 # R.I.C.E. Tables (I.C.E. Tables) and Equilibrium Constant Calculations Tutorial

## Key Concepts

• R.I.C.E. tables are also known as I.C.E. tables, ICE boxes, RICE boxes, ICE charts or RICE charts.
• R.I.C.E. tables are a way of organising data about equilibrium systems in order to make problem solving easier.
• R.I.C.E. stands for:
Reaction
Initial concentration
Change in concentration
Equilibrium concentration
• I.C.E. stands for:
Initial concentration
Change in concentration
Equilibrium concentration
• For the general reaction in which reactant A decomposes to produce products B and C:

aA(aq) ⇋ bB(aq) + cC(aq)

R.I.C.E. Table
Reaction: aA(aq) bB(aq) + cC(aq)
Initial concentration: [A(i)]   [B(i)]   [C(i)]
Change in concentration: ax   bx   cx
Equilibrium concentration: [A(eq)]   [B(eq)]   [C(eq)]

The equilibrium constant for this general reaction can be calculated using the following:

 Kc = [B(eq)]b[C(eq)]c[A(eq)]a

and the values for the concentration of each species at equilibrium, that is, the values of [A(eq)], [B(eq)], and [C(eq)], found using the R.I.C.E. table

Note that the same R.I.C.E. table and expression for K could be used to calculate the equilibrium constant for the same reaction occurring as a mixture of gases, that is:

aA(g) ⇋ bB(g) + cC(g)

 Kc = [B(eq)]b[C(eq)]c[A(eq)]a

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## Setting Up the R.I.C.E. Table

Step 1: Write the balanced chemical equation for the chemical reaction at equilibrium (include states of matter):

Example: Decomposition of A(g)
aA(g) bB(g) + cC(g)

Step 2: Draw the R.I.C.E. Table using the chemical formula of each reactant and product as a heading across the top and use the following headings vertically down:

• Reaction
• Initial concentration
• Change in concentration
• Equilibrium concentration

Example: Decomposition of A(g)
Reaction: aA(g) bB(g) + cC(g)
Initial concentration:
Change in concentration:
Equilibrium concentration:

Step 3: Insert values for the initial concentration for any species for which the initital concentration has been given.
Note that it may be necessary to calculate the initial concentration if you have been given the mass or moles of a species and the volume (see molarity calculations).

Example: Decomposition of A(g)
given: [A(initial)] = 1.0 mol L-1
Reaction: aA(g) bB(g) + cC(g)
Initial concentration:
mol L-1
1.0   0   0
Change in concentration:
Equilibrium concentration:

Step 4: Let x be the common factor in the concentration change.
Write an expression for the change in concentration for each species using the stoichiometric coefficients (mole ratios) in the balanced chemical equation:
for the reaction in which a known amount of A(g) decomposes:

aA(g)bB(g) + cC(g)

• the concentration of A(g) will change by a x x = ax
• the concentration of B(g) will change by b x x = bx
• the concentration of C(g) will change by c x x = cx

Insert the values for change in concentration into the R.I.C.E. table:

Example: Decomposition of A(g)
given: [A(initial)] = 1.0 mol L-1
Reaction: aA(g) bB(g) + cC(g)
Initial concentration:
mol L-1
1.0   0   0
Change in concentration:
mol L-1
-ax +bx +cx
Equilibrium concentration:

Note: since initially ONLY reactant A was present (no B, no C), clearly the reaction must proceed in the forward direction to produce products B and C.
Therefore the concentration of A must decrease by the amount ax (-ax), and, the concentrations of B and C must increase by bx (+bx) and cx (+cx) respectively.

Step 5: Use the initial concentration and concentration change expressions to a write an expression for the concentration of each species at equilibrium:

In our example for the decomposition of A(g):

• concentration of A(g) will decrease from 1.0 mol L-1 to (1.0 - ax) because A(g) is being consumed during the reaction to produce products B(g) and C(g)
• concentration of B(g) will increase from 0 mol L-1 to (0 + bx) because B(g) is being produced by the decomposition of A(g)
• concentration of C(g) will increase from 0 mol L-1 to (0 + cx) because C(g) is being produced by the decomposition of A(g)

Insert these expressions for equilibrium concentrations into the R.I.C.E. table:

