RICE Tables

RICE Tables equilibrium

Before diving into RICE table, let's review the basics of dynamic equilibrium and equilibrium constant. When a chemical reaction reaches a point where the reactant concentrations and the product concentrations are no longer changing (remains constant), it has reached equilibrium.

When a reaction reaches equilibrium it does not mean that the reaction has stopped, it only means that it is reacting in both directions at the same rate. In other words, the forward rate and the reverse rate of the reaction are equal!

$$ A \rightleftharpoons B $$

Figure 1 shows a Concentration vs. time graph for reactant A and product B. Notice that reactant A went from having a 0.7 M concentration to a 0.3 M concentration, whereas product B went from having 0 M concentration to having a concentration of 0.2 M.

Figure 1. Concentration vs. Time graph for A and B, Isadora Santos - StudySmarter Originals.

So, the change in concentration of A is equal to - 0.4 M, while the change in concentration of B is + 0.2 M. Since A, has decreased twice as much as B, has increases, then it means that to produce one B, we need two A's.

$$ \color {orchid} 2 \color {black} A_{(aq)} \to B_{(aq)} $$

Now, the equilibrium constant (K_eq) is simply the ratio of concentration products to concentration of reactants at equilibrium.

• If K is greater than 1, then products are favored (and the forward direction is favored).
• If K is less than 1, then reactants are favored (and the reserve direction is favored).

$$ K_{eq} = \frac{[Products]}{[Reactants]} $$

In this case, the equilibrium constant is:

$$ K_{eq} = \frac{[Products]}{[Reactants]} = \frac{[B]}{[A]^{\color {orchid} \textbf{2}}} = \frac{[0.2]}{[0.3]^{\color {orchid}2}}=2.2 $$

Let's look at an example:

When dealing which chemical equilibrium calculations, we can use RICE tables to show the initial concentration, the change in concentration and the concentration at equilibrium of reactants and products in a chemical reaction.

RICE Tables: Meaning

Let's start by looking at the meaning of RICE tables. In the term RICE, R stands for "reaction", I stands for "initial concentrations", C stands for "change in concentration", and E stands for "equilibrium concentration".

As an example, let's say that we have a chemical reaction where 1.50 mol of N₂ and 3.50 mol of H₂ are allowed to come to equilibrium in a 1 L container at 700 °C to form NH₃ as product. Let's calculate the equilibrium constant for this reaction if the equilibrium concentration of 2NH₃ is 0.540 M.

$$ N_{2} (g) + \color {orchid}3 \color{black }\text{ }H_{2} (g)\rightleftharpoons \color {orchid}2 \color{black}\text{ }NH_{3}(g) $$

Step 1: Construct a RICE table to find the equilibrium concentration of CO₂.

The first thing we need to do is draw the RICE table, as seen in the figure below. The best place to construct it is under the chemical reaction.

Figure 2. An empty RICE table, Isadora Santos- StudySmarter Originals.

Step 2. Fill out the rows with the values for initial concentration, change in concentration, and equilibrium concentrations.

In the I row, we need to add the initial concentrations of the reactants and products. The initial concentration of N₂ is 1.50 mol/L whereas the initial concentration of H₂ is 3.50 mol/L. For the product NH₃, we can just add 0 because at the start of the reaction, no product has been formed yet.

In the C row, we need to write the change in concentration due to reaction using the given reaction stoichiometric coefficients. "x" represents the change in concentration.

• N₂ has a coefficient of 1.
• H₂ has a coefficient of 3.
• NH₃ has a coefficient of 2.

Lastly, we need to fill out the E row by adding the reactant and product concentrations at equilibrium in terms of the initial concentration and its change in x.

Figure 3. How to fill out a RICE table, Isadora Santos - StudySmarter Originals.

Step 3. Determine the equilibrium expression for the reaction.

Now, we need to write down the equilibrium expression for the chemical reaction.

$$ K_{eq} = \frac{[Products]}{[Reactants]} = \frac{[NH_{3}]^{\color {orchid}\textbf{2}}}{[N_{2}][H_{2}]^{\color {orchid}\textbf{3}}} $$

Step 4. Solve for the equilibrium constant (K_eq)

Now that we know the equilibrium equation for this chemical reaction, we can solve for K_eq. But first, notice that the problem tells us the equilibrium concentration of NH₃ is 0.540 M. So, we can use it to solve for "x".

$$ [NH_{3}]_{eq} = + 2x = 0.540 \text { }M $$ $$x = 0.270\text{ } M$$

Now that we know the value for x, we can finally solve for the equilibrium constant (K_eq).

$$ K_{eq} = \frac{[NH_{3}]^{\color {orchid}\textbf{2}}}{[N_{2}][H_{2}]^{\color {orchid}\textbf{3}}} = \frac{[0.540 ]^{\color {orchid}\textbf{2}}}{[1.50 - 0.270][3.50 -0.270]^{\color {orchid}\textbf{3}}} = 0.0122 $$

RICE Tables: examples

Now, let's solve another example using RICE tables.

In a 1.0 L container, 4 moles of NO are added and allowed to reach equilibrium for the reaction below. At equilibrium, there are 1.98 moles of N₂ present. Determine how many moles of NO are present in the mixture at equilibrium.

$$ \color {orchid} 2 \color {black}\text{ NO} (g) \rightleftharpoons \text{N}_{2} (g)\text{ + O}_{2} (g) $$

First, we need to construct an ice table and fill it with the information given by the question. Notice that the only initial concentration that we have is that for NO because, at this point in the reaction, no products are present!

Figure 4. RICE table example, Isadora Santos - StudySmarter Originals.

Since the amount of N₂ at equilibrium is given to us, we can use it to solve for x.

$$ 0 +x = 1.98 $$ $$ x = 1.98 $$

Now that we have the value for x, we can determine the concentration of NO at equilibrium, which we determined was 4 - 2x.

$$ [NO]_{eq} = 4 - 2x = 4-2(1.98) $$ $$ x = 0.04\text{ } M $$

RICE Table: Practice Problems

Now that you have learned how to fill out a RICE table, let's look at a practice problem involving partial pressures and equilibrium.

Suppose that you have a container charged with 2 atm H₂ and 2 atm Cl₂. At equilibrium, the partial pressure of HCl is said to be 3 atm. What is the equilibrium constant for the following reaction? Let's find out!

$$H_{2}(g)\text{ + }Cl_{2} (g) \rightleftharpoons \text{ } \color {orchid} 2\color{black} \text{ }HCl (g)$$

As with all other problems we looked at, we need to fill out a RICE table.

Figure 5. RICE table for an example using partial pressures, Isadora Santos - StudySmarter Originals.

Since we were given the equilibrium partial pressure of HCl (3 atm), we can use it to solve for "x".

$$2x=3\text{ atm}$$ $$x = 1.5 \text{ atm}$$

Now that we have "x", we can solve for the equilibrium constant of the reaction.

$$ K_{P}=\frac{[P_{HCl}]^{\color {orchid}2}}{[P_{H_{2}}][P_{Cl_{2}}]}=\frac{[3 ]^{\color {orchid}2}}{[2-1.5][2-1.5]} = 36 $$

RICE Tables: Review

To make things simpler, let's make a simple review on RICE tables to help you revise for your chemistry exam.

1. When dealing with equilibrium concentrations and equilibrium constants, you should always use a RICE table to help you with your calculations.
2. When writing the equilibrium expression for a chemical reaction, ignores compounds/molecules in the solid or liquid states.

Now, I hope that you feel more confident in your ability to tackle problems involving RICE tables!