CHEMICAL EQUILIBRIUM

HOW TO SOLVE EQUILIBRIUM PROBLEMS
Silvia Kolchens, Pima Community College



iCe table | Calculating equilibrium concentrations | Exercise 1 | Calculating equilibrium constants | Exercise 2 |


There are generally two types of equilibrium problems:
Both type of problems can be solved using reaction tables (iCe tables)

1. Using equilibrium constant and initial quantities (concentrations) to solve for the equilibrium quantities.

When hydrogen iodide (HI) is placed in a closed container it will decompose into its elements hydrogen and iodine and an equilibrium will be established:

2 HI (g) ó H2(g) + I2(g)

If the initial concentration of HI and the equilibrium constant for the system are known, then we can calculate the equilibrium concentration for HI, H2, and I2. These type of calculations are generally performed using iCe tables (or reaction tables).

Initial concentration of HI: 0.025 mol/L

Kc=2.2x10-3

iCe -table:

 
2 HI
H2
I2
i (initial)
0.025
0.00
0.00
C (Change)
-2x
+x
+x
e (equilibrium)
0.025-2x
x
x

"x" is the change in concentration : reactants decrease and products increase by an amount "x".

We can write the equilibrium constant for the reaction

We need to solve for "x", the change in concentration. This can easily be done by taking the square root:

Solving for "x" yields

Now we can calculate the equilibrium concentrations:

 
2 HI
H2
I2
i (initial)
0.025
0.00
0.00
C (Change)
-2 x 2.8x10-3
+2.8x10-3
+2.8x10-3
e (equilibrium)
0.0222
2.8x10-3
2.8x10-3

Solving for "x" was particularly easy in this example since the equation could be solved by taking the square root on both sides of the equation. However, this is not always the case. An example is the decomposition of ammonia into its elements hydrogen and notrogen:

2 NH3 ó N2 + 3 H2

 
2 NH3
N2
3 H2
i (initial)
initial
0.00
0.00
C (Change)
-2 x 
+ x
+3 x
e (equilibrium)
0.025-x
x
x

We can solve this equation by either

Now we can easily solve for "x". The assumption that x<<intial concentration can generally be used if the change in concentration is smaller than 5% of the initial concentration.
 
 

Exercise 1: Use equilibrium constant and initial concentrations to solve for the equilibrium quantities:

The extent of the change from CO and H2O to CO2 and H2 is used to regulate the proportions syngas fuel mixtures. If 0.250 mol CO and 0.250 mol H2O are placed in a 125 ml flask at 900 K, what is the composition of the equilibrium mixture? Kc = 1.56.

CO(g) + H2O (g) ó CO2(g) + H2(g)
Concentration [M] CO H2O CO2 H2
I        
C        
E        

Equilibrium Constant Kc = ----------------------------------------

Solving for "x":

Calculate equilibrium concentrations:

[CO] =

[H2O] =

[CO2] =

[H2] =

Answ.: initial concentration [CO] = [H2O] = 2.00 M

x = 1.11 M;

[CO] = [H2] = 0.89 M
 

2. Use concentration to solve for the equilibrium constant

If we know the intial concentration of the reactants and determine the concentration of one of the products at equilibrium, then we can calculate the equilibrium constant K. This is shown in the following example, the decomposition of hydrogen fluoride

2 HF (g) ó H2(g) + F2(g)

If the initial concentration of HI is 0.075 M, and the equilibrium concentration if I2 is determined to be 0.0084, then set up an iCe table:

 
2 HF
H2
F2
i (initial)
0.075
0.00
0.00
C (Change)
-2 x
+ x
+ x
e (equilibrium)
0.075-2x
x
0.0084

It is apparent that "x" equals 0.0084, therefore we can calculate the equilibrium concentrations for HF and H2:

 
2 HF
H2
I2
i (initial)
0.075
0.00
0.00
C (Change)
-2 x 0.0084
+ 0.0084
+ 0.0084
e (equilibrium)
0.0582
0.0084
0.0084

 

Now we can calculate the equilibrium constant K:

Exercise 2: Using partial pressure data to determine the equilibrium constant:

An evacuated vessel containing a small amount of powdered graphite is heated to 1080 K and then CO2 is added to a pressure of 0.458 atm. Once the CO2 is added, the system starts to produce CO. After equilibrium is reached, the total pressure inside the vessel is 0.757 atm. Calculate Kp.
Hint: Use partial pressures (P) in place of concentrations, and Kp in place of Kc.

CO2(g) + C (s) ó 2 CO (g)


Pressure [atm] CO2(g) C(s) 2 CO
initial 0.458 0 0
Change -x   + 2x
equlibrium 0.458 - x   2 x

Ptotal = 0.757 atm = PCO2(eq) + P CO(eq) = 0.458 atm x + 2x

X = _________________

PCO2(eq) = _____________

PCO (eq) = ______________

Kp = -------------------------------
Answ.: x = 0.299 atm; PCO2(eq) = 0.159 atm; PCO (eq) =0.598 atm; Kp = 2.25