Example: Decomposition of A(g)
given: [A(initial)] = 1.0 mol L-1
Reaction: aA(g) bB(g) + cC(g)
Initial concentration:
mol L-1
1.0   0   0
Change in concentration:
mol L-1
-ax +bx +cx
Equilibrium concentration:
mol L-1
1.0 - ax 0 + bx 0 + cx

Step 6: Insert the value for the equilibrium concentration of any species for which the equilibrium concentration has been given into the R.I.C.E. table:

Example: Decomposition of A(g)
given: [A(initial)] = 1.0 mol L-1
and [Cequilibrium] = 0.5 mol L-1
Reaction: aA(g) bB(g) + cC(g)
Initial concentration:
mol L-1
1.0   0   0
Change in concentration:
mol L-1
-ax +bx +cx
Equilibrium concentration:
mol L-1
1.0 - ax 0 + bx 0.5
= 0 + cx

Step 7: Determine the value of x

In our example we have the equation:

 0.5 = 0 + cx

Since 0 + cx is the same as cx, we can write:

 0.5 = cx

Rearrange this equation by dividing both sides of the equation by c:

 0.5 c = cxc

will allow us to calculate the value of x:

 x = 0.5 c

Modify your R.I.C.E. table to include this calculation:

Example: Decomposition of A(g)
given: [A(initial)] = 1.0 mol L-1
and [C(equilibrium)] = 0.5 mol L-1
Reaction: aA(g) bB(g) + cC(g)
Initial concentration:
mol L-1
1.0   0   0
Change in concentration:
mol L-1
-ax +bx +cx
Equilibrium concentration:
mol L-1
1.0 - ax 0 + bx 0.5
 0.5 = 0 + cx x = 0.5 c

Step 8: Use the calculated value of x to determine the equilibrium concentrations of all the other species:

Complete the R.I.C.E. table by substituting the calculated value of x into the expressions for the equilibrium concentration of all the other species.

## Calculating the Equilibrium Constant, Kc

The R.I.C.E. table enabled you to calculate the concentrations of all the species present in the reaction at equilibrium.

You can now use these equilibrium concentrations to calculate the value of the equilibrium constant, Kc, for this reaction under these conditions.

Step 1: Write the expression for the equilibrium constant using the balanced chemical equation:

For the reaction: aA(g)bB(g) + cC(g)
the expression for the
equilibrium constant is:
 Kc = [B(eq)]b[C(eq)]c[A(eq)]a

Step 2: Substitute the values for the equilibrium concentration of each species into the equation and solve.

The concentration of each species at equilibrium is given in the bottom row of the R.I.C.E. table.

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## Calculating Equilibrium Constant : Initial Concentrations and Equilibrium Concentration of One Species Known

Example: A mixture of 5.0 mol H2(g) and 10.0 mol I2(g) are placed in a 5.0 L container at 450°C and allowed to come to equilibrium.
At equilibrium the concentration of HI(g) is 1.87 mol L-1.
Calculate the value for the equilibrium constant, K, for this reaction under these conditions.

1. Set Up the R.I.C.E. table.

Step 1: Write the balanced chemical equation for the chemical reaction at equilibrium (include states of matter):

Synthesis of HI(g)
H2(g) + I2(g) 2HI(g)

Step 2: Draw the R.I.C.E. Table:

Synthesis of HI(g)
Reaction: H2(g) + I2(g) 2HI(g)
Initial concentration:
Change in concentration:
Equilibrium concentration:

Step 3: Insert values for the initial concentration for any species for which the initital concentration has been given.
Note that it may be necessary to calculate the initial concentration if you have been given the mass or moles of a species and the volume (see molarity calculations).

Synthesis of HI(g)
Reaction: H2(g) + I2(g) 2HI(g)
Initial concentration:
mol L-1
1.0
(5.0 mol ÷ 5.0 L = 1.0)
2.0
(10.0 mol ÷ 5.0 L = 2.0)
0.0
Change in concentration:
Equilibrium concentration:

Step 4: Let x be the common factor in the concentration change.
Write an expression for the change in concentration for each species using the stoichiometric coefficients (mole ratios) in the balanced chemical equation:

Synthesis of HI(g)
Reaction: H2(g) + I2(g) 2HI(g)
Initial concentration:
mol L-1
1.0
(5.0 mol ÷ 5.0 L = 1.0)
2.0
(10.0 mol ÷ 5.0 L = 2.0)
0.0
Change in concentration:
mol L-1
-x -x +2x
Equilibrium concentration:

