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 value of \([\mathrm{Z}]\) at equilibrium is \(2 \times 10^{-4} \, \text{mol L}^{-1}\).

Step 1: Write the Equilibrium Expression

For the reaction \( \mathrm{X}(\mathrm{g})+\mathrm{Y}(\mathrm{g}) \rightleftharpoons \mathrm{Z}(\mathrm{g}) \), the equilibrium constant expression \( \mathrm{Kc} \) is given by: \[ \mathrm{Kc} = \frac{[\mathrm{Z}]}{[\mathrm{X}][\mathrm{Y}]} \] where \([\mathrm{Z}]\), \([\mathrm{X}]\), and \([\mathrm{Y}]\) are the molar concentrations of \(\mathrm{Z}\), \(\mathrm{X}\), and \(\mathrm{Y}\) at equilibrium, respectively.

Step 2: Substitute Given Relationships

We know that \([\mathrm{X}] = \frac{1}{2}[\mathrm{Y}] = \frac{1}{2}[\mathrm{Z}]\). Let's assume \([\mathrm{Z}] = x \). Therefore, \([\mathrm{X}] = \frac{x}{2}\) and \([\mathrm{Y}] = x\).

Step 3: Substitute in the Equilibrium Expression

Substitute the expressions for \([\mathrm{X}]\), \([\mathrm{Y}]\), and \([\mathrm{Z}]\) back into the equilibrium expression: \[ \mathrm{Kc} = \frac{x}{\left(\frac{x}{2}\right)(x)} = \frac{x}{\frac{x^2}{2}} = \frac{x}{\frac{x^2}{2}} = \frac{2}{x} \]

Step 4: Solve for \(\mathbf{x}\)

Since \( \mathrm{Kc} = 10^{4} \), set \( \frac{2}{x} = 10^{4} \) and solve for \( x \): \[ x = \frac{2}{10^{4}} = 2 \times 10^{-4} \] Thus, \([\mathrm{Z}] = x = 2 \times 10^{-4} \, \text{mol L}^{-1} \).

Key Concepts

Equilibrium Constant (Kc)

The Equilibrium Constant, denoted as \(K_c\), is a fascinating and integral concept in chemistry. Put simply, it provides a snapshot of the relation between the concentrations of reactants and products at equilibrium of a chemical reaction, at a given temperature. Unlike initial concentrations, \(K_c\) is constant for a specific reaction at a specific temperature. This means that no matter how much of the reactants or products you start with, the ratio defined by \(K_c\) will remain the same once equilibrium is reached.
For a general reaction, \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant expression is given by the formula:

• \[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]

In this equation, the square brackets denote the concentrations of the substances, raised to the power of their respective stoichiometric coefficients in the balanced chemical equation. In our specific example, the reaction is \(X(g)+Y(g) \rightleftharpoons Z(g)\), hence the \(K_c\) expression will be \( K_c = \frac{[Z]}{[X][Y]} \).
The significance of \(K_c\) lies in its ability to allow chemists to predict the direction in which a reaction will proceed under different conditions. If \(K_c\) is greater than one, it implies a product-favored reaction; if less than one, it's reactant-favored. In our case, with \(K_c\) of \(10^4\), the reaction strongly favors the formation of product \(Z\).

Reaction Stoichiometry

Understanding Reaction Stoichiometry is essential for solving equilibrium problems, as it determines how the concentrations of reactants and products relate. Stoichiometry in our context refers to the quantitative relationship between reactants and products in a chemical reaction based on their balanced chemical equation.
In our given reaction \(X(g)+Y(g) \rightleftharpoons Z(g)\), the stoichiometry dictates that one mole of \(X\) reacts with one mole of \(Y\) to produce one mole of \(Z\). This 1:1:1 ratio is key to figuring out the equilibrium concentrations from known quantities. The equality in moles means changes in reaction quantities result uniformly in all species involved, allowing us to derive the expressions needed in calculations.
In practice, stoichiometry can be viewed as conversion factors in calculations, making it pivotal when we apply given concentration relations, such as \([X] = \frac{1}{2}[Y] = \frac{1}{2}[Z]\). This tells us directly how to adjust concentrations as the reaction progresses to equilibrium, revealing the interconnectedness of all components due to their stoichiometric coefficients.

Equilibrium Concentration Calculation

Equilibrium Concentration Calculation involves computing the precise amounts of each reactant and product at equilibrium. Begin by expressing known relationships and applying them to the equilibrium constant equation.
With our provided condition \([X] = \frac{1}{2}[Y] = \frac{1}{2}[Z]\), let's assume \([Z] = x\). This means \([X] = \frac{x}{2}\) and \([Y] = x\).
Plugging these into the equilibrium expression:

• \[ K_c = \frac{x}{\left(\frac{x}{2}\right) \cdot x} = \frac{2}{x} \]

To solve for \(x\), set \(\frac{2}{x} = 10^4\), as given by the problem. This equation resolves to:

• \[ x = \frac{2}{10^4} = 2 \times 10^{-4} \]

Hence, the equilibrium concentration of \([Z]\) is \(2 \times 10^{-4} \text{ mol L}^{-1}\).
This step-by-step approach using stoichiometric relationships and the \(K_c\) formula enables accurate calculation of equilibrium concentrations, demonstrating the practical application of these core chemical principles.