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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:
Since 0 + cx is the same as cx, we can write:
Rearrange this equation by dividing both sides of the equation by c:
will allow us to calculate the value of x:
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

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.
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: 



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)} 
+ 
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: 



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)} 
+ 
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 