Step 5: Use the initial concentration and concentration change expressions to a write an expression for the concentration of each species at equilibrium:
Note: in this reaction H2(g) and I2(g) are reacting to produce HI(g), therefore:

concentration of H2(g) will decrease, that is,
at equilibrium [H2(g)] < 1.0 mol L-1

concentration of I2(g) will decrease, that is,
at equilibrium [I2(g)] < 2.0 mol L-1

concentration of HI(g) will increase, that is,
at equilibrium [HI(g)] > 0.0 mol L-1

Synthesis of HI(g)
Reaction: H2(g) + I2(g) 2HI(g)
Initial concentration:
mol L-1
1.0
(5.0 mol ÷ 5.0 L = 1.0)
2.0
(10.0 mol ÷ 5.0 L = 2.0)
0.0
Change in concentration:
mol L-1
-x -x +2x
Equilibrium concentration:
mol L-1
1.0 - x 2.0 - x 0 + 2x

Step 6: Insert the value for the equilibrium concentration of any species for which the equilibrium concentration has been given into the R.I.C.E. table:

Synthesis of HI(g)
Reaction: H2(g) + I2(g) 2HI(g)
Initial concentration:
mol L-1
1.0
(5.0 mol ÷ 5.0 L = 1.0)
2.0
(10.0 mol ÷ 5.0 L = 2.0)
0.0
Change in concentration:
mol L-1
-x -x +2x
Equilibrium concentration:
mol L-1
1.0 - x 2.0 - x 1.87
1.87 = 0 + 2x

Step 7: Determine the value of x

Synthesis of HI(g)
Reaction: H2(g) + I2(g) 2HI(g)
Initial concentration:
mol L-1
1.0
(5.0 mol ÷ 5.0 L = 1.0)
2.0
(10.0 mol ÷ 5.0 L = 2.0)
0.0
Change in concentration:
mol L-1
-x -x +2x
Equilibrium concentration:
mol L-1
1.0 - x 2.0 - x 1.87
1.87 = 0 + 2x
2x = 1.87
2x/2 = 1.87/2
x = 0.935

Step 8: Use the calculated value of x to determine the equilibrium concentrations of all the other species:

Synthesis of HI(g)
Reaction: H2(g) + I2(g) 2HI(g)
Initial concentration:
mol L-1
1.0
(5.0 mol ÷ 5.0 L = 1.0)
2.0
(10.0 mol ÷ 5.0 L = 2.0)
0.0
Change in concentration:
mol L-1
-x
= -0.935
-x
= -0.935
+2x
Equilibrium concentration:
mol L-1
0.065
= 1.0 - x
= 1.0 - 0.935
= 0.065
1.065
= 2.0 - x
= 2.0 - 0.935
= 1.065
1.87
1.87 = 0 + 2x
2x = 1.87
2x/2 = 1.87/2
x = 0.935

2. Calculate K

Step 1: Write the expression for the equilibrium constant using the balanced chemical equation:

For the reaction: H2(g) + I2(g)2HI(g)
the expression for the
equilibrium constant is:
 Kc = [HI(g)]2   [H2(g)][I2(g)]

Step 2: Substitute the values for the equilibrium concentration of each species into the equation and solve.

The concentration of each species at equilibrium is given in the bottom row of the R.I.C.E. table.

 Kc = [HI(g)]2   [H2(g)][I2(g)] Kc = [1.87]2   [0.065][1.065] Kc = 3.4969   0.069225 Kc = 50.5

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## Calculating Equilibrium Constant : Initial Reactant Concentration and %Dissociation Known

Example: 4.0 moles of NOCl(g) were placed in a 2.0 L vessel at 460°C.

The following dissociation reaction occurred:

2NOCl(g) ⇋ 2NO(g) + Cl2(g)

At equilibrium the NOCl(g) was 33% dissociated.

Calculate the equilibrium constant, K, for this reaction under these conditions.

1. Set up the R.I.C.E. table:

Step 1: Write the balanced chemical equation for the chemical reaction at equilibrium (include states of matter):

Example: Dissociation of NOCl(g)
2NOCl(g) 2NO(g) + Cl2(g)

Step 2: Draw the R.I.C.E. Table :

Example: Dissociation of NOCl(g)
Reaction: 2NOCl(g) 2NO(g) + Cl2(g)
Initial concentration:
Change in concentration:
Equilibrium concentration:

Step 3: Insert values for the initial concentration for any species for which the initital concentration has been given.
Note that it may be necessary to calculate the initial concentration if you have been given the mass or moles of a species and the volume (see molarity calculations).

