Problem 86 - A Complete Resource Book in Chemistry for JEE Main

Question

One mole of \(\mathrm{A}(\mathrm{g})\) is heated to \(300^{\circ} \mathrm{C}\) in a closed one litre vessel till the following equilibrium is reached. \(\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})\) The equilibrium constant of this reaction at \(300{ }^{\circ} \mathrm{C}\) is 4 . What is the concentration of \(\mathrm{B}\) (in \(\mathrm{mol} \mathrm{L}^{-1}\) ) at equilibrium? (a) \(0.2\) (b) \(0.6\) (c) \(0.75\) (d) \(0.1\)

Step-by-Step Solution

Answer

None of the given options match; the calculated concentration of B at equilibrium is 0.8 mol/L.

Step 1: Write the balanced equation and initial conditions

The chemical reaction given is \( \mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g}) \). Initially, one mole of \( \mathrm{A} \) is present in a 1-liter vessel, so the initial concentration of \( \mathrm{A} \) is 1 M (since concentration is defined as moles per liter).

Step 2: Set up the equilibrium expression

According to the reaction \( \mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g}) \), the equilibrium constant \( K_c \) is given by the expression:\[K_c = \frac{[B]}{[A]}\]where \([B]\) and \([A]\) are the equilibrium concentrations of \( \mathrm{B} \) and \( \mathrm{A} \), respectively. We are given that \( K_c = 4 \).

Step 3: Define the change in concentration at equilibrium

Let \( x \) be the equilibrium concentration of \( \mathrm{B} \) in mol/L, meaning that \( x \) amount of \( \mathrm{A} \) has been converted to \( \mathrm{B} \). Therefore, the concentration of \( \mathrm{A} \) at equilibrium will be \( 1 - x \).

Step 4: Substitute into the equilibrium expression

Substitute \( [B] = x \) and \( [A] = 1 - x \) into the equation for the equilibrium constant:\[4 = \frac{x}{1-x}\]

Step 5: Solve for x

Rearrange the equation to solve for \( x \):\[4(1-x) = x\]\[4 - 4x = x\]\[4 = 5x\]\[x = \frac{4}{5} = 0.8\]mol/L.

Step 6: Verify the possible options

The calculated concentration of \( \mathrm{B} \) is 0.8 mol/L, which does not match any of the given options exactly. Reevaluate to ensure no calculation errors have occurred or check the problem statement for potential discrepancies.

Key Concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K_c \), is a crucial concept in chemical equilibrium. It provides a quantitative measure of the position of equilibrium in a chemical reaction. Specifically, it relates the concentrations of products to the concentrations of reactants at equilibrium. For a general reaction:

• \( aA + bB \rightleftharpoons cC + dD \)

The equilibrium constant \( K_c \) is given by:\[K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\]Here, the square brackets denote the concentration of each species, and the exponents (\(a, b, c, d\)) correspond to their stoichiometric coefficients.
The value of \( K_c \) provides insight into the reaction's tendencies:

• If \( K_c \gg 1 \), the equilibrium lies to the right, favoring products.
• If \( K_c \ll 1 \), the equilibrium lies to the left, favoring reactants.
• If \( K_c \approx 1 \), neither products nor reactants are favored.

In the exercise, \( K_c \) is given as \(4\), suggesting the reaction favors the formation of \( B \) over \( A \) at equilibrium.

Concentration of Gases

When dealing with gaseous reactions, the concentration of gases plays a foundational role. Gas concentration is measured in molarity, the number of moles of a gas per liter of volume (mol/L). In a closed system, like the one in the exercise, the initial concentration can be easily determined by dividing the number of moles by the volume of the container.
In the exercise, \(1\) mole of \( A \) is introduced into a \(1\)-liter vessel, resulting in an initial concentration of \(1\) M.
As the system reaches equilibrium, some of the \( A \) is converted to \( B \). If we denote the equilibrium concentration of \( B \) as \( x \), then the equilibrium concentration of \( A \) will be \( 1 - x \). Knowing how gases convert at equilibrium allows us to make use of the equilibrium constant to solve for the concentration of each species.

Reaction Stoichiometry

Reaction stoichiometry is all about the quantitative relationships between reactants and products in a chemical reaction. It provides the necessary coefficients that allow chemists to predict the amounts of different substances consumed and produced.
For the reaction \( A(g) \rightleftharpoons B(g) \), the stoichiometry is straightforward since both sides of the reaction have the same coefficients. This means for every mole of \( A \) that reacts, one mole of \( B \) is produced.

• Start by establishing a change variable, \( x \), which represents the moles of \( A \) turned into \( B \).
• The equilibrium concentration of each is: \([B] = x\), and \([A] = 1 - x\).

Using \( K_c = 4 \), the stoichiometry helps set up the equation \( \frac{x}{1-x} = 4 \). Solving this provides \( x \), the equilibrium concentration of \( B \), emphasizing the simplicity of stoichiometry in calculations.