R.I.C.E. Tables (I.C.E. Tables) and Equilibrium Constant Calculations Chemistry 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:

K_c =

[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)

K_c =

[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 [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 + 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

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

0.5
c

cx
c

will allow us to calculate the value of 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

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, K_c

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, K_c, 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:

K_c =

[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 H_2(g) and 10.0 mol I_2(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.

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)	+	I_2(g)	⇋	2HI_(g)

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

Synthesis of HI_(g)
Reaction:	H_2(g)	+	I_2(g)	⇋	2HI_(g)
Initial concentration:
Change in concentration:
Equilibrium concentration:

Synthesis of HI_(g)
Reaction:	H_2(g)	+	I_2(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:

Synthesis of HI_(g)
Reaction:	H_2(g)	+	I_2(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 H_2(g) and I_2(g) are reacting to produce HI_(g), therefore:

concentration of H_2(g) will decrease, that is,
at equilibrium [H_2(g)] < 1.0 mol L^-1
concentration of I_2(g) will decrease, that is,
at equilibrium [I_2(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:	H_2(g)	+	I_2(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:	H_2(g)	+	I_2(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:	H_2(g)	+	I_2(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:	H_2(g)	+	I_2(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

Calculate K

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

For the reaction:

H_2(g) + I_2(g) ⇋ 2HI_(g)

the expression for the
equilibrium constant is:

K_c =

[HI_(g)]²
[H_2(g)][I_2(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.

K_c =	[HI_(g)]² [H_2(g)][I_2(g)]
K_c =	[1.87]² [0.065][1.065]
K_c =	3.4969 0.069225
K_c =	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) + Cl_2(g)

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

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

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)	+	Cl_2(g)

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

Example: Dissociation of NOCl_(g)
Reaction:	2NOCl_(g)	⇋	2NO_(g)	+	Cl_2(g)
Initial concentration:
Change in concentration:
Equilibrium concentration:

Example: Dissociation of NOCl_(g)
Reaction:	2NOCl_(g)	⇋	2NO_(g)	+	Cl_2(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:

Example: Dissociation of NOCl_(g)
Reaction:	2NOCl_(g)	⇋	2NO_(g)	+	Cl_2(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 Cl_2(g) therefore at equilibrium:

[NOCl_(g)] < 2.0 mol L^-1
[NO_(g)] > 0.0 mol L^-1
[Cl_2(g)] > 0.0 mol L^-1

Example: Dissociation of NOCl_(g)
Reaction:	2NOCl_(g)	⇋	2NO_(g)	+	Cl_2(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)	+	Cl_2(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)	+	Cl_2(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)	+	Cl_2(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

Calculate the equilibrium constant, K_c

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

For the reaction:

2NOCl_(g) ⇋ 2NO_(g) + Cl_2(g)

the expression for the
equilibrium constant is:

K_c =

[NO_(g)]²[Cl_2(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.

K_c =	[NO_(g)]²[Cl_2(g)] [NOCl_(g)]²
K_c =	[0.66]²[0.33] [1.34]²
K_c =	0.4356 × 0.33 1.7956
K_c =	0.143748 1.7956
K_c =	0.080

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Problem Solving using StoPGoPS method

Question: Consider the reaction described by the following equation: H_2(g) + CO_2(g) ⇋ H₂O_(g) + CO_(g) One mole of H_2(g) and 1 mole of CO_2(g) are allowed to react in a 1 litre container until equilibrium is attained. If the equilibrium concentration of H_2(g) is 0.44 mol L^-1, determine the value of the equilibrium constant, K_c, under these conditions.

STOP

STOP! State the Question.

What is the question asking you to do?
Calculate the equilibrium constant, K_c
K_c = ?

PAUSE

PAUSE to Prepare a Game Plan

(1) What information (data) have you been given in the question?
H_2(g) + CO_2(g) ⇋ H₂O_(g) + CO_(g)
n(H_2(g))_(i) = moles H_2(g) initially = 1 mol
n(CO_2(g))_(i) = moles CO_2(g) initially = 1 mol
V = volume of vessel = 1 L
n(H_2(g))_(eq) = moles H_2(g) at equilibrium = 0.44 mol

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

K_c =

[H₂O_(g)][CO_(g)]
[H_2(g)][CO_2(g)]

where:
[H₂O_(g)] = equilibrium concentration of H₂O_(g) in mol L^-1
[CO_(g)] = equilibrium concentration of CO_(g) in mol L^-1
[H_2(g)] = equilibrium concentration of H_2(g) in mol L^-1
[CO_2(g)] = equilibrium concentration of CO_2(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 K_c

GO with the Game Plan

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

Reaction:	H_2(g)	+	CO_2(g)	⇋	H₂O_(g)	+	CO_(g)
Initial concentration: mol L^-1	1 c = n ÷ V c = 1 mol ÷ 1 L c = 1 mol L^-1		1 c = n ÷ V c = 1 mol ÷ 1 L c = 1 mol L^-1		0 No H₂O_(g) is present initially		0 No CO_(g) is present initially
Change in concentration: mol L^-1	-x		-x		+x		+x
Equilibrium concentration: mol L^-1	0.44 (given) 1 - x 0.44 = 1 - x x = 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 K_c

K_c =

[H₂O_(g)][CO_(g)]
[H_2(g)][CO_2(g)]

[0.56][0.56]
[0.44][0.44]

0.3136
0.1936

1.6

PAUSE

PAUSE to Ponder Plausibility

Is your answer plausible?

Work backwards:

K_c = 1.6
[H_2(g)] = 0.44 mol L^-1
[CO_2(g)] = 1 - x
[H₂O_(g)] = x
[CO_(g)] = x

K_c	=	[H₂O_(g)][CO_(g)] [H_2(g)][CO_2(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 ± 1.8198 2
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 K_c is reasonable.

STOP

STOP! State the Solution

State your solution to the problem.

K_c = 1.6

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

Key Concepts

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

Calculating the Equilibrium Constant, Kc

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

Calculating Equilibrium Constant : Initial Reactant Concentration and %Dissociation Known

Problem Solving using StoPGoPS method

Calculating the Equilibrium Constant, K_c