Problem 136 - A Complete Resource Book in Chemistry for JEE Main

Question

At \(550 \mathrm{~K}\), the \(\mathrm{Kc}\) for the following reaction is \(10^{4}\) \(\mathrm{mol}^{-1} \mathrm{~L}\) \(\mathrm{X}(\mathrm{g})+\mathrm{Y}(\mathrm{g}) \rightleftharpoons \mathrm{Z}(\mathrm{g})\) At equilibrium, it was observed that \([\mathrm{X}]=\frac{1}{2}[\mathrm{Y}]=\frac{1}{2}[\mathrm{Z}]\) What is the value of \([\mathrm{Z}]\) at equilibrium? (a) \(10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\) (b) \(2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\) (c) \(2 \times 10^{4} \mathrm{~mol} \mathrm{~L}^{-1}\) (d) \(10^{4} \mathrm{~mol} \mathrm{~L}^{-1}\)

Step-by-Step Solution

Answer

The calculated \([\mathrm{Z}]\) is \(4 \times 10^{-4} \, \text{mol} \, \text{L}^{-1}\).

Step 1: Understand the relationship between concentrations

According to the problem, \([\mathrm{X}] = \frac{1}{2}[\mathrm{Y}] = \frac{1}{2}[\mathrm{Z}]\). Let's introduce a variable \([\mathrm{Z}] = z\). This implies \([\mathrm{X}] = \frac{z}{2}\) and \([\mathrm{Y}] = \frac{z}{2}\).

Step 2: Write the expression for Kc

For the reaction \(\mathrm{X} + \mathrm{Y} \rightleftharpoons \mathrm{Z}\), the equilibrium constant \(K_c\) is given by: \[ K_c = \frac{[\mathrm{Z}]}{[\mathrm{X}][\mathrm{Y}]} = 10^4 \, \text{mol}^{-1} \text{L} \]

Step 3: Substitute the concentration expressions

Substitute the expressions \([\mathrm{X}] = \frac{z}{2}\), \([\mathrm{Y}] = \frac{z}{2}\), and \([\mathrm{Z}] = z\) into the equilibrium expression:\[ K_c = \frac{z}{\left(\frac{z}{2}\right)^2} = 10^4 \]

Step 4: Solve for z

Simplify and solve the equation:\[ \frac{z}{\left(\frac{z}{2}\right)^2} = 10^4 \]First simplify:\[ \frac{z}{\frac{z^2}{4}} = 10^4 \]Simplifying further gives:\[ \frac{4z}{z^2} = 10^4 \]Cross-multiply:\[ 4 = 10^4z \]Finally, solve for \(z\):\[ z = \frac{4}{10^4} = 4 \times 10^{-4} \, \text{mol} \, \text{L}^{-1} \]

Step 5: Verify the solution with given choices

Now, compare \(z = 4 \times 10^{-4} \, \text{mol} \, \text{L}^{-1}\) with the options:- (a) \(10^{-4} \, \text{mol} \, \text{L}^{-1}\)- (b) \(2 \times 10^{-4} \, \text{mol} \, \text{L}^{-1}\)- (c) \(2 \times 10^{4} \, \text{mol} \, \text{L}^{-1}\)- (d) \(10^{4} \, \text{mol} \, \text{L}^{-1}\)The value \(4 \times 10^{-4} \, \text{mol} \, \text{L}^{-1}\) does not exactly match any of the choices, suggesting a reevaluation of the problem statement or options might be necessary.

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. This balance results in no net change in the concentration of the reactants and products over time.
In our specific example, the reaction is: \( \mathrm{X}(\mathrm{g}) + \mathrm{Y}(\mathrm{g}) \rightleftharpoons \mathrm{Z}(\mathrm{g}) \). At equilibrium, concentrations remain constant even though reactions continue to occur.

• The concentrations of \( \mathrm{X} \), \( \mathrm{Y} \), and \( \mathrm{Z} \) maintain a fixed ratio as described by the equilibrium constant, \( K_c \).
• The equation reflects this stability: \[ K_c = \frac{[\mathrm{Z}]}{[\mathrm{X}][\mathrm{Y}]} \]

Understanding equilibrium helps predict how changes in concentration, temperature, or pressure can affect the system. It also helps in making scientific predictions about the outcomes of chemical processes.

Reaction Concentration

Reaction concentration plays a pivotal role in determining the state of equilibrium. In equilibrium scenarios, each reactant and product concentration is intimately tied to others according to the reaction's stoichiometry.
In the exercise you presented, the relationship between the reactants and product was given as:

• \([\mathrm{X}] = \frac{1}{2}[\mathrm{Y}] = \frac{1}{2}[\mathrm{Z}]\).

This means we can express the concentrations of \( \mathrm{X} \) and \( \mathrm{Y} \) in terms of \( [\mathrm{Z}] \), making it easier to substitute into the equilibrium expression.

• If \([\mathrm{Z}] = z\), then \([\mathrm{X}] = \frac{z}{2}\) and \([\mathrm{Y}] = \frac{z}{2}\).

These concentrations must satisfy the equilibrium condition, where altering any concentration will shift the equilibrium to counteract the change and restore balance.

Equilibrium Calculations

Equilibrium calculations involve using given conditions to find unknown concentration values. In this example, with a known \( K_c \) value of \( 10^4 \ \text{mol}^{-1} \ \text{L} \), we must find the value of \( [\mathrm{Z}] \) such that the equilibrium condition is satisfied.
Substituting the variable concentrations into the equation:

• \[ K_c = \frac{z}{\left(\frac{z}{2}\right)^2} = 10^4 \]

We simplify to solve for \( z \):

• \[ \frac{z}{\frac{z^2}{4}} = 10^4 \]
• \[ \frac{4z}{z^2} = 10^4 \]
• Finally, \( 4 = 10^4z \) and \( z = \frac{4}{10^4} \)
• The result is \( z = 4 \times 10^{-4} \ \text{mol} \ \text{L}^{-1} \).

Though this answer doesn't match the given options exactly, comprehension of this process enables better understanding and identification of errors or needed reevaluation.