Example: Dissociation of NOCl(g)
Reaction: 2NOCl(g) 2NO(g) + Cl2(g)
Initial concentration:
mol L-1
2.0
(4.0 mol ÷ 2.0 L = 2.0 mol L-1)
0.0   0.0
Change in concentration:
Equilibrium concentration:

Step 4: Let x be the common factor in the concentration change.
Write an expression for the change in concentration for each species using the stoichiometric coefficients (mole ratios) in the balanced chemical equation:

Example: Dissociation of NOCl(g)
Reaction: 2NOCl(g) 2NO(g) + Cl2(g)
Initial concentration:
mol L-1
2.0
(4.0 mol ÷ 2.0 L = 2.0 mol L-1)
0.0   0.0
Change in concentration:
mol L-1
-2x +2x +x
Equilibrium concentration:

Step 5: Use the initial concentration and concentration change expressions to a write an expression for the concentration of each species at equilibrium:

Note that NOCl(g) is dissociating to produce NO(g) and Cl2(g) therefore at equilibrium:

[NOCl(g)] < 2.0 mol L-1
[NO(g)] > 0.0 mol L-1
[Cl2(g)] > 0.0 mol L-1

Example: Dissociation of NOCl(g)
Reaction: 2NOCl(g) 2NO(g) + Cl2(g)
Initial concentration:
mol L-1
2.0
(4.0 mol ÷ 2.0 L = 2.0 mol L-1)
0.0   0.0
Change in concentration:
mol L-1
-2x +2x +x
Equilibrium concentration:
mol L-1
2.0 - 2x 0.0 + 2x 0.0 + x

Step 6: Insert the value for the equilibrium concentration of any species for which the equilibrium concentration has been given into the R.I.C.E. table:

Note that if 33% of the NOCl(g) dissociated, then :

(a) the change in concentration of NOCl(g) must be 33%
(b) and at equilibrium 100 - 33 = 67% of the NOCl(g) must be present

Example: Dissociation of NOCl(g)
Reaction: 2NOCl(g) 2NO(g) + Cl2(g)
Initial concentration:
mol L-1
2.0
(4.0 mol ÷ 2.0 L = 2.0 mol L-1)
0.0   0.0
Change in concentration:
mol L-1
33/100 × 2.0 = -2x
0.66 = -2x
+2x +x
Equilibrium concentration:
mol L-1
1.34
67/100 x 2.0 = 2.0 - 2x
1.34 = 2.0 - 2x
0.0 + 2x 0.0 + x

Step 7: Determine the value of x

Example: Dissociation of NOCl(g)
Reaction: 2NOCl(g) 2NO(g) + Cl2(g)
Initial concentration:
mol L-1
2.0
(4.0 mol ÷ 2.0 L = 2.0 mol L-1)
0.0   0.0
Change in concentration:
mol L-1
33/100 x 2.0 = 2x
0.66 = -2x
0.66/2 = -x
-x = 0.33
+2x +x
Equilibrium concentration:
mol L-1
1.34
67/100 x 2.0 = 2.0 - 2x
1.34 = 2.0 - 2x
2x = 2.0 - 1.34
2x = 0.66
x = 0.66/2
x = 0.33
0.0 + 2x 0.0 + x

Step 8: Use the calculated value of x to determine the equilibrium concentrations of all the other species:

Example: Dissociation of NOCl(g)
Reaction: 2NOCl(g) 2NO(g) + Cl2(g)
Initial concentration:
mol L-1
2.0
(4.0 mol ÷ 2.0 L = 2.0 mol L-1)
0.0   0.0
Change in concentration:
mol L-1
33/100 x 2.0 = 2x
0.66 = -2x
0.66/2 = -x
-x = 0.33
+2x +x
Equilibrium concentration:
mol L-1
1.34
67/100 x 2.0 = 2.0 - 2x
1.34 = 2.0 - 2x
2x = 2.0 - 1.34
2x = 0.66
x = 0.66/2
x = 0.33
0.66
0.0 + 2x
= 0.0 + 2×0.33
= 0.0 + 0.66
0.33
0.0 + x
= 0.0 + 0.33

2. Calculate the equilibrium constant, Kc

Step 1: Write the expression for the equilibrium constant using the balanced chemical equation:

For the reaction: 2NOCl(g)2NO(g) + Cl2(g)
the expression for the
equilibrium constant is:
 Kc = [NO(g)]2[Cl2(g)][NOCl(g)]2

Step 2: Substitute the values for the equilibrium concentration of each species into the equation and solve.

