Now you are quite familiar with the term Equilibrium. We know that at equilibrium the rate of forward reaction and the rate of backward reaction are equal to each other. So, the composition of reaction mixture also has a fixed value which means the reactant and product will be present in a fixed ratio. This ratio is the characteristics of that particular reaction and it is governed by the law of equilibrium. In this post we will learn about it. I want to tell you one thing that Equilibrium is possible only in reversible reactions and not all reactions are reversible. In the coming posts we will study reversible reactions.
Let's take an example of a reversible reaction in which reactant ‘A’ reacts with ‘B’ to form products ‘C’ and ‘D’. Now write the balanced equation of this reaction.
Let's take an example of a reversible reaction in which reactant ‘A’ reacts with ‘B’ to form products ‘C’ and ‘D’. Now write the balanced equation of this reaction.
aA+ bB ↔ cC + dD
To get the equilibrium constant Kcwe have to determine the molar concentration of all species at equilibrium. And this is shown by enclosing the specie in square bracket.
Kc = [C]c [D]d/ [A]a [B]b
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| Equilibrium Constant Kc |
Kc is known as equilibrium constant of the reaction. And subscript c indicates that the concentrations are expressed in moles per litre (it is also termed as Molarity and shown by symbol M).
To get the equilibrium constant Kcwe have to determine the molar concentration of all species at equilibrium. Molar concentration is shown by enclosing the species in square bracket. Let's take an example of a real reaction.
H2(g) + I2(g) ↔ HI(g)
Now write the balanced equation in which number of moles of each element on both sides of arrow becomes equal. In the above equation if we multiply 2 on right side, then both sides will have 2 moles of ‘H’ and 2 moles of ‘I’. So the balanced equation will be:
H2(g) + I2(g) ↔ 2HI(g) -----------(1)
and Kc will be:
Kc = [HI]2/ [H] [I]
Let's find the unit of Equilibrium constant:
Kc = [moles/L]2/ [moles/L] [moles/L]
Kc = [moles/L]
Kc = [moles/L]
Now try a different reaction.
HI(g) ↔ H2(g) + I2(g)
Now write the balance equation:
2HI(g) ↔ H2(g) + I2(g) -----------(2)
And if we name Kc for this reaction is K'c , then:
K'c = [H] [I]/ [HI]2
Now find out the unit of K'c
K'c = [moles/L] [moles/L] / [moles/L]2
K'c = 1/ [moles/L]
K'c = 1/ [moles/L]
So, you see the Equilibrium constant has different units for different reactions.
One more thing, did you notice that 2nd reaction is the reverse reaction to the 1stone? And did you find any relation between equilibrium constants of both reactions?
One more thing, did you notice that 2nd reaction is the reverse reaction to the 1stone? And did you find any relation between equilibrium constants of both reactions?
Kc = 1/ K'c
If we reverse a reaction, equilibrium constant also gets reversed for that reaction.
And what will happen if we multiply the reaction 1? Like:
And what will happen if we multiply the reaction 1? Like:
n H2(g) + n I2(g) ↔ n 2HI(g)
K"c = [HI]2n/ [H]n [I]n
K"c = (Kc)n
K"c = (Kc)n
Let's try it with real data for reaction 1. Calculate Kc if at equilibrium, concentration of H2 is 1.4×10-2 mol /L , I2 is 0.12×10-2 mol/L and HI is 2.52×10-2 mol/L.
Kc = (2.52×10-2)2/ ( 1.4×10-2) ( 0.12×10-2)
Kc = 46.4 mol/L
Kc = 46.4 mol/L
Let's take another data set. Calculate Kcif take initially 2.4×10-2 M of H2 and 1.38×10 -2M of I2 and at equilibrium concentration of H2 becomes 1.4×10-2 M , I2 becomes 0.12×10-2 M and HI is 2.52×10-2 M.
Kc = (2.52×10-2)2/ ( 1.4×10-2) ( 0.12×10-2)
Kc = 46.4mol/L
Now you have seen that initial concentration doesn't affect the equilibrium constant. Let's take another example.
Calculate Kc if at equilibrium concentration of H2 is 0.77×10-2 mol /L , I2is 0.31×10-2 mol/L and HI is 3.34×10-2 mol/L.
Kc = (3.34×10-2)2/( 0.77×10-2)( 0.31×10-2)
Kc = 46.4 mol/L
Kc = 46.4 mol/L
You might get surprised, even after changing the equilibrium concentration, equilibrium constant doesn't change. Equilibrium constant is the characteristic of a given reaction at a particular temperature. If you change the reaction temperature of the same reaction you will get a different value of Kc.
What are the other factors which affect equilibrium and equilibrium constant? In the next post we will discuss it in detail.
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