Question

Two moles of \(\mathrm{NH}_{3}\) when put into a previously evacuated vessel (one litre), partially dissociated into \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\). If at equilibrium one mole of \(\mathrm{NH}_{3}\) is present, the equilibrium constant is : (a) \(3 / 4 \mathrm{~mol}^{2}\) litre \(^{-2}\) (b) \(27 / 64 \mathrm{~mol}^{2}\) litre \(^{-2}\) (c) \(27 / 32 \mathrm{~mol}^{2}\) litre \(^{-2}\) (d) \(27 / 16 \mathrm{~mol}^{2}\) litre \(^{-2}\)

Short answer

(c) \(27 / 32 \mathrm{~mol}^{2}\) litre \(^{-2}\)

Step-by-step solution

Write the Balanced Reaction

First, write the balanced chemical equation for the dissociation of ammonia (\(\mathrm{NH}_3\)) into nitrogen (\(\mathrm{N}_2\)) and hydrogen (\(\mathrm{H}_2\)). The balanced reaction is \(2 \mathrm{NH}_3 \rightleftharpoons \mathrm{N}_2 + 3 \mathrm{H}_2\).

Determine the Change in Mole Quantity

Let the change in moles of \(\mathrm{NH}_3\) at equilibrium be \(x\). Since initially there were 2 moles of \(\mathrm{NH}_3\), and at equilibrium, there is 1 mole, \(x = 1\) mole of \(\mathrm{NH}_3\) has dissociated.

Calculate Equilibrium Mole Quantities

If 1 mole of \(\mathrm{NH}_3\) has dissociated, 0.5 mole of \(\mathrm{N}_2\) and 1.5 moles of \(\mathrm{H}_2\) would be produced according to the stoichiometry of the balanced equation. Therefore, at equilibrium, the mole quantity of \(\mathrm{NH}_3\) is 1, \(\mathrm{N}_2\) is 0.5, and \(\mathrm{H}_2\) is 1.5.

Write the Expression for Equilibrium Constant (Kc)

The equilibrium constant expression for the reaction is given by \(K_c = \frac{[\mathrm{N}_2] \cdot [\mathrm{H}_2]^3}{[\mathrm{NH}_3]^2}\).

Plug in the Equilibrium Concentrations

Substitute the equilibrium concentrations into the equilibrium constant expression: \(K_c = \frac{(0.5) \cdot (1.5)^3}{(1)^2}\).

Calculate the Equilibrium Constant

Calculate the value of \(K_c\): \(K_c = \frac{0.5 \cdot 3.375}{1}\) which simplifies to \(K_c = \frac{27}{32} = 0.84375 \mathrm{~mol}^2 \mathrm{L}^{-2}\).

Choose the Correct Answer

Match the calculated value of the equilibrium constant to the given options. The correct option is (c) \(27 / 32 \mathrm{~mol}^2\) litre \(^{-2}\).

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is essential when studying how reactions reach a state of balance. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, meaning that the concentrations of reactants and products remain constant over time. This doesn't mean the reactants and products are in equal concentration, but rather that their rates of production and consumption are equal.

For a reaction such as the dissociation of ammonia, we can represent this dynamic yet stable state with an equilibrium constant, denoted as K_c. This constant is a numerical value that quantifies the relationship between the concentrations of reactants and products at equilibrium. In general, a large K_c value implies a reaction favoring the formation of products, while a small K_c suggests the reaction favors the reactants.

Dissociation of Ammonia

Ammonia (H₃) dissociates reversibly into nitrogen (N₂) and hydrogen (H₂) gases—this reaction is a classic example of a reversible reaction reaching chemical equilibrium. The balanced equation (2 NH₃ ightleftharpoons N₂ + 3 H₂) shows the stoichiometry of the reactants and products. Using this stoichiometry is crucial when calculating the changes in concentrations or moles of each reactant and product as the reaction progresses toward equilibrium.

The extent of dissociation also depends on environmental factors such as temperature and pressure. Understanding the conditions under which ammonia dissociates helps predict the behavior of the equilibrium system.

Equilibrium Concentrations

Equilibrium concentrations are the amounts of reactants and products present when a chemical reaction has reached equilibrium. These can be expressed in molarity (moles per liter) for solutions or in partial pressures for gases. In our example with ammonia dissociation, we begin with two moles in a one-liter vessel, and at equilibrium, we find there's one mole of ammonia, half a mole of nitrogen, and one and a half moles of hydrogen.

To calculate these, we start by identifying the initial moles or concentrations, determine the change in mole quantity (often represented by the variable 'x'), and then deduce the final concentrations at equilibrium. These concentrations are plugged into the equilibrium constant expression to find the equilibrium constant, K_c. It is important to always maintain the stoichiometric ratios when doing these conversions, as they will affect the final calculated values.