Problem 6 - A Complete Resource Book in Chemistry for JEE Main

Question

If equilibrium constant for the reaction \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\) is \(\mathrm{K}_{e}\), then the equilibrium con- stant for the reaction \(\mathrm{NH}_{3} \longrightarrow 1 / 2 \mathrm{~N}_{2}+\frac{3}{2} \mathrm{H}_{2}\) will be (a) \(\frac{1}{\mathrm{~K}_{c}}\) (b) \(\frac{1}{\mathrm{~K}^{2}}\) (c) \(\sqrt{K}_{c}\) (d) \(\frac{1}{\sqrt{K_{t}}}\)

Step-by-Step Solution

Answer

The correct answer is (d) \( \frac{1}{\sqrt{K_{t}}} \).

Step 1: Write the Equilibrium Constant Expression

For the reaction \( \mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} \), the equilibrium constant \( K_{e} \) is given by the expression: \[ K_{e} = \frac{[\mathrm{NH}_{3}]^2}{[\mathrm{N}_{2}][\mathrm{H}_{2}]^3} \].

Step 2: Define Reverse Reaction

The reverse reaction is \( \mathrm{NH}_{3} \rightarrow \frac{1}{2} \mathrm{N}_{2} + \frac{3}{2} \mathrm{H}_{2} \) and we want to find the equilibrium constant for this reaction.

Step 3: Reverse Reaction Equilibrium Constant

The equilibrium constant for the reverse reaction is the reciprocal of the original equilibrium constant. Therefore, if the equilibrium constant for the forward reaction is \( K_{e} \), the reverse reaction will have an equilibrium constant of \( \frac{1}{K_{e}} \).

Step 4: Adjust Coefficients

In the reverse reaction, coefficients are multiplied by a factor of 1/2. When you multiply the stoichiometric coefficients of a reaction by a number, the equilibrium constant is taken to that power. Here, since the original reaction is halved, the equilibrium constant is \( \left(\frac{1}{K_{e}}\right)^{1/2} \).

Step 5: Select the Correct Answer

Therefore, the correct expression for the equilibrium constant of \( \mathrm{NH}_{3} \rightarrow \frac{1}{2} \mathrm{N}_{2} + \frac{3}{2} \mathrm{H}_{2} \) is \( \frac{1}{\sqrt{K_{e}}} \). This makes option (d) \( \frac{1}{\sqrt{K_{t}}} \) the correct choice assuming \( K_{t} = K_{e} \).

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry, describing the state at which a reversible chemical reaction occurs at an equal rate in both the forward and backward directions. When a system reaches equilibrium, the concentrations of reactants and products remain constant over time, though they are not necessarily equal. This means there is no net change in the concentrations, even though the reactions continue to occur.
A critical aspect of chemical equilibrium is the equilibrium constant, denoted as \( K \). This constant is specific to a particular reaction at a given temperature and provides a quantifiable measure of the position of equilibrium.
- When \( K \) is much larger than 1, the equilibrium position is towards the products.
- When \( K \) is much smaller than 1, the equilibrium position is towards the reactants.
- When \( K \) is around 1, significant amounts of both reactants and products are present.In the example of the reaction \( \mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} \), the given equilibrium constant \( K_{e} \) indicates how the concentrations of nitrogen, hydrogen, and ammonia relate to each other at equilibrium.

Reversible Reactions

Reversible reactions are chemical reactions where the products can react to recreate the original reactants. These reactions do not go to completion as the products are constantly converting back into the reactants in the equilibrium state.
In reversible reactions, while the forward and reverse reactions occur simultaneously, the rates become equal at equilibrium. This dynamic balance is crucial as it allows for the equilibrium constant to stay constant for the reaction under constant conditions.
A good example is the synthesis of ammonia:

• The forward reaction combines nitrogen and hydrogen to form ammonia: \( \mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3} \).


• The reverse reaction involves breaking ammonia back into nitrogen and hydrogen: \( 2 \mathrm{NH}_{3} \rightarrow \mathrm{N}_{2} + 3 \mathrm{H}_{2} \).

By discussing reversible reactions, chemists can better understand products distribution and manipulate conditions to shift equilibrium to favor desired outcomes.

Stoichiometry

Stoichiometry is the practice of using balanced chemical equations to calculate the quantities of reactants and products involved in a chemical reaction. It serves as a foundation for understanding chemical reactions quantitatively and uses the concept of molar ratios derived from the balanced equations.
Balanced chemical equations display the stoichiometric coefficients that indicate the number of moles of each substance involved in the reaction.
In our ammonia synthesis example, the balanced reaction is \( \mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} \). From this equation:

• 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.


• The stoichiometric coefficients have a significant impact when dealing with equilibrium constants.


• Any change in these coefficients, such as halving or doubling, affects the expression for the equilibrium constant accordingly.

Understanding stoichiometry is essential when computing equilibrium constants, especially for reactions that are adjusted, as in the case where the coefficients are divided by a factor such as 2, impacting the equilibrium expression and constant calculation.