Equilibrium constant calculations

The three rows. Every ICE table has the same shape: write the balanced equation across the top, then one row each for Initial concentration, Change (always $\pm x$ times the stoichiometric coefficient — negative for reactants being consumed, positive for products being formed), and Equilibrium (Initial plus Change, algebraically). For a generic reaction $aA + bB \rightleftharpoons cC + dD$, the equilibrium concentrations read straight off the bottom row, and those are the only numbers allowed into the $K_c$ expression:

where: square brackets denote equilibrium molar concentrations (mol L⁻¹), and the exponents $a,b,c,d$ are the balanced-equation stoichiometric coefficients — not arbitrary powers, and not the coefficients of the initial or change rows, which must first be combined into the equilibrium row before substitution.

Two directions of the same table. Nearly every numerical splits into one of two directions. In a forward calculation you are given $K_c$ and the initial concentrations, and you solve the resulting algebraic equation for $x$ — sometimes a simple linear equation, sometimes a quadratic that needs the quadratic formula or a careful small-$x$ approximation. In a reverse calculation you are given some or all of the equilibrium concentrations directly (often from experiment — titration, spectroscopy, or a mass-balance measurement) and you substitute straight into the $K_c$ expression to compute a number. Both directions use the identical ICE table; only the direction of substitution changes.

Pressures instead of concentrations. When a reaction is entirely gaseous and pressures are the natural measurement, the same logic runs with an ICE table in partial pressures instead of concentrations, feeding into $K_p$ instead of $K_c$. The two constants are related by

where: $R = 0.0821\ \text{L atm mol}^{-1}\text{K}^{-1}$ (matching the units of $K_c$ and pressures in atm), $T$ is the absolute temperature in kelvin, and $\Delta n_g$ is the change in moles of gas, (moles of gaseous products) minus (moles of gaseous reactants), counted only from species that are actually gases.

The small-$x$ shortcut. When $K_c$ is very small compared with the initial concentrations, the extent of reaction $x$ often turns out to be negligible next to $[A]_0$, so $[A]_0 - x \approx [A]_0$ simplifies a quadratic into a one-line calculation. This shortcut is never assumed blindly: after solving for $x$ under the approximation, you check that $x$ is genuinely small (a common rule of thumb is under about $5\%$ of $[A]_0$); if it fails the check, the exact quadratic must be solved instead. This approximation is used heavily for weak-electrolyte ionisation elsewhere in the equilibrium chapter, but the logic is identical here.

Extending the Pass-1 case: finding equilibrium composition, then checking the small-$x$ shortcut would have failed. For $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$, $K_c = 64$ at a given temperature, and $1.0\ \text{mol}$ each of $\text{H}_2$ and $\text{I}_2$ are placed in a $1.0\ \text{L}$ vessel. Find $[\text{HI}]$ at equilibrium, and comment on whether the small-$x$ approximation could have been used.

Step 1. Set up the ICE table (concentrations in $\text{mol L}^{-1}$, since $V = 1.0\ \text{L}$):

Step 2. Substitute the equilibrium row into $K_c$:

Step 3. Both sides are perfect squares, so take the square root of both sides (valid since $0 < x < 1$ keeps every term positive): $\dfrac{2x}{1.0-x} = 8 \implies 2x = 8 - 8x \implies 10x = 8 \implies x = 0.80$.

Step 4. Equilibrium concentrations: $[\text{H}_2] = [\text{I}_2] = 1.0 - 0.80 = 0.20\ M$; $[\text{HI}] = 2(0.80) = 1.60\ M$. Check: $\dfrac{(1.60)^2}{(0.20)(0.20)} = \dfrac{2.56}{0.04} = 64$ ✓. Equilibrium $[\text{HI}] = 1.60\ M$.

Step 5 (the shortcut check). Could the small-$x$ approximation ($1.0 - x \approx 1.0$) have saved algebra here? No — with $K_c = 64$ being large rather than small, $x = 0.80$ turns out to be $80\%$ of the initial $1.0\ M$, nowhere near the "under about $5\%$" threshold that justifies dropping $x$. The exact quadratic (here solvable by a square root) had to be used. This is the same reasoning, run in reverse, as the small-$x$ check flagged in the Pass-1 article: always verify the size of $x$ before trusting the shortcut, whichever direction the problem runs.

Competitive stretch

If the same vessel were instead $2.0\ \text{L}$ (with the same $1.0\ \text{mol}$ each of $\text{H}_2$ and $\text{I}_2$), the initial concentrations would each halve to $0.50\ M$, but $K_c$ for this particular reaction has $\Delta n_g = 2 - 2 = 0$, so scaling every concentration by the same factor leaves the ratio in $K_c$ unchanged — the equilibrium mole fractions come out identical, only the volume differs. This is a quick sanity check worth knowing: for a $\Delta n_g = 0$ reaction, diluting does not shift the position of equilibrium at all.

• Substituting initial concentrations into $K_c$ instead of equilibrium concentrations. Only the bottom (Equilibrium) row of the ICE table belongs in the $K_c$ expression; the Initial row is just the starting point.
• Getting the stoichiometric scaling of $x$ wrong. A coefficient of $2$ means the change is $\pm 2x$, not $\pm x$; check every species against the balanced equation before writing the Change row.
• Using the small-$x$ approximation without checking it afterwards. If $x$ turns out to be more than about $5\%$ of the initial concentration, the approximation is invalid and the exact (usually quadratic) equation must be solved instead.
• Confusing $K_c$ and $K_p$, or mixing their units. Convert with $K_p = K_c(RT)^{\Delta n_g}$, keeping $R$, $T$, and pressure/concentration units consistent, and count $\Delta n_g$ only from gaseous species.

Ask the tutor: Want to try an ICE-table problem with a genuine quadratic (where the square-root shortcut does not work), or check whether the small-$x$ approximation applies to a specific $K_c$ value?