Equilibrium RICE Tables
When you are doing more complicated equilibrium calculations, it helps to draw up a RICE table (shown below). A RICE table is an easy way of organising information in equilibrium calculations.
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R |
Reaction |
the balanced chemical equation |
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I |
Initial quantity |
the moles of reactants and products at the beginning of the reaction |
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C |
Change |
how much the moles of the reactants and products changed between the beginning of the reaction and equilibrium |
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E |
Equilibrium quantity |
the moles of the reactants and products at equilibrium |
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E |
Equilibrium concentration |
to calculate \(\text{K}_{\text{c}}\) you need the concentration of the reactants and products at equilibrium |
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Reaction |
\(\qquad\) | \(\qquad\) | \(\qquad\) | \(\qquad\) |
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Initial quantity (mol) |
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Change (mol) |
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Equilibrium quantity (mol) |
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Equilibrium concentration (\(\text{mol.dm$^{-3}$}\)) |
Table: A RICE table for mole values.
Here are some guidelines on how to use a RICE table:
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Fill in the balanced chemical equation: \(\color{blue}{\text{aA}} + \color{blue}{\text{bB}} \rightleftharpoons \color{red}{\text{cC}} + \color{red}{\text{dD}}\)
Reaction
\(\color{blue}{\textbf{aA}}\)
\(\color{blue}{\textbf{bB}}\)
\(\color{red}{\textbf{cC}}\)
\(\color{red}{\textbf{dD}}\)
Initial quantity (mol)
Change (mol)
Equilibrium quantity (mol)
Equilibrium concentration
(\(\text{mol.dm$^{-3}$}\))
-
In the initial row fill in the number of moles of each substance present at the beginning of the reaction. For example, if there are \(\text{0.3}\) \(\text{moles}\) of A, \(\text{0.7}\) \(\text{moles}\) of B and no moles of product initially:
Reaction
\(\color{blue}{\textbf{aA}}\)
\(\color{blue}{\textbf{bB}}\)
\(\color{red}{\textbf{cC}}\)
\(\color{red}{\textbf{dD}}\)
Initial quantity (mol)
\(\text{0.3}\)
\(\text{0.7}\)
\(\text{0}\)
\(\text{0}\)
Change (mol)
Equilibrium quantity (mol)
Equilibrium concentration
(\(\text{mol.dm$^{-3}$}\))
-
In the change row fill in the number of moles of reactant used, or the number of moles of product formed, in terms of x. Use the balanced equation for this. For example, if dx moles of D is produced, bx moles of B will be used:
Reaction
\(\color{blue}{\textbf{aA}}\)
\(\color{blue}{\textbf{bB}}\)
\(\color{red}{\textbf{cC}}\)
\(\color{red}{\textbf{dD}}\)
Initial quantity (mol)
\(\text{0.3}\)
\(\text{0.7}\)
\(\text{0}\)
\(\text{0}\)
Change (mol)
-ax
-bx
+cx
+dx
Equilibrium quantity (mol)
Equilibrium concentration
(\(\text{mol.dm$^{-3}$}\))
-
In the equilibrium row fill in the number of moles of each substance present at equilibrium in terms of x. For example, there will be \(\text{0.3}\) - ax moles of A at equilibrium.
Reaction
\(\color{blue}{\textbf{aA}}\)
\(\color{blue}{\textbf{bB}}\)
\(\color{red}{\textbf{cC}}\)
\(\color{red}{\textbf{dD}}\)
Initial quantity (mol)
\(\text{0.3}\)
\(\text{0.7}\)
\(\text{0}\)
\(\text{0}\)
Change (mol)
-ax
-bx
+cx
+dx
Equilibrium quantity (mol)
\(\text{0.3}\) - ax
\(\text{0.7}\) - bx
+cx
+dx
Equilibrium concentration
(\(\text{mol.dm$^{-3}$}\))
-
Use any extra information you have been given to calculate x, and then finish filling in the table.
For example, if the number of moles of B at equilibrium is \(\text{0.2}\):
\(\text{0.7}\) - bx = \(\text{0.2}\) and you can solve for x.
-
In the last row fill in the concentration of each substance at equilibrium. You will have to calculate this value.
