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

Question

\(\mathrm{K}_{\mathrm{c}}\) for the reaction \(\mathrm{SO}_{2}(\mathrm{~g})+\mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g})\) \+ NO (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 \(\operatorname{mol} \mathrm{L}^{-1}\) 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 correct answer is option (d): 0.4 mol/L for \(\mathrm{SO}_{2}\) and 1.6 mol/L for \(\mathrm{SO}_{3}\).

Step 1: Write the balanced chemical equation

The balanced chemical equation for the reaction is: \[ \mathrm{SO}_{2}(\mathrm{~g}) + \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g}) + \mathrm{NO}(\mathrm{~g}) \] This indicates that 1 mole of \(\mathrm{SO}_{2}\) reacts with 1 mole of \(\mathrm{NO}_{2}\) to produce 1 mole of \(\mathrm{SO}_{3}\) and 1 mole of \(\mathrm{NO}\).

Step 2: Set the initial concentrations

Start with 1 mole each of \(\mathrm{SO}_{2}\), \(\mathrm{NO}_{2}\), \(\mathrm{SO}_{3}\), and \(\mathrm{NO}\) in a 1-litre vessel. Thus, initial concentrations are: \([\mathrm{SO}_{2}]=1\), \([\mathrm{NO}_{2}]=1\), \([\mathrm{SO}_{3}]=1\), \([\mathrm{NO}]=1\) \text{mol/L}.

Step 3: Determine change in concentration at equilibrium

Let \(x\) be the number of moles of \(\mathrm{SO}_{2}\) and \(\mathrm{NO}_{2}\) that react to form \(x\) moles of \(\mathrm{SO}_{3}\) and \(\mathrm{NO}\). At equilibrium, the concentrations are: \([\mathrm{SO}_{2}]=1-x\), \([\mathrm{NO}_{2}]=1-x\), \([\mathrm{SO}_{3}]=1+x\), \([\mathrm{NO}]=1+x\).

Step 4: Write the expression for \(K_c\)

\(K_c\) is given by: \[ K_c = \frac{[\mathrm{SO}_{3}][\mathrm{NO}]}{[\mathrm{SO}_{2}][\mathrm{NO}_{2}]} \] Substituting the equilibrium concentrations gives: \[ K_c = \frac{(1+x)(1+x)}{(1-x)(1-x)} = \frac{(1+x)^2}{(1-x)^2} \]

Step 5: Solve for x using the given value of \(K_c\)

Given \(K_c = 16\), solve \[ \frac{(1+x)^2}{(1-x)^2} = 16 \] Taking square roots on both sides gives \( \frac{1+x}{1-x} = 4 \). Solving for \(x\), we get: \[ 1+x = 4-4x \] \[ 5x = 3 \] \[ x = 0.6 \]

Step 6: Calculate equilibrium concentrations of \(\mathrm{SO}_{2}\) and \(\mathrm{SO}_{3}\)

Substitute \(x=0.6\) back into the expressions for equilibrium concentrations. For \([\mathrm{SO}_{2}]=1-x\), we have: \([\mathrm{SO}_{2}]=1-0.6=0.4\). For \([\mathrm{SO}_{3}]=1+x\), we have: \([\mathrm{SO}_{3}]=1+0.6=1.6\). Therefore, the equilibrium concentrations are 0.4 mol/L for \(\mathrm{SO}_{2}\) and 1.6 mol/L for \(\mathrm{SO}_{3}\).

Key Concepts

Equilibrium Constant (Kc)

The equilibrium constant, abbreviated as \(K_c\), is a numerical value that provides insight into the ratio of concentrations of products to reactants at chemical equilibrium. It is crucial for understanding the extent of a reaction. For a given formula: \[ K_c = \frac{[C]^c \cdot [D]^d}{[A]^a \cdot [B]^b} \] Where \([A]\) and \([B]\) are reactants, \([C]\) and \([D]\) are products, and the lowercase letters \(a\), \(b\), \(c\), and \(d\) are their respective stoichiometric coefficients.
In our exercise, the reaction is \(\mathrm{SO}_{2}(\mathrm{~g}) + \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g}) + \mathrm{NO}(\mathrm{~g})\). The equilibrium constant, \(K_c\), has been provided as 16. This implies that at equilibrium, the concentration of the products \(\mathrm{SO}_{3}\) and \(\mathrm{NO}\) is significantly greater than that of the reactants \(\mathrm{SO}_{2}\) and \(\mathrm{NO}_{2}\).
It's important to note that \(K_c\) is temperature-dependent. This means if you alter the temperature, \(K_c\) will change, indicating a shift in equilibrium in favor of either the reactants or the products, depending on the reaction's nature.

Le Chatelier's Principle

Le Chatelier's Principle is an essential concept in chemical equilibrium. It provides us with a way to predict how a change in conditions will affect the system in equilibrium. According to this principle, if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts in such a way as to counteract the change.
The main factors that can disturb equilibrium include changes in concentration, pressure, and temperature. For instance, if you increase the concentration of a reactant, the system reacts by consuming more of it to form more products. Conversely, if a product is removed, equilibrium shifts towards the formation of more products from reactants.
Let's apply this to our exercise: if the concentration of \(\mathrm{SO}_{3}\) were increased, the system would shift to form more \(\mathrm{SO}_{2}\) and \(\mathrm{NO}_{2}\) to restore balance. If the temperature were increased and the reaction is exothermic, like many combustion reactions, then \(\mathrm{SO}_{2}\) and \(\mathrm{NO}_{2}\) formation would increase to absorb the excess heat, shifting equilibrium to the left. This principle is a strong predictor of behavior at equilibrium and helps chemists control reactions to favor specific outcomes.

Reaction Quotient

The reaction quotient, denoted as \(Q\), shares a similar form with the equilibrium constant, \(K_c\), but is used to assess the current state of a reaction not necessarily at equilibrium. It is calculated using the same formula as \(K_c\): \[ Q = \frac{[C]^c \cdot [D]^d}{[A]^a \cdot [B]^b} \]
Unlike \(K_c\), \(Q\) can take any positive value because it reflects the ratio of products to reactants at any point in time. It is crucial for predicting the direction a reaction will proceed to achieve equilibrium.
In the context of our exercise, you can compare \(Q\) to \(K_c\). If \(Q < K_c\), the reaction will proceed in the forward direction, towards products, to reach equilibrium. If \(Q > K_c\), the reaction will shift backward, towards reactants.
Therefore, calculating \(Q\) helps determine whether the system is at equilibrium or needs to shift to achieve it, and in which direction.