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

Question

\(\mathrm{K}_{c}\) 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 \(\operatorname{mol} \mathrm{L}^{-1}\) 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 equilibrium concentrations are 0.4 mol/L for SO2 and 1.6 mol/L for SO3. Option (d) is correct.

Step 1: Set up Initial Concentrations

Since one mole of each of the four gases is placed in a 1 L vessel, initial concentrations of \(\text{SO}_2\), \(\text{NO}_2\), \(\text{SO}_3\), and \(\text{NO}\) are each \(1 \text{ mol/L}\).

Step 2: Establish Change at Equilibrium

Let \(x\) be the amount (in moles per liter) of \(\text{SO}_2\) and \(\text{NO}_2\) that react to form \(\text{SO}_3\) and \(\text{NO}\). Therefore, at equilibrium, the concentrations will be: \([\text{SO}_2] = 1 - x\), \([\text{NO}_2] = 1 - x\), \([\text{SO}_3] = 1 + x\), and \([\text{NO}] = 1 + x\).

Step 3: Write Equilibrium Expression

Using the expression for \(K_c\), we have:\[K_c = \frac{[\text{SO}_3][\text{NO}]}{[\text{SO}_2][\text{NO}_2]} = 16\].Substitute the equilibrium concentrations: \[16 = \frac{(1+x)(1+x)}{(1-x)(1-x)}\].

Step 4: Simplify and Solve the Equation

Expand and simplify the equation: \[16 = \frac{(1 + 2x + x^2)}{(1 - 2x + x^2)}\].Cross-multiply to solve for \(x\):\[16(1 - 2x + x^2) = 1 + 2x + x^2\].Further simplification gives:\[16 - 32x + 16x^2 = 1 + 2x + x^2\].Re-arrange and simplify:\[15x^2 - 34x + 15 = 0\].

Step 5: Solve the Quadratic Equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find \(x\). Here, \(a = 15\), \(b = -34\), and \(c = 15\):\[x = \frac{-(-34) \pm \sqrt{(-34)^2 - 4\cdot15\cdot15}}{2\cdot15}\].Calculate \(x\):\[x = \frac{34 \pm \sqrt{1156 - 900}}{30} = \frac{34 \pm \sqrt{256}}{30} = \frac{34 \pm 16}{30}\].

Step 6: Determine Valid Values of x

Calculate possible values for \(x\): \[x = \frac{34 + 16}{30} = \frac{50}{30} = \frac{5}{3} \approx 1.67\]\[x = \frac{34 - 16}{30} = \frac{18}{30} = 0.6\].Since \(x\) must be less than or equal to the initial concentrations (1), \(x = 0.6\) is valid.

Step 7: Calculate Equilibrium Concentrations

Given \(x = 0.6\): Calculate \([\text{SO}_2] = 1 - x = 0.4 \text{ mol/L}\) and \([\text{SO}_3] = 1 + x = 1.6 \text{ mol/L}\).

Key Concepts

Equilibrium Constant (Kc)

Chemical equilibrium represents the state in which the concentrations of reactants and products remain constant over time. At this point, the forward and reverse reactions occur at equal rates. A central concept in understanding equilibrium is the Equilibrium Constant, denoted as \(K_c\). The \(K_c\) is a ratio of the concentration of the products to the concentration of the reactants, each raised to the power of their stoichiometric coefficients. It is calculated at a constant temperature and provides insight into the position of equilibrium. A larger \(K_c\) value indicates a reaction favoring the products, while a smaller \(K_c\) suggests more reactants are present.

• For the reaction \(\text{SO}_2(\text{g}) + \text{NO}_2(\text{g}) \rightleftharpoons \text{SO}_3(\text{g}) + \text{NO}(\text{g})\), the equilibrium expression is:
• \[K_c = \frac{[\text{SO}_3][\text{NO}]}{[\text{SO}_2][\text{NO}_2]}\]

Here, \(K_c\) equals 16, suggesting that at equilibrium, the reaction heavily leans toward the formation of \(\text{SO}_3\) and \(\text{NO}\). By initially knowing the \(K_c\), one can predict how far a reaction will proceed and the expected concentrations at equilibrium.

Quadratic Equation in Chemistry

Quadratic equations commonly appear in chemical equilibrium problems. Whenever changes in concentrations involve a type of reaction that can be represented by a quadratic equation, solving the equation becomes necessary to find the equilibrium concentrations.

• In this particular problem, the expression for \(K_c\) leads to a quadratic equation that looks like:
• \[16(1 - 2x + x^2) = 1 + 2x + x^2\]
• Simplifying further, we get: \[15x^2 - 34x + 15 = 0\]

Such equations can be solved using the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\], where \(a\), \(b\), and \(c\) are coefficients of the quadratic equation. Here, the calculated roots for \(x\) provide possible equilibrium concentrations.

• When applying this in chemistry, remember only physically meaningful solutions satisfy the constraints of the problem. Meaning, a root higher than the initial concentration is not valid chemically.

Reaction Stoichiometry

Reaction stoichiometry is vital in calculating relationships between reactants and products in a chemical reaction. It involves the quantitative relationship drawn from the reaction equation itself, providing a simplified method for finding how much of a substance is consumed or produced.

• For this reaction, the stoichiometric coefficients are 1:1, indicating equal molar relationships between all reactants and products.

When we let \(x\) be the amount of \(\text{SO}_2\) and \(\text{NO}_2\) that reacts, it directly links to the amount of \(\text{SO}_3\) and \(\text{NO}\) produced.

• For example, if \(x\) moles of \(\text{SO}_2\) are consumed, then \(x\) moles of \(\text{NO}_2\) are also consumed, and \(x\) moles of both \(\text{SO}_3\) and \(\text{NO}\) are produced.

This simplifies tremendously when calculating equilibrium concentrations, reflecting the principles that guide chemical reactions, making it highly systematic and predictable.