Equilibrium
ICE Tables
ICE tables are used to solve equilibrium problems. They stand for Initial, Change, and Equilibrium.
Format:
| Reactant A | Reactant B | Product C | |
|---|---|---|---|
| Initial | $ x_A $ | $ x_B $ | $ x_C $ |
| Change | $ \Delta x_A $ | $ \Delta x_B $ | $ \Delta x_C $ |
| Equilibrium | $ x_A + \Delta x_A $ | $ x_B + \Delta x_B $ | $ x_C + \Delta x_C $ |
Where $ x_A $ , $ x_B $ , and $ x_C $ are the initial concentrations (or partial pressures) of A, B, and C respectively. $ \Delta x_A $ , $ \Delta x_B $ , and $ \Delta x_C $ represent the change in concentration (or partial pressure) during the reaction to reach equilibrium. The equilibrium concentrations are found by summing the initial and change values.
Example:
Consider the reaction: $ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) $
Let’s say we start with 1.0 M $ N_2 $ and 1.5 M $ H_2 $ , and at equilibrium, we have 0.6 M $ NH_3 $ . We can set up the ICE table as follows:
| $ N_2 $ | $ 3H_2 $ | $ 2NH_3 $ | |
|---|---|---|---|
| Initial | 1.0 M | 1.5 M | 0 M |
| Change | $ -x $ | $ -3x $ | $ +2x $ |
| Equilibrium | $ 1.0 - x $ | $ 1.5 - 3x $ | $ 2x $ |
Since we know the equilibrium concentration of $ NH_3 $ is 0.6 M, we can solve for $ x $ :
$ 2x = 0.6 M $ $ \implies x = 0.3 M $
Therefore, the equilibrium concentrations are:
$ [N_2 = 1.0 - 0.3 = 0.7 M $ $ [H_2 = 1.5 - 3(0.3) = 0.6 M $ $ [NH_3 = 0.6 M $
Important Considerations:
- The stoichiometric coefficients are crucial in determining the changes in concentration.
- The change in concentration is always negative for reactants and positive for products.
- The equilibrium constant ( $ K $ ) expression can be used to solve for unknown values in the ICE table. Equilibrium Constants
- For problems involving small $ K $ values, we can often use approximations to simplify the calculations. Approximations in Equilibrium Calculations