated balanced equation and unit convention. It does not directly give product amounts for arbitrary stoichiometric powers or dimensionful constants. For A(aq) ⇌ B(aq), $K_c$ = $[B]$/$[A]$ is dimensionless and a large $K_c$ directly means much more B than A. For the same equation and units, increasing K with temperature means products become more favoured relative to reactants.

An equilibrium constant is a number calculated from the equilibrium amounts of products and reactants for a particular reversible equation at a fixed temperature. For $K_c$, the expression uses equilibrium concentrations. For $K_p$, the expression uses equilibrium partial pressures.

The exact expression depends on the balanced equation, but the meaning is always similar: products are compared with reactants. For a simple dimensionless 1:1 expression, a large constant directly gives a large product-to-reactant concentration ratio, as in the example below. For more complicated expressions, interpret the powers and units rather than assuming the numerical constant equals an amount ratio.

If the same expression is evaluated before equilibrium is restored, it is often called the reaction quotient, $Q$. At equilibrium $Q=K$. You can use this notation, or simply call it the current concentration or pressure ratio.

Think of $K$ as a target ratio that an equilibrium mixture must satisfy at a particular temperature. The individual concentrations or partial pressures can change, but when the system has settled back to equilibrium at the same temperature, the expression must again have the same value of $K$.

Worked reasoning: estimating position from $K$

For a simple equilibrium:

the expression is:

If $K_c = 50$, then at equilibrium $[\mathrm{B}]$ is much larger than $[\mathrm{A}]$. The position is to the right and products are favoured.

If $K_c = 0.020$, then at equilibrium $[\mathrm{B}]$ is much smaller than $[\mathrm{A}]$. The position is to the left and reactants are favoured.

This is an estimate, not a full composition calculation. OCR may ask you to decide whether a change makes products or reactants more favoured, or whether $K$ increases, decreases or stays the same.

Temperature Changes K

Temperature is different from concentration, pressure and catalysts because temperature changes the relative favourability of the forward and reverse reactions. Therefore it can change the value of an equilibrium constant.

Use the enthalpy change for the forward reaction to decide the direction.

The reverse happens when temperature is decreased.

The reason links position and $K$. If the new temperature makes products more favoured, the products-over-reactants ratio at equilibrium is larger, so $K$ is larger. If the new temperature makes reactants more favoured, the products-over-reactants ratio is smaller, so $K$ is smaller.

Worked reasoning: using $K_p$ values to infer temperature

Sulfur trioxide can decompose:

The forward reaction is endothermic. If the temperature is increased, the equilibrium shifts to the right because the forward direction absorbs heat. More $\mathrm{SO_2}$ and $\mathrm{O_2}$ are present relative to $\mathrm{SO_3}$, so the $K_p$ value increases.

If two experiments give $K_p = 3.3 \times 10^{-5}$ and $K_p = 7.7 \times 10^{-2}$, the larger $K_p$ corresponds to the higher temperature for this endothermic forward reaction.

Concentration and Pressure Restore K

Changing concentration or pressure can change the position of equilibrium, but it does not change the value of $K_c$ or $K_p$ at a fixed temperature.

The clean OCR-style explanation is not just "Le Chatelier says so". The equilibrium constant controls the new position. After a concentration or pressure change, the expression no longer has the value $K$. The mixture then shifts until the expression is restored to the same $K$.

Worked reasoning: concentration

For:

the expression is:

If more $\mathrm{A}$ is added at constant temperature, the denominator of the expression initially increases. The expression is now too small compared with $K_c$. To restore the same $K_c$, the equilibrium shifts to the right: $\mathrm{A}$ and $\mathrm{B}$ are used up and more $\mathrm{C}$ forms.

If more $\mathrm{C}$ is added, the numerator initially increases. The expression is now too large compared with $K_c$. To restore the same $K_c$, the equilibrium shifts to the left.

Worked reasoning: pressure

Pressure changes matter for gaseous equilibria. For the Haber equilibrium:

the pressure expression is:

If the total pressure is increased by compressing the mixture at constant temperature, all partial pressures initially increase. In this expression, the denominator contains four pressure factors in total: one from $p(\mathrm{N_2})$ and three from $p(\mathrm{H_2})^3$. The numerator contains two pressure factors from $p(\mathrm{NH_3})^2$.