Problem 85 - A Complete Resource Book in Chemistry for JEE Main

Question

Equilibrium constant for the reaction \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})\) is \(81 .\) If the velocity constant of the forward reaction is \(162 \mathrm{~L}\) \(\mathrm{mol}^{-1} \mathrm{sec}^{-1}\), what is the velocity constant (in \(\mathrm{L} \mathrm{mol}^{-1}\) \(\mathrm{sec}^{-1}\) ) for the backward reaction? (a) 13122 (b) 2 (c) 261 (d) 243

Step-by-Step Solution

Answer

The velocity constant for the backward reaction is 2 L mol⁻¹ sec⁻¹ (Option b).

Step 1: Understand the Relationship

The equilibrium constant \( K_c \) for a reaction is defined as the ratio of the velocity constant of the forward reaction \( k_{f} \) to the velocity constant of the backward reaction \( k_{b} \). Mathematically, this is represented as: \[ K_c = \frac{k_f}{k_b} \] Given that the equilibrium constant \( K_c = 81 \) and the forward reaction velocity constant \( k_f = 162 \), we can use this relationship to find \( k_b \).

Step 2: Solve for the Backward Reaction Constant

To find the velocity constant of the backward reaction \( k_b \), rearrange the formula to solve for \( k_b \): \[ k_b = \frac{k_f}{K_c} \] Substitute the known values into the equation: \[ k_b = \frac{162}{81} \] Simplify the expression to get the answer.

Step 3: Perform the Calculation

Perform the division \( \frac{162}{81} \) to obtain \( k_b \): \[ k_b = 2 \] Therefore, the velocity constant of the backward reaction is \( 2 \ \mathrm{L} \ \mathrm{mol}^{-1} \ \mathrm{sec}^{-1} \).

Key Concepts

Velocity Constant

In chemistry, the velocity constant, also known as the rate constant, is an essential parameter in reaction kinetics. It provides insights into how quickly a reaction proceeds. Each reaction has its own value, depending on various factors including temperature, pressure, and the presence of catalysts. The velocity constant is represented by the symbol \( k \) and is utilized in the rate laws of chemical reactions. For instance, in the reaction given in the exercise, the velocity constant for the forward reaction is 162 L mol\(^{-1} \) sec\(^{-1} \). Understanding the velocity constant is crucial for determining how efficiently reactions move towards completion.

• The velocity constant is central to calculating reaction speed.
• It helps relate the concentrations of reactants to the rate of reaction.
• The units of a velocity constant vary based on the reaction order.

The velocity constant is unique for every reaction, so knowing how to measure and apply it is vital for any chemist aiming to understand chemical dynamics deeply.

Forward Reaction

A forward reaction describes the process where reactants are transformed into products. In the context of equilibrium, the forward reaction competes in parallel with the backward reaction. For the chemical equation \(\mathrm{H}_{2}\mathrm{O}(\text{g})+\mathrm{CO}(\text{g})\rightarrow\mathrm{H}_{2}(\text{g})+\mathrm{CO}_{2}(\text{g})\), the forward reaction involves the conversion of water vapor and carbon monoxide into hydrogen gas and carbon dioxide. The speed of the forward reaction is given by its velocity constant \( k_f \).

• The forward reaction is essential in establishing the route of product formation.
• It determines how product concentrations increase over time.
• Conditions such as temperature and pressure impact the rate of the forward reaction.

Recognizing the characteristics of a forward reaction enables scientists to predict how a system reaches equilibrium and how long it will remain there.

Backward Reaction

The backward reaction is the reverse of the forward reaction. In an equilibrium scenario, it involves the transformation of products back into reactants. For the given exercise, the backward reaction can be described as the formation of water vapor and CO from hydrogen and CO\(_2\). The velocity constant for the backward reaction, \( k_b \), showcases the rate at which this reverse process occurs.

• Backward reactions compete with forward reactions in reaching equilibrium.
• They are significant in understanding the full dynamics of chemical equilibrium.
• The equilibrium constant, \( K_c \), is derived from the ratio of velocity constants \( k_f \) and \( k_b \).

Analyzing backward reactions is central to chemical kinetics as it illustrates not only the sustainability of products but also how quickly a reaction might revert to its starting reactants. This knowledge is vital for controlling product yields in industrial and laboratory settings.