Equilibrium constants K1 and K2 for the following equilibria: NO(g)+\(\bigg(\frac{1}{2}\bigg)\)O2 \(^{K_1}_\rightleftharpoons\) NO2(g) and 2NO2(g) \(^{K_2}_\rightleftharpoons\) 2NO(g)+O2(g) are related as:
\(K_2\) = \(\frac{1}{K_1}\)
K2 = \(K^2_1\)
K2 = \(K^1_2\)
K2 = \(\frac{1}{K^2_1}\)
The Correct Option is D
Solution and Explanation
By changing the first reaction in reverse order and multiplying it by 2 we will get the equation similar to second equation.
Thus, K1 and K2 can be related in the above equations as:
K2 = \(\frac{1}{K^2_1}\)
\(\Rightarrow\) K1 = \(\frac{1}{\sqrt K_2}\)
Therefore, the correct option is (D): K2 = \(\frac{1}{K^2_1}\)
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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.