Example: Equilibrium Calculations
Question
\(\text{1.4}\) \(\text{moles}\) of \(\text{NH}_{3}\)(g) is introduced into a sealed \(\text{2.0}\) \(\text{dm$^{3}$}\) reaction vessel. The ammonia decomposes when the temperature is increased to \(\text{600}\) \(\text{K}\) and reaches equilibrium as follows:
\(2\text{NH}_{3}(\text{g})\) \(\leftrightharpoons\) \(\text{N}_{2}(\text{g}) + 3\text{H}_{2}(\text{g})\)
When the equilibrium mixture is analyzed, the concentration of \(\text{NH}_{3}\)(g) is \(\text{0.3}\) \(\text{mol.dm$^{-3}$}\).
-
Calculate the concentration of \(\text{N}_{2}\)(g) and \(\text{H}_{2}\)(g) in the equilibrium mixture.
-
Calculate the equilibrium constant for the reaction at \(\text{600}\) \(\text{K}\).
Step 1: Draw a RICE table
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Reaction |
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Initial quantity (mol) |
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Change (mol) |
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Equilibrium quantity (mol) |
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Equilibrium concentration (\(\text{mol.dm$^{-3}$}\)) |
Step 2: Fill in the balanced chemical equation on the table:
|
Reaction |
\(\color{blue}{\textbf{2NH}_{3}}\) |
\(\color{red}{\textbf{1N}_{2}}\) |
\(\color{red}{\textbf{3H}_{2}}\) |
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Initial quantity (mol) |
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Change (mol) |
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Equilibrium quantity (mol) |
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Equilibrium concentration (\(\text{mol.dm$^{-3}$}\)) |
Make sure you know which compounds are the \(\color{blue}{\textbf{reactants}}\), and which are the \(\color{red}{\textbf{products}}\).
Step 3: Fill in the number of moles of \(\text{NH}_{3}\), \(\text{N}_{2}\) and \(\text{H}_{2}\) present at the beginning of the reaction
There are \(\text{1.4}\) \(\text{mol}\) of \(\text{NH}_{3}\), and no moles of \(\text{N}_{2}\) or \(\text{H}_{2}\) at the beginning of the reaction.
|
Reaction |
\(\color{blue}{\textbf{2NH}_{3}}\) |
\(\color{red}{\textbf{1N}_{2}}\) |
\(\color{red}{\textbf{3H}_{2}}\) |
|
Initial quantity (mol) |
\(\text{1.4}\) |
0 |
0 |
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Change (mol) |
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Equilibrium quantity (mol) |
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Equilibrium concentration (\(\text{mol.dm$^{-3}$}\)) |
Step 4: Fill in the change in the number of moles of \(\text{NH}_{3}\), \(\text{N}_{2}\) and \(\text{H}_{2}\) in terms of x
The mole ratio of \(\text{NH}_{3}(\text{g})\) : \(\text{N}_{2}\) : \(\text{H}_{2}\) is \(2:1:3\).
Therefore, for every \(\text{2}\) moles of \(\text{NH}_{3}\) used, \(\text{1}\) mole of \(\text{N}_{2}\) and \(\text{3}\) moles of \(\text{H}_{2}\) will be formed. If the number of moles of \(\text{NH}_{3}\) decreases by 2x (\(-\text{2}\)x), then the number of moles of \(\text{N}_{2}\) increases by 1x (\(\text{+1}\)x) and the number of moles of \(\text{H}_{3}\) increases by 3x (\(\text{+3}\)x).
|
Reaction |
\(\color{blue}{\textbf{2NH}_{3}}\) |
\(\color{red}{\textbf{1N}_{2}}\) |
\(\color{red}{\textbf{3H}_{2}}\) |
|
Initial quantity (mol) |
\(\text{1.4}\) |
0 |
0 |
|
Change (mol) |
\(-\text{2}\)x |
+x |
\(\text{+3}\)x |
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Equilibrium quantity (mol) |
|||
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Equilibrium concentration (\(\text{mol.dm$^{-3}$}\)) |
Step 5: Fill in the number of moles of \(\text{NH}_{3}\), \(\text{N}_{2}\) and \(\text{H}_{2}\) at equilibrium in terms of x
To get the number of moles of a substance at equilibrium you take the number of moles initially, and add the change in number of moles in terms of x.
|
Reaction |
\(\color{blue}{\textbf{2NH}_{3}}\) |
\(\color{red}{\textbf{1N}_{2}}\) |
\(\color{red}{\textbf{3H}_{2}}\) |
|
Initial quantity (mol) |
\(\text{1.4}\) |
0 |
0 |
|
Change (mol) |
\(-\text{2}\)x |
+x |
\(\text{+3}\)x |
|
Equilibrium quantity (mol) |
\(\text{1.4}\) \(-\text{2}\)x |
+x |
\(\text{+3}\)x |
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Equilibrium concentration (\(\text{mol.dm$^{-3}$}\)) |
Step 6: Calculate the value of x
You know that the concentration of \(\text{NH}_{3}\) is \(\text{0.3}\) \(\text{mol.dm$^{-3}$}\) at equilibrium and the reaction vessel has a volume of \(\text{2}\) \(\text{dm$^{3}$}\).
