In: Chemistry
A solution is made by mixing 30.0 mL of 0.150M (mol/L) compound A with 25.0 mL of 0.200M (mol/L) compound B. At equilibrium, the concentration of C is 0.0454M. Calculate the equilibrium constant, K, for this reaction.
A (aq) + 2B (aq) <---> C (aq)
Reactions which go for completion are called irreversible reaction. While there are few reactions which do not go for completion and hence, known as reversible reaction.
In these reversible reactions, the reactants form products in a forward reaction and the products formed can get change back to reactants in the backward reaction.
A State known as equilibrium state is reached wherein reactants and products are present but their concentrations do not seem to change with time. This state is known as state of chemical equilibrium.
For a chemical reaction, chemical equilibrium is achieved when rate of forward reaction becomes equals to rate of background reaction. In this state concentrations of both reactants and products do not change with time if reaction conditions remain same.
For a given elementary reaction,
Rate of forward reaction = k1[A]a[B]b...............(1)
Rate of backward reaction = k2[C]c[D]d...............(2)
Where,
k1 = Rate constant of forward reaction
k2 = Rate constant of backward reaction
On dividing equation (1) by (2), we get
At equilibrium,
Rate of forward reaction = Rate of background reaction
Where K is equilibrium constant
In the given reaction, one mole of A reacts with 2 moles of B to form C.
For the above reaction, expression for equilibrium constant is:
Given,
Amount of A = 30.0 mL of 0.150 M
Amount of B = 25.0 mL of 0.200 M
Equilibrium concentration of C = 0.045 M
Volume of the solution = 30.0 mL + 25.0 mL = 55 mL
Given equilibrium concentration of C is 0.0454M
Concentration of A at equilibrium = (0.08181 – 0.0454)M = 0.0364 M
Concentration of B at equilibrium = (0.09090 – 2*0.0454)M
=(0.09090 – 0.0908)M = 0.0001 M
Using the expression,
Equilibrium constant for the given reaction is 1.25 *108.