Problem 117 - A Complete Resource Book in Chemistry for JEE Main

Question

The equilibrium constant value for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) is \(1.48 \times 10^{-5}\), the value for the reaction \(1 / 2 \mathrm{~N}_{2}(\mathrm{~g})+3 / 2 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})\) is \(\mathrm{n} \times 10^{-3}\) where \(\mathrm{n}\) is (a) \(1.85\) (b) \(3.85\) (c) \(4.85\) (d) 10

Step-by-Step Solution

Answer

The value of \(\mathrm{n}\) is 3.85, so the correct option is (b).

Step 1: Understand the Problem

The problem is to find the equilibrium constant for the reaction \(\frac{1}{2} \mathrm{N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})\) using the known equilibrium constant for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\). The given equilibrium constant for the latter reaction is \(1.48 \times 10^{-5}\).

Step 2: Relationship Between Reactions

Recognize that the given reaction is essentially the original reaction scaled down by a factor. This means that the stoichiometry is divided by 2 (i.e., original coefficients halved). For reaction equilibria, when a balanced equation is multiplied or divided by any coefficient, the equilibrium constant is raised to the power of that coefficient. Here, it is divided by 2.

Step 3: Applying the Mathematical Rule

The original equation's equilibrium constant is \(K_c = 1.48 \times 10^{-5}\). For the reaction \(\frac{1}{2} \mathrm{N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})\), the equilibrium constant \(K'_c\) is calculated as \(\sqrt{K_c}\) because the equation is halved. This implies: \[K'_c = (K_c)^{\frac{1}{2}} = (1.48 \times 10^{-5})^{\frac{1}{2}}\].

Step 4: Calculate the New Equilibrium Constant

Calculate \(K'_c = \sqrt{1.48 \times 10^{-5}}\). Using a calculator, we find \(K'_c \approx 3.85 \times 10^{-3}\).

Step 5: Identify the Correct Option

Compare the calculated equilibrium constant with the given options. Here \(\mathrm{n} = 3.85\), corresponding to option (b).

Key Concepts

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It informs us about the proportions of substances involved, which is crucial when altering reaction equations.
For example, in the reaction \(_2 + 3H_2 \rightleftharpoons 2NH_3\), the coefficients (1, 3, and 2) tell us the exact mole ratio in which substances react and are produced.
When you modify these coefficients, as in the second reaction \(\frac{1}{2}N_2 + \frac{3}{2}H_2 \rightleftharpoons NH_3\), these new ratios directly impact the calculation of the equilibrium constant.
Such changes in stoichiometric coefficients are important in understanding how the equilibrium constant transforms based on reaction scale, which is foundational to following thermodynamic and equilibrium principles.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the backward reaction, meaning that the concentrations of reactants and products remain constant over time.
At this point, the reaction doesn't stop; instead, it proceeds in both directions at the same rate.
This dynamic state is represented by the equilibrium constant \(K_c\), which provides a snapshot of the ratio of product concentrations to reactant concentrations applied for a balanced reaction equation.

• If \(K_c > 1\), products are favored.
  
• If \(K_c < 1\), reactants are predominant.
  

This delicate balance can shift due to changes in temperature, pressure, or concentration, according to Le Chatelier’s Principle, to find a new equilibrium state.

Equilibrium Law

The equilibrium law, or the law of mass action, states that for a chemical reaction at equilibrium, the ratio of product concentrations raised to the power of their stoichiometric coefficients is constant, and this constant is the equilibrium constant \(K_c\).
This law allows the prediction of the effect of concentration changes on the system at equilibrium.
For example, in our altered equation \(\frac{1}{2}N_2 + \frac{3}{2}H_2 \rightleftharpoons NH_3\), the law aids in understanding how halving the coefficients affects the equilibrium constant.
When coefficients are changed in a balanced equation, the exponents in the \(K_c\) expression change, requiring recalculation of the constant, showcasing the precision needed in equilibrium calculations.

Thermodynamic Principles

Thermodynamic principles offer broad insight into energy transformations during chemical reactions, including the surroundings' response and inherent spontaneity.
At its core, thermodynamics helps clarify why the equilibrium state is chosen and how energy aspects like enthalpy, and entropy, play crucial roles.
Although the focus in equilibrium is on concentration, thermodynamics explains why certain reactions are more favorable or faster to reach equilibrium.

• Enthalpy: An understanding of heat exchange during the reaction.
  
• Entropy: Insight into disorder and energy distribution.
  
• Free Energy: Determines the spontaneity and feasibility of reaching equilibrium.
  

Connecting thermodynamics to equilibrium gives you a full view of a reaction's direction and extent, especially under varying conditions.