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Making an ICE Chart
An Aid in Solving Equilibrium Problems

An useful tool in solving equilibrium problems is an ICE chart.

How to Make an ICE Chart

In making an ICE chart the following items should be noted:

The following is a "how to" make an ICE chart using the example to illustrate the process.

Example:  A mixture consisting initially of 3.00 moles NH3, 2.00 moles of N2, and 5.00 moles of H2, in a 5.00 L container was heated to 900 K, and allowed to reach equilibrium.  Determine the equilibrium concentration for each species present in the equilibrium mixture.

2 NH3(g) double arrows  N2(g) + 3 H2(g)    Kc = 0.0076 @ 900 K

Solving for x and the concentrations

The value of x was determined using the method of successive approximations.

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Sample Ice Charts

The following examples illustrating writing ICE charts for the problems given.  Although each problem appears to be "different" the process for creating the ICE chart is the same.

Only Reactant Species Are Present Initially

Example:  4.00 moles of HI are placed in an evacuated 5.00 L flask and then heated to 800 K.  The system is allowed to reach equilibrium.  What will be the equilibrium concentration of each species?

2 HI(g) double arrows  H2(g) + I2(g)    Kc = 0.016 @ 800 K

Let "x" represent the change in concentration of the hydrogen gas.  Since we start with HI alone, the reaction must proceed to the right resulting in an increase in the amount of hydrogen gas.
 

 
HI(g)
H2(g)
I2(g)
Initial Concentration (M)
0.800
0
0
Change in Concentration (M)
- 2 x
+ x
+ x
Equilibrium Concentration (M)
0.800 - 2 x
0 + x
0 + x

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Only Product Species Are Present Initially

Example:  0.600 moles of NO and 0.750 moles of O2 are placed in an empty 2.00 L flask.  The system is allowed to establish equilibrium.  What will be the equilibrium concentration of each species in the flask?

2 NO2(g) double arrows  2 NO(g) + O2(g)     Kc = 0.50

Let "x" represent the change in concentration of the oxygen gas.  Since only NO and O2 are present, the reaction must proceed to the left in order to establish equilibrium.  The  O2 gas will decrease in concentration over time.
 

 
NO2(g)
NO(g)
O2(g)
Initial Concentration (M)
0
0.300
0.375
Change in Concentration (M)
+ 2 x
- 2 x
- x
Equilibirum Concentration (M)
2 x
0.300 - 2 x
0.375 - x

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Species Added to a System Initially in a State of Equilibrium

Example:  The concentrations of an equilibrium mixture of O2 , CO, and CO2  were 0.18 M, 0.35 M, and 0.029 M respectively.  Enough CO was added to the flask containing the equilibrium mixture to momentarily raise its concentration to 0.60 M.  What will be the concentration of each species in the flask once equilibrium has been re-established after the additional carbon monoxide was added?

2 CO2(g) double arrows  2 CO(g) + O2(g)

After the addition of more CO the system is no longer in equilibrium.  The once equilibrium quantities of the other three substances are now initial quantities.  Let "x" represent the change in the amount of O2 gas.  Adding the CO will force the reaction to proceed in the reverse direction (K < Q) causing the amount of O2 to decrease.  (Note:  The equilibrium quantities given in the problem, before the addition of more CO, are also used to calculate the value of the equilibrium constant, Kc.)
 

CO2(g)
CO(g)
O2(g)
Initial Concentration (M)
0.029
0.60
0.18
Change in Concentration (M)
+ 2 x
- 2 x
- x
Equilibrium Concentration (M)
0.029 + 2 x
0.60 - 2 x
0.18 - x

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Gaseous Species and Kp

Example:  Cl2 gas undergoes homolytic cleavage into chlorine atoms at 1100oC.  Kp at 1100oC for this process is 1.13 x 10-4.  If a sample with an initial Cl2 gas pressure of 0.500 atm was allowed to reach equilibrium, what is the total pressure in the flask?

Cl2(g) double arrows  2 Cl(g)

Let "x" represent the change in the pressure of the Cl2 gas.  Since the reaction will proceed forwards to establish equilibrium the pressure of the Cl2 gas will decrease.  The total pressure at equilibrium will equal the sum of the partial pressures of each gas at equilibrium.
 

 
Cl2(g)
Cl(g)
Initial Pressure (atm)
0.500
0
Change in Pressure (atm)
- x
+ 2 x
Pressure at Equilibrium (atm)
0.500 - x
0 + 2 x

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