The equilibrium constant for the reaction $N_{2} +3H_{2} {\rightleftharpoons} 2NH_{3}$ is ?K?. Then, the equilibrium constant for the reaction
$2N_{2}+6H_{2} {\rightleftharpoons} 4NH_{3}$ will be
- K
- $K^{2}$
- $\sqrt{K}$
- 2K
The Correct Option is B
Solution and Explanation
$N_{2}+3H_{2} {\rightleftharpoons} \,2NH_{3},$ equilibrium constant = K. $\therefore K=\frac{\left[NH_{3}\right]^{2}}{\left[N_{2}\right]\left[H_{2}\right]^{3}}$ Now, for $2N_{2}+6H_{2} {\rightleftharpoons}\,4NH_{3},\, K^{'}=\frac{\left[NH_{3}\right]^{4}}{\left[N_{2}\right]^{2}\left[H_{2}\right]^{6}}=K^{2}$
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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.