Question:

$9.2\, g\, N_2O_4$ is heated in a $1L$ vessel till equilibrium state is established $N_2O_{4(g)} {<=>}2NO_{2(g)}$ In equilibrium state $50\% N_2O_4$ was dissociated, equilibrium constant will be (mol. wt. of $N_2O_4 = 92$)

Updated On: Jul 28, 2022

$0.3$
$0.2$
$0.1$
$0.4$

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The Correct Option is B

Solution and Explanation

Moles of $N_{2}O_{4} =\frac{9.2}{92.0} =0.1$ mole $\begin{matrix}&N_{2}O_{4}& {<=>}&2NO_{2}(g)&&\\ \text{Initiailly }&0.1 {\text{moles}}&&0&\\ \text{At eqm.}&0.05&&0.05 \times 2 \end{matrix}$ $K=\frac{\left[NO_{2}\right]^{2}}{\left[N_{2}O_{4}\right]}$ $=\frac{0.1\times0.1}{0.05}$ $=0.2$

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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;

R_f = K_f [A]^a [B]^b

The rate of backward reaction;

R_b = K_b [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.