What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO2]= 0.60 M, [O2] = 0.82 M and [SO3] = 1.90 M ?
\(2SO_2(g) + O_2(g) ⇋ 2SO_3(g)\)
\(2SO_2(g) + O_2(g) ⇋ 2SO_3(g)\)
Solution and Explanation
The equilibrium constant (Kc) for the give reaction is:
\(K_c =\frac {[SO_3]^2}{[SO_2]^2[O_2]}\)
\(K_c= \frac {(1.90)^2 M^2}{(0.60)^2(0.821) M^3 }\)
\(K_c = 12.239 \ M^{-1}\) (approximately)
Hence, Kc for the equilibrium is 12.239 M-1.
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Concepts Used:
Law of Chemical Equilibrium
Law of Chemical Equilibrium states that at a constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants each raised to a power equal to the corresponding stoichiometric coefficients as represented by the balanced chemical equation.
Let us consider a general reversible reaction;
A+B ↔ C+D
After some time, there is a reduction in reactants A and B and an accumulation of the products C and D. As a result, the rate of the forward reaction decreases and that of backward reaction increases.
Eventually, the two reactions occur at the same rate and a state of equilibrium is attained.
By applying the Law of Mass Action;
The rate of forward reaction;
Rf = Kf [A]a [B]b
The rate of backward reaction;
Rb = Kb [C]c [D]d
Where,
[A], [B], [C] and [D] are the concentrations of A, B, C and D at equilibrium respectively.
a, b, c, and d are the stoichiometric coefficients of A, B, C and D respectively.
Kf and Kb are the rate constants of forward and backward reactions.
However, at equilibrium,
Rate of forward reaction = Rate of backward reaction.
Kc is called the equilibrium constant expressed in terms of molar concentrations.
The above equation is known as the equation of Law of Chemical Equilibrium.