The concentration of each species at equilibrium is given in the bottom row of the R.I.C.E. table.

 Kc = [NO(g)]2[Cl2(g)][NOCl(g)]2 Kc = [0.66]2[0.33][1.34]2 Kc = 0.4356 × 0.331.7956 Kc = 0.1437481.7956 Kc = 0.080

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## Problem Solving using StoPGoPS method

 Question: Consider the reaction described by the following equation: H2(g) + CO2(g) ⇋ H2O(g) + CO(g) One mole of H2(g) and 1 mole of CO2(g) are allowed to react in a 1 litre container until equilibrium is attained. If the equilibrium concentration of H2(g) is 0.44 mol L-1, determine the value of the equilibrium constant, Kc, under these conditions.

STOP STOP! State the Question.
What is the question asking you to do?
Calculate the equilibrium constant, Kc
Kc = ?
PAUSE PAUSE to Prepare a Game Plan
(1) What information (data) have you been given in the question?
H2(g) + CO2(g) ⇋ H2O(g) + CO(g)
n(H2(g))(i) = moles H2(g) initially = 1 mol
n(CO2(g))(i) = moles CO2(g) initially = 1 mol
V = volume of vessel = 1 L
n(H2(g))(eq) = moles H2(g) at equilibrium = 0.44 mol

(2) What is the relationship between what you know and what you need to find out?

 Kc = [H2O(g)][CO(g)][H2(g)][CO2(g)]

where:
[H2O(g)] = equilibrium concentration of H2O(g) in mol L-1
[CO(g)] = equilibrium concentration of CO(g) in mol L-1
[H2(g)] = equilibrium concentration of H2(g) in mol L-1
[CO2(g)] = equilibrium concentration of CO2(g) in mol L-1

Step 1: Use a R.I.C.E. Table to calculate the equilibrium concentrations of all species

Step 2: Use the equilibrium concentrations to calculate the value of Kc

GO GO with the Game Plan

Step 1: Use a R.I.C.E. Table to calculate the equilibrium concentrations of all species

 Reaction: Initial concentration:mol L-1 Change in concentration:mol L-1 Equilibrium concentration:mol L-1 H2(g) + CO2(g) ⇋ H2O(g) + CO(g) 1c = n ÷ Vc = 1 mol ÷ 1 Lc = 1 mol L-1 1c = n ÷ Vc = 1 mol ÷ 1 Lc = 1 mol L-1 0No H2O(g) is present initially 0No CO(g) is present initially -x -x +x +x 0.44(given)1 - x0.44 = 1 - xx = 1 - 0.44 = 0.56 0.44(calculated) 1 - x= 1 - 0.56= 0.44 0.56(calculated) 0 + x= 0 + 0.56 = 0.56 0.56(calculated) 0 + x= 0 + 0.56 = 0.56

Step 2: Use the equilibrium concentrations to calculate the value of Kc

 Kc = [H2O(g)][CO(g)][H2(g)][CO2(g)] = [0.56][0.56][0.44][0.44] = 0.31360.1936 = 1.6

PAUSE PAUSE to Ponder Plausibility

Work backwards:

Kc = 1.6
[H2(g)] = 0.44 mol L-1
[CO2(g)] = 1 - x
[H2O(g)] = x
[CO(g)] = x

 Kc = [H2O(g)][CO(g)][H2(g)][CO2(g)] 1.6 = [x][x]     [0.44][1 - x] 1.6 = x2     [0.44 - 0.44x] 1.6[0.44 - 0.44x] = x2 0.704 - 0.704x = x2 0 = x2 + 0.704x - 0.704

 x = -b ± √(b2 -4 × a × c)2 × a x = -0.704 ± √(0.7042 -4 × 1 × -0.704)2 × 1 x = -0.704 ± 1.81982 x = 0.56

Which is the same as the value of x calculated in the R.I.C.E. table, so we will get the same initial and equilibrium concentrations for each species.
We are therefore confident that our value of Kc is reasonable.

STOP STOP! State the Solution
State your solution to the problem.

Kc = 1.6