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Equilibrium Constant Equations: Calculations of Kc, Kp, and Equilibrium Concentrations, Study Guides, Projects, Research of Chemistry

Examples and explanations for calculating equilibrium constant (kc and kp), determining equilibrium concentrations (molar concentrations or partial pressures) for given initial concentrations, and finding kc or kp given kp or kc. Both homogeneous and heterogeneous reactions.

Typology: Study Guides, Projects, Research

2021/2022

Uploaded on 09/27/2022

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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
32 4
() () () ()NH aq H O l NH aq OH aq
+−
++
ZZX
YZZ
If [NH3]= 0.02M, [NH4+] =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
KNH
+−
=]
Substitute the given concentrations to evaluate Kc. Hence
4
3
[][]
(0.05)(0.6) 1.5
[ ] 0.02
c
NH OH
KNH
+−
===
Example
Consider the following heterogerneous equailibrium reaction:
32
() () ( )CaCO s CaO s CO g+
ZZX
YZZ
At 8000C, the pressure of CO2 gas is 0.236 atm. Calculate (a) Kp and (b) Kc for the
reaction at this temperature.
1
pf3
pf4

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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 )

  • ZZX +^ + − YZZ

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:

cisstilbene ZYZZZX transstilbene

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