Training Chemistry Math Chemical Equilibrium and the ICE Table
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Chemical Equilibrium and the ICE Table

24 min Chemistry Math

Chemical Equilibrium

When a reversible reaction reaches equilibrium, the rates of the forward and reverse reactions are equal. The relationship between concentrations is described by the equilibrium constant.

Equilibrium Constant

For $aA + bB \rightleftharpoons cC + dD$:

$$K = \frac{[C]^c[D]^d}{[A]^a[B]^b}$$

ICE Table Method

Initial → Change → Equilibrium. Set up a table with concentrations, let $x$ be the change, then substitute into the $K$ expression and solve for $x$.

Example 1

For $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$, if $K = 50$ and we start with $[\text{H}_2] = [\text{I}_2] = 1.0$ M, find equilibrium concentrations.

ICE table (change = $-x$ for reactants, $+2x$ for product):

$$K = \frac{(2x)^2}{(1-x)(1-x)} = \frac{4x^2}{(1-x)^2} = 50$$

Take square root: $\frac{2x}{1-x} = \sqrt{50} \approx 7.07$

$2x = 7.07(1-x)$ → $2x = 7.07 - 7.07x$ → $9.07x = 7.07$ → $x \approx 0.78$

$[\text{H}_2] = [\text{I}_2] = 0.22$ M, $[\text{HI}] = 1.56$ M.

Example 2

If $K = 4.0$ for $A \rightleftharpoons 2B$, and we start with $[A] = 1.0$ M, find $[B]$ at equilibrium.

$K = \frac{(2x)^2}{1-x} = \frac{4x^2}{1-x} = 4$

$4x^2 = 4 - 4x$ → $x^2 + x - 1 = 0$ → $x = \frac{-1 + \sqrt{5}}{2} \approx 0.618$

$[B] = 2(0.618) = 1.236$ M.

Practice Problems

1. Write the $K$ expression for $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$.
2. If $[\text{NH}_3] = 0.5$ M, $[\text{N}_2] = 0.1$ M, $[\text{H}_2] = 0.2$ M at equilibrium, find $K$.
3. For $A \rightleftharpoons B$ with $K = 9$, start with $[A] = 1$ M. Find $[B]$ at equilibrium. (Hint: let $x$ shift.)
4. If $K \gg 1$, does the equilibrium favor products or reactants?
5. What happens to $K$ when you reverse the reaction?
6. Le Chatelier's principle: if you add more reactant to a system at equilibrium, which direction does the reaction shift?
Show Answer Key

1. $K = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$

2. $K = \frac{0.5^2}{0.1 \times 0.2^3} = \frac{0.25}{0.0008} = 312.5$

3. $K = x/(1-x) = 9$ → $x = 9/10 = 0.9$ → $[B] = 0.9$ M

4. Products

5. The new constant is $1/K$.

6. Toward products (right).

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