Problem 203 - A Complete Resource Book in Chemistry for JEE Main

Question

The equilibrium constant \(\left(\mathrm{K}_{0}\right)\) for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})\) at temperature \(\mathrm{T}\) is \(4 \times 10^{-4}\). The value of \(\mathrm{K}_{\mathrm{c}}\) for the reaction, \(\mathrm{NO}(\mathrm{g}) \rightarrow \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) at the same temperature is \([\mathbf{2 0 1 2}]\) (a) \(4 \times 10^{-4}\) (b) \(0.02\) (c) \(2.5 \times 10^{2}\) (d) \(50.0\)

Step-by-Step Solution

Answer

Correct option (d): 50.0

Step 1: Identify the Given Reaction and Its Equilibrium Constant

The given reaction is \( \mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2\mathrm{NO}(\mathrm{g}) \) with an equilibrium constant \( K_0 = 4 \times 10^{-4} \). This equilibrium constant corresponds to the direct formation of NO from N2 and O2.

Step 2: Write the Reverse Reaction and Determine Its Equilibrium Constant

The reverse reaction of the given is \( 2\mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \). The equilibrium constant for the reverse reaction (\( K_r \)) is the reciprocal of the equilibrium constant for the forward reaction. Thus, \( K_r = \frac{1}{K_0} = \frac{1}{4 \times 10^{-4}} = 2.5 \times 10^{3} \).

Step 3: Adjust for the Stoichiometry of the Given Reaction

In the problem, the reaction of interest is \( \mathrm{NO}(\mathrm{g}) \rightarrow \frac{1}{2} \mathrm{N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \). Notice that this involves just half of each molecule of N2 and O2 compared to the previous step, or equivalently, half of the NO per the forward-backward reaction pairs. This reaction can be derived by raising the equilibrium constant from the previous step to the power of 1/2 to conform to the proposed stoichiometry: \[ K_c = \left( 2.5 \times 10^{3} \right)^{\frac{1}{2}} = 50.0 \]

Step 4: Choose the Correct Option

From the calculations in the previous steps, the equilibrium constant \( K_c \) for the reaction \( \mathrm{NO}(\mathrm{g}) \rightarrow \frac{1}{2} \mathrm{N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \) is calculated as 50.0. The correct option matching this \( K_c \) value is option (d).

Key Concepts

Reaction Stoichiometry

Stoichiometry is the part of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It tells you how much of each substance is needed to react or is produced in a balanced equation. For our given problem, stoichiometry is used when we consider the amounts of nitrogen (N extsubscript{2}), oxygen (O extsubscript{2}), and nitrogen monoxide (NO) in the reactions.
In the first reaction, two molecules of nitrogen and oxygen combine to produce two molecules of nitrogen monoxide. Thus, the stoichiometry is straightforward:

• 1 molecule of N extsubscript{2} + 1 molecule of O extsubscript{2} = 2 molecules of NO

For the reverse reaction, stoichiometry plays a crucial role in adjusting equilibrium calculations. Here, the problem’s stoichiometry is adjusted by breaking down the produced NO into equality for half-molecules:

• 1 molecule of NO = 1/2 molecule of N extsubscript{2} + 1/2 molecule of O extsubscript{2}

This adjustment must be reflected in the equilibrium constant calculations.

Reverse Reactions

In chemical reactions, understanding the forward and reverse reactions is essential. The concept of reverse reactions is concerned with the reverse process of a given chemical reaction. If a reaction can proceed in one direction, it can often proceed in the opposite direction.
In our exercise, after identifying the forward reaction's equilibrium constant, the reverse reaction is presented as:

• 2 NO (g) → N extsubscript{2} (g) + O extsubscript{2} (g)

The reverse reaction involves the breaking down of NO back into its diatomic constituents. The equilibrium constant for the reverse reaction is the reciprocal of the forward constant. For example, if the forward equilibrium constant ( K extsubscript{0}) is 4 x 10⁻⁴, the reverse constant is calculated by 1/ K extsubscript{0}, showing the intrinsic connection between reciprocal processes. Upon solving, K extsubscript{r} becomes 2.5 x 10³.

Equilibrium Calculations

Equilibrium calculations are vital when evaluating how a reaction behaves under specific conditions. Equilibrium is established when the rate of the forward reaction equals the reverse reaction, meaning there is no net change in the concentration of reactants and products.
To solve such problems, one must adjust their calculations based on stoichiometry. After computing the reverse equilibrium constant, consider necessary adjustments due to different stoichiometries. In this example, the given problem further complicates the equilibrium constant by incorporating half-stoichiometric coefficients:

• NO (g) → 1/2 N extsubscript{2} (g) + 1/2 O extsubscript{2} (g)

To calculate the equilibrium constant for this setup ( K extsubscript{c}), one must take the square root of the reverse constant due to halving the stoichiometry. Therefore, the calculation is: K extsubscript{c} = ( 2.5 x 10³)⁰⋅⁵, resulting in 50.0.
This approach allows for accurate predictability and understanding of chemical equilibria in varied reaction conditions.