1. 11-5 RICE Problems And you

2. A common problem in chemistry is to mix several chemicals, allow them to reach equilibrium, and then calculate the final concentrations. The steps involved in calculating the equilibrium concentrations are affectionately termed “RICE”. To illustrate the process, here’s an example:

3. : A mixture of 0.100 M NO, 0.050 M H 2, 0.100 M H 2 O was allowed to reach equilibrium (initially there was no N 2 ). At equilibrium, the concentration of NO was found to be 0.062 M. Determine the value of the equilibrium constant, K c, for the reaction: Step 1: Write the Reaction “R” 2 NO(g) + 2 H 2 (g) ↔ N 2 (g) + 2 H 2 O(g)

4. Step 2: Create an chart with initial concentrations “I”, the change in concentrations “C”, and the concentrations at equilibrium “E”. R:2 NO 2 H 2 N 2 2 H 2 O I:0.100 0.050000.100 C: E:0.062. Note how we have assigned the change in the concentration (x) based on the appearance (+) of nitrogen, since its mole coefficient is 1. The concentration of the reactants will then decrease (-) by 2x and the water will increase (+) by 2x. +X+2X-2X

5. R:2 NO 2 H 2 N 2 2 H 2 O I:0.100 0.05000 0.100 C:-2x -2x +x +2x E:0.062. The change in [NO] was 0.100 – 2x = 0.062. 2x = 0.0338 x =0.019

6. R:2 NO 2 H 2 N 2 2 H 2 O I:0.100 0.050000.100 C:-2x -2x +x +2x E:0.062. x =0.019 We can now calculate the remaining “E” values: H 2 = 0.0500 – 0.038 = 0.012M N 2 = 0.019M H 2 O = 0.100 + 0.038 = 0.138M 0.012M0.019M 0.138M

7. 2 NO 2 H 2 N 2 2 H 2 O 0.062M 0.012M 0.019M 0.138M Solving for K c [N 2 ][H 2 O] 2 [NO] 2 [H 2 ] 2 = (0.019)(0.138) 2 (0.062) 2 (0.012) 2 = 650 or 6.5 x 10 2

8. Cool huh! I have posted an additional powerpoint that walks through the steps sequentially if you need added examples to follow.