ICE Table: Equilibrium Concentrations of H2, I2, and HI

Medium

Problem

A flask initially contains 0.100 M H2 and 0.100 M I2 with no HI. For the reaction H2(g) + I2(g) ⇌ 2HI(g) at this temperature, Kc = 50.2. Calculate the equilibrium concentrations of all three species.

Given

Reaction: H2(g) + I2(g) ⇌ 2HI(g)
Initial [H2] = 0.100 M, initial [I2] = 0.100 M, initial [HI] = 0 M
Kc = 50.2 at the reaction temperature
The system starts with no product, so the reaction shifts right (x > 0).

Find

The equilibrium concentrations [H2], [I2], and [HI].

Principle

An ICE table tracks concentrations through three rows: Initial, Change, and Equilibrium. The change row uses one variable x scaled by the stoichiometric coefficients (reactants decrease, products increase). Substituting the equilibrium row into the Kc expression gives an equation in x alone; here taking the square root of both sides avoids solving a quadratic.

Formula

K_c = \frac{[HI]^2}{[H_2][I_2]}

Kc = [HI]^2 / ([H2][I2]), with ICE: Initial, Change, Equilibrium rows

Solution

Set up the ICE table. Let x be the amount of H2 consumed. Initial: [H2] = 0.100, [I2] = 0.100, [HI] = 0. Change: [H2] = -x, [I2] = -x, [HI] = +2x. Equilibrium: [H2] = 0.100 - x, [I2] = 0.100 - x, [HI] = 2x.
Write Kc: Kc = [HI]^2 / ([H2][I2]) = (2x)^2 / (0.100 - x)^2 = 4x^2 / (0.100 - x)^2 = 50.2.
Take the square root of both sides to avoid a quadratic: 2x / (0.100 - x) = sqrt(50.2) = 7.086.
Multiply out: 2x = 7.086(0.100 - x) = 0.7086 - 7.086x.
Collect x terms: 2x + 7.086x = 9.086x = 0.7086, so x = 0.7086 / 9.086 = 0.0780 M.
Read off equilibrium concentrations: [HI] = 2x = 0.156 M; [H2] = [I2] = 0.100 - 0.0780 = 0.0220 M.
Verify in Kc: (0.156)^2 / (0.0220)^2 = 0.024336 / 0.000484 = 50.3, which matches Kc = 50.2 within rounding. Note x < 0.100, so the result is physically valid.

[HI] = 0.156 M, [H2] = [I2] = 0.0220 M

Unit check

Kc for this reaction has units (M^2)/(M x M), which cancel to dimensionless, and all concentrations substituted are in mol/L. The variable x = 0.0780 M has concentration units, and the check calculation returns Kc with no units, consistent with the given value.

What this means physically

Because Kc = 50.2 is much larger than 1, the equilibrium lies far to the right: most of the H2 and I2 convert, leaving only 0.0220 M of each reactant and building 0.156 M of product HI. The ICE table turned the word problem into one unknown, x, which the equilibrium expression solved directly.

Common mistake

Students often forget the stoichiometric coefficient 2 on HI and write [HI] = x instead of 2x, or forget to square the (0.100 - x) denominator. A second classic error is plugging initial concentrations directly into Kc instead of the equilibrium expressions.