C = \(\dfrac{\text{n}}{\text{V}}\), therefore n = C \(\times\) V
n(\(\text{NH}_{3}\) at equilibrium) = \(\text{0.3}\) \(\text{mol.dm$^{-3}$}\) \(\times\) \(\text{2.0}\) \(\text{dm$^{3}$}\) = \(\text{0.6}\) \(\text{mol}\)
From the RICE table we can see that there are \(\text{1.4}\) \(-\text{2}\)x moles of \(\text{NH}_{3}\) at equilibrium. Therefore:
\(\text{0.6}\) \(\text{mol}\) = \(\text{1.4}\) \(\text{mol}\) - 2x
2x = \(\text{1.4}\) \(\text{mol}\) - \(\text{0.6}\) \(\text{mol}\) = \(\text{0.8}\) \(\text{mol}\)
Therefore, x = \(\text{0.4}\) \(\text{mol}\)
Step 7: Calculate the concentration of \(\text{N}_{2}\) formed (at equilibrium).
From the RICE table we can see that the number of moles of \(\text{N}_{2}\) at equilibrium = x
therefore n(\(\text{N}_{2}\)) = \(\text{0.4}\) \(\text{mol}\)
C(\(\text{N}_{2}\))\(= \dfrac{\text{n}}{\text{V}} = \dfrac{\text{0.4}\text{ mol}}{\text{2.0}\text{ dm$^{3}$}} =\) \(\text{0.2}\) \(\text{mol.dm$^{-3}$}\)
Step 8: Calculate the concentration of \(\text{H}_{2}\) formed (at equilibrium).
From the RICE table we can see that the number of moles of \(\text{H}_{2}\) at equilibrium = 3x
therefore n(\(\text{H}_{2}\)) = \(\text{3}\) x \(\text{0.4}\) \(\text{mol}\) = \(\text{1.2}\) \(\text{mol}\)
C(\(\text{H}_{2}\)) \(= \dfrac{\text{n}}{\text{V}} = \dfrac{\text{1.2}\text{ mol}}{\text{2.0}\text{ dm$^{3}$}} =\) \(\text{0.6}\) \(\text{mol.dm$^{-3}$}\)
Step 9: Complete the RICE table
|
Reaction |
\(\color{blue}{\textbf{2NH}_{3}}\) |
\(\color{red}{\textbf{1N}_{2}}\) |
\(\color{red}{\textbf{3H}_{2}}\) |
|
Initial quantity (mol) |
\(\text{1.4}\) |
0 |
0 |
|
Change (mol) |
\(-\text{2}\)x |
+x |
\(\text{+3}\)x |
|
Equilibrium quantity (mol) |
\(\text{1.4}\) \(-\text{2}\)x |
+x |
\(\text{+3}\)x |
|
Equilibrium concentration (\(\text{mol.dm$^{-3}$}\)) |
\(\text{0.3}\) |
\(\text{0.2}\) |
\(\text{0.6}\) |
Step 10: Write the expression for \(\text{K}_{\text{c}}\) for this reaction
\(\text{NH}_{3}\) is the reactant, \(\text{N}_{2}\) and \(\text{H}_{2}\) are the products. They are all in the gas phase, and so are included in the expression for \(\text{K}_{\text{c}}\).
\(\text{K}_{\text{c}}\) \(= \dfrac{\text{[N}_{2}\text{]}\text{[H}_{2}\text{]}^{3}}{\text{[NH}_{3}\text{]}^{2}}\)
Step 11: Calculate \(\text{K}_{\text{c}}\)
\(\text{K}_{\text{c}}\) \(= \dfrac{\text{(}\text{0.2}\text{)}\text{(}\text{0.6}\text{)}^{3}}{\text{(}\text{0.3}\text{)}^{2}} =\) \(\text{0.48}\)
This lesson is part of:
Chemical Equilibrium