Problem 122 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{K}_{\text {c for the reaction } \mathrm{SO}_{2}}(\mathrm{~g})+\mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g})\) \(+\mathrm{NO}(\mathrm{g})\) is 16 at a given temperature. If we take one mole each of all the four gases in one litre vessel, the equilibrium concentration of \(\mathrm{SO}_{2}\) and \(\mathrm{SO}_{3}\) respectively in \(\mathrm{mol} \mathrm{L}^{-1}\) are (a) \(0.4,0.8\) (b) \(0.8,1.6\) (c) \(1.6,0.8\) (d) \(0.4,1.6\)

Step-by-Step Solution

Answer

The equilibrium concentrations are 0.4 M for \( \mathrm{SO}_2 \) and 1.6 M for \( \mathrm{SO}_3 \), making the answer (d) 0.4, 1.6.

Step 1: Write the balanced chemical equation

The given chemical reaction is \( \mathrm{SO}_2(\mathrm{~g}) + \mathrm{NO}_2(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_3(\mathrm{~g}) + \mathrm{NO}(\mathrm{g}) \). This is a 1:1:1:1 reaction with equal number of moles on both sides of the equation.

Step 2: Define initial concentrations

Since we have 1 mole of each gas in a 1 L vessel, the initial concentrations for \( \mathrm{SO}_2, \mathrm{NO}_2, \mathrm{SO}_3, \text{ and } \mathrm{NO} \) are all 1 M (1 mole/L).

Step 3: Express equilibrium concentrations

Let the change in concentration of \( \mathrm{SO}_2 \) and \( \mathrm{NO}_2 \) that reacts at equilibrium be \( x \). At equilibrium, the concentration of \( \mathrm{SO}_2 \) and \( \mathrm{NO}_2 \) will be \( 1-x \), while \( \mathrm{SO}_3 \) and \( \mathrm{NO} \) will have a concentration of \( 1+x \).

Step 4: Write the expression for the equilibrium constant

The equilibrium constant \( K_c \) is given by the expression: \[ K_c = \frac{[\mathrm{SO}_3][\mathrm{NO}]}{[\mathrm{SO}_2][\mathrm{NO}_2]} \]. Since \( K_c = 16 \), we can substitute the equilibrium concentrations to get: \[ 16 = \frac{(1+x)^2}{(1-x)^2} \].

Step 5: Solve for x

Solve the equation \( \frac{(1+x)^2}{(1-x)^2} = 16 \). Taking the square root on both sides gives \( \frac{1+x}{1-x} = 4 \). Therefore, solve \( 4(1-x) = 1+x \) which simplifies to \( 4 - 4x = 1 + x \). This results in \( 3 = 5x \) or \( x = 0.6 \).

Step 6: Calculate equilibrium concentrations

The equilibrium concentration of \( \mathrm{SO}_2 \) is \( 1-x = 1-0.6 = 0.4 \) M and the concentration of \( \mathrm{SO}_3 \) is \( 1+x = 1+0.6 = 1.6 \) M.

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. Consequently, the concentrations of the products and reactants remain constant over time, although they are not necessarily equal. This balance occurs when the dynamic processes of forming products and reforming reactants are occurring at the same rate, maintaining a stable ratio over time.
Understanding chemical equilibrium is crucial for predicting the concentrations of reactants and products in a reaction at any given time, which helps in identifying the conditions under which a reaction is most efficient.

• It's important to remember that at equilibrium, no reactants and products are added or removed; it's a closed system.
• The concept does not imply that the amounts of reactants and products are equal, but that the rates of their formation are equal.
• An equilibrium tends to shift to counteract changes in conditions (Le Chatelier's principle).

Grasping chemical equilibrium is like understanding a seesaw—both sides balance each other, even if the weight and distances from the pivot are different.

Balanced Chemical Equation

A balanced chemical equation is fundamental in solving equilibrium-related problems. It involves making sure that the number of atoms for each element is equal on both sides of the equation, reflecting the law of conservation of mass.
For the equation given in the exercise: \[\mathrm{SO}_2(\mathrm{~g}) + \mathrm{NO}_2(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_3(\mathrm{~g}) + \mathrm{NO}(\mathrm{g})\]This is a simple equation representing a 1:1:1:1 ratio, which means that one mole of each reactant enters the reaction to form one mole of each product. The symmetry in numbers ensures that the mass is conserved throughout the reaction.

• Balancing equations is essential before moving on to calculate or deduce any equilibrium properties.
• It assists in predicting how changes in conditions can affect the reaction, such as when calculations for equilibrium constants are involved.
• A balanced chemical equation provides a clear foundation for understanding how the reaction progresses over time.

Mastering this concept ensures clarity and precision in carrying out further calculations and experimental predictions.

Equilibrium Concentrations

Equilibrium concentrations refer to the amounts of reactants and products in a chemical reaction once it has reached equilibrium. Calculating these concentrations is key in understanding how the reaction behaves under certain conditions.
In the original problem, the initial concentrations of the gases were 1 M for each due to the volume of the vessel being 1 L and having 1 mole of each reactant. With the change in concentration described by \[x\], the equilibrium concentrations can be expressed as:- \(\mathrm{SO}_2\) and \(\mathrm{NO}_2\): \(1 - x\) - \(\mathrm{SO}_3\) and \(\mathrm{NO}\): \(1 + x\)By substituting these expressions into the equilibrium constant equation, students can solve for \(x\) and plug that value back in to find the actual concentrations at equilibrium.

• This approach highlights that equilibrium is not static but involves reaching a particular dynamic balance.
• Understanding equilibrium concentrations aid in determining the extent of the reaction and predicting the yield of products.
• Practical applications include tasks in various industries, such as manufacturing and pharmaceuticals, where being able to control product formation is crucial.

By mastering this concept, one will be well-equipped to analyze reactions and adjust conditions to shift the equilibrium as desired.