Problem 81 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(2 \mathrm{NO} \rightleftharpoons \mathrm{N}_{2}+\mathrm{O}_{2}\) \(\mathrm{K}_{c_{1}}=2.5 \times 10^{\mathrm{s}}\) \(\mathrm{NO}+\frac{1}{2} \mathrm{Br}_{\mathrm{g}} \rightleftharpoons \mathrm{NOBr}\) \(\mathrm{K}_{\mathrm{c}_{2}}=1.6\) find \(K_{C}\) for the reaction given below \(\frac{1}{2} \mathrm{~N}_{2}+\frac{1}{2} \mathrm{O}_{2}+\frac{1}{2} \mathrm{Br}_{2} \rightleftharpoons \mathrm{NOBr}\) (a) \(1.01 \times 10^{-15}\) (b) \(2.02 \times 10^{-15}\) (c) \(1.01 \times 10^{30}\) (d) \(2.02 \times 10^{15}\)

Step-by-Step Solution

Answer

The equilibrium constant \( K_c \) for the reaction is \( 1.01 \times 10^{-15} \).

Step 1: Understand the Reaction Directions

First, we need to identify the relationship between the given reactions and the target reaction. The first reaction involving NO goes in the reverse direction to form N2 and O2, while in the second reaction, NO reacts with Br2 to form NOBr.

Step 2: Relating Given Constants to Target Reaction

The target reaction can be considered as the combination of the given reactions to form NOBr from N2, O2, and Br2. We need to find the equilibrium constant for this new reaction.

Step 3: Manipulate First Reaction Equation

Reverse reaction (1), which forms 2NO from N2 and O2. Reversing the equation involves taking the reciprocal of Kc1. Adjust the stoichiometry by halving the reaction: \[ N_2 + O_2 ightleftharpoons 2NO, \ K_{c1}' = \frac{1}{K_{c1}} \] Halving the stoichiometric coefficients, \[ \frac{1}{2}N_2 + \frac{1}{2}O_2 \rightleftharpoons NO, \ K_{c1}'' = (K_{c1}')^{\frac{1}{2}} \]

Step 4: Calculate Kc1' and Kc1''

First, find Kc1' by inverting Kc1: \[ K_{c1}' = \frac{1}{2.5 \times 10^5} \] Then, find Kc1'' by taking the square root:\[ K_{c1}'' = \left( \frac{1}{2.5 \times 10^5} \right)^{\frac{1}{2}} \]

Step 5: Combine the Adjusted Reactions

Combine both adjusted reactions to reflect the target reaction. Since the second reaction provides the formation of NOBr from NO and Br2, the final reaction can be:\[ \frac{1}{2}N_2 + \frac{1}{2}O_2 + \frac{1}{2}Br_2 \rightleftharpoons NOBr \]The equilibrium constant for this combined reaction is \[ K_c = K_{c1}'' \times K_{c2} \].

Step 6: Calculate Final Equilibrium Constant

Substitute the values from earlier steps to find the effective Kc for the target equation:\[ K_c = \left( \frac{1}{2.5 \times 10^5} \right)^{\frac{1}{2}} \times 1.6 \].Simplify to find the numerical value of Kc.

Key Concepts

Equilibrium Constant

The equilibrium constant, represented as \(K_c\), quantifies the balance between reactants and products in a chemical reaction at equilibrium. It's derived from the concentrations of the involved species, mathematically capturing this state of balance. For reversible reactions, the value of \(K_c\) is fundamental, as it shows how far the reaction proceeds before reaching equilibrium.
In our given problem, we start with two separate equilibrium systems, each with their own \(K_c\) values. The task is to find \(K_c\) for a new reaction derived from these systems. It's crucial to understand how transformations, such as reversing reaction direction or changing stoichiometry, relate to \(K_c\).
For example, flipping a reaction requires taking the reciprocal of \(K_c\), and altering stoichiometric coefficients calls for taking appropriate roots or powers of \(K_c\). This manipulation of \(K_c\) is key in solving problems involving a series of related reactions.

Reaction Stoichiometry

Reaction stoichiometry refers to the quantitative relationship among reactants and products in a chemical reaction. It shows how molecules or moles of substances interact, foundational in calculating equilibrium constants.
The stoichiometric coefficients within a balanced chemical equation signal the necessity of maintaining proportions during reactions. They become essential when computing equilibrium constants, because altering these coefficients directly impacts the mathematically derived \( K_c \).

• For instance, halving the stoichiometric coefficients in our target equation involves taking the square root of \( K_{c1}' \).
• This ensures the calculated equilibrium constant mirrors the adjusted stoichiometry, in terms of moles and concentration.

The precise manipulation of these coefficients through multiplication, division, or other operations can change how \( K_c \) is determined, playing an essential role in reaction calculations.

Reversible Reactions

Reversible reactions are those that proceed in both forward and reverse directions, reaching a state of equilibrium. In equilibrium, the rates of the forward and reverse reactions are equal, maintaining constant concentrations of species.
In the given exercise, the reactions under consideration are reversible, necessitating mindful manipulations to derive the equilibrium constant for the target reaction. Recognizing these reactions' reversible nature helps in discerning how changes to one side affect the other.

• Reversing any reaction direction during calculations requires taking the reciprocal of the equilibrium constant, such as inverting \( K_{c1} \) for the reaction \( 2NO \rightarrow N_2 + O_2 \).
• Additionally, combining such reversible reactions aligns with equilibrium principles, where the overall \( K_c \) for a multistep reaction becomes a product of its component \( K_c \) values.

Understanding the concept of reversible reactions allows you to comprehend fully how reactants and products balance within a system.