Question

Given \(\left[\mathrm{CS}_{2}\right]=0.120 \mathrm{M},\left[\mathrm{H}_{2}\right]=0.10,\left[\mathrm{H}_{2} \mathrm{~S}\right]=0.20\) and \(\left[\mathrm{CH}_{4}\right]=8.40 \times 10^{-5} M\) for the following reaction at \(900^{\circ} \mathrm{C}\), at eq. Calculate the equilibrium constant \(\left(K_{c}\right)\). $$ \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) $$ (a) \(0.0120\) (b) \(0.0980\) (c) \(0.280\) (d) \(0.120\)

Short answer

0.280

Step-by-step solution

Write the equilibrium expression

For the reaction \[\[\begin{align*}\text{CS}_{2}(g) + 4 \text{H}_{2}(g) &\rightleftharpoons \text{CH}_{4}(g) + 2 \text{H}_{2} \text{S}(g)\end{align*}\]\]the equilibrium constant expression, based on the law of mass action, is $$K_{c} = \frac{[\text{CH}_{4}][\text{H}_{2}S]^{2}}{[\text{CS}_{2}][\text{H}_{2}]^{4}}$$

Plug in the equilibrium concentrations

Substitute the given equilibrium concentrations into the expression to find the value of the equilibrium constant. $$K_{c} = \frac{(8.40 \times 10^{-5} M)(0.20 M)^{2}}{(0.120 M)(0.10 M)^{4}}$$

Calculate the equilibrium constant

Perform the calculations, being careful to adhere to the correct order of operations and to manage the exponents properly. $$K_{c} = \frac{(8.40 \times 10^{-5})(0.04)}{(0.120)(0.0001)}$$ $$K_{c} = \frac{3.36 \times 10^{-6}}{1.2 \times 10^{-5}}$$ $$K_{c} = 0.280$$

Key concepts

Chemical Equilibrium

The concept of chemical equilibrium is crucial when studying chemical reactions. It refers to the state in which the concentration of reactants and products in a chemical reaction do not change anymore, despite the reaction continuing at a molecular level. In other words, the rate at which the reactants transform into products is equal to the rate at which the products revert back to reactants.

This does not mean the reactants and products are present in equal amounts; instead, it represents a balance between the forward and reverse reactions. The point of equilibrium is critical because many chemical reactions tend to reach such a state under a set of given conditions, including temperature and pressure. For students preparing for examinations like the JEE in Physical Chemistry, understanding this dynamic yet static state can be a game changer for both theoretical understanding and practical applications.

Law of Mass Action

The Law of Mass Action is a principle that comes into play when we discuss chemical equilibrium. It states that for a reversible reaction at equilibrium and a constant temperature, the rate of the reaction is directly proportional to the product of the concentrations of the reactants, each raised to a power equal to their stoichiometric coefficients in the balanced chemical equation.

To apply this law quantitatively, we use what is known as an equilibrium constant, denoted as \( K_c \) for concentrations or \( K_p \) for partial pressures. The law is helpful in determining the direction and extent of a chemical reaction and therefore is a staple in problems under JEE Physical Chemistry. By understanding how this law relates the concentrations of chemicals in a reaction to the equilibrium constant, students can predict the behavior of a system under various conditions, which is essential for solving numerical problems effectively.

JEE Physical Chemistry

JEE Physical Chemistry is one of the three major subject areas covered in the Joint Entrance Examination (JEE), an engineering entrance assessment in India. It involves studying the physical properties and transformations of chemical substances. In this context, concepts like chemical equilibrium and the Law of Mass Action are of paramount importance.

These topics not only form the core of numerous problems in JEE Physical Chemistry but also help students develop a profound understanding of how chemical reactions occur and reach equilibrium. To master these concepts, students should focus on solving a variety of problems, such as calculating equilibrium constants or predicting the direction of a reaction shift when external conditions change. Regular practice with these concepts will build the proficiency needed for solving complex problems and achieving success in both the JEE and future scientific endeavors.