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Calculations Involving Equilibrium Constant Equation
Basically, there are three types of calculations involved in equilibrium constant equation:
(a) Calculation of equilibrium constant Kc or Kp (b) Calculation of Kc or Kp given Kp or Kc (c) Calculation of equilibrium concentrations (molar concentrations or partial pressures)
Let us illustrate these three types of calculations with examples.
Calculation of Equilibrium Constant
Example
Consider the following equilibrium reaction
NH (^) 3 ( aq ) H O l 2 ( ) NH 4 ( aq ) OH ( aq )
If [NH 3 ]= 0.02M, [NH 4 +^ ] =0.05 M, and [OH - ] = 0.6 M, what is the equilibrium constant Kc for this reaction?
Answer
The equilibrium constant equation excluding the water (remember pure liquids do not appear in equilibrium constant expression) for the above reaction is
4 3
[ ][
c [ ]
NH OH
K
NH
]
Substitute the given concentrations to evaluate Kc. Hence
4 3
[ ][ ] (0.05)(0.6)
[ ] 0.
c
NH OH
K
NH
Example
Consider the following heterogerneous equailibrium reaction:
CaCO 3 (^) ( ) s ZZYZZX CaO s ( ) + CO 2 ( g )
At 800^0 C, the pressure of CO 2 gas is 0.236 atm. Calculate (a) Kp and (b) Kc for the reaction at this temperature.
Answer
Since CaCO 3 and CaO are solids, their concentrations do not enter into the equilibrium constant expression. Also, the pressure of CO 2 is given, and therefore, first we solve for Kp and then solve for Kc.
(a) Kp = PCO2 = 0.
(b) The relation between KP and Kc is
Kp = Kc (0.0821 x T)Δn
Here T = 273 +800 = 1073 K, and Δn = 1. So, we substitute these values into the above equation to yield,
0.236 = Kc (0.0821 x 1073)
Kc = 2.68 x 10-
Calculation of Kc or Kp given Kp or Kc
Example
The equilibrium constant ( Kc ) for the reaction
N O 2 4 (^) ( g ) ZZYZZX 2 NO 2 ( g )
is 4.63x10 -3^ at 25 0 C. What is the value of Kp for this reaction at this temperature?
Answer
Relationship between Kp and Kc is
Kp = Kc (0.0821 x T) Δn
Here T = 25 + 273 = 298 K, and Δn = 2 – 1 = 1. Thus
Kp = 4.63 x 10-3^ ( 0.0821 x 298) = 0.
Example
For the reaction
N 2 (^) ( g ) + 3 H (^) 2 ( g ) ZZYZZX 2 NH 3 ( g )
that equilibrium concentration is the difference between initial concentration and the amount(concentration) disappeared at equilibrium]. It is very useful to set up the problem in the following manner:
cis − stilbene ZYZZZX trans − stilbene
Initial concentration (M): 0.850 0 Change in concentration (M) -x +x
Equilibrium concentration (M) 0.850 - x x
A negative (-) change indicates a decrease in concentration and a positive (+) change indicates an increase in concentration at equilibrium. Next, we set up an equilibrium constant expression and substitute these values and solve for x:
[ ] c [ ]
trans stilbene K cis stilbene
x x
x = 0.816 M
The equilibrium concentrations of cis- and trans-stilbene are
[trans-stilben] = x = 0.816 M [cis-stilebene] = (0.850 –x ) = 0.850-0.816 = 0.034 M
