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) \(0.1\)

Short answer

The concentration of \( \mathrm{B} \) at equilibrium is 0.8 mol L^{-1}.

Step-by-step solution

Setting up the Initial Concentration

Initially, we have one mole of \( \mathrm{A(g)} \) in a one-liter vessel, so the initial concentration of \( \mathrm{A} \) is 1 M. \( \mathrm{B} \) is not formed yet, so its initial concentration is 0 M.

Define Change in Concentration with Variable x

Let \( x \) be the change in concentration of \( \mathrm{A(g)} \) and \( \mathrm{B(g)} \) at equilibrium. When \( \mathrm{A} \) converts to \( \mathrm{B} \), \( \mathrm{A} \) decreases by \( x \) and \( \mathrm{B} \) increases by \( x \).

Express Concentrations at Equilibrium

At equilibrium, the concentration of \( \mathrm{A(g)} \) is \( 1-x \) and the concentration of \( \mathrm{B(g)} \) is \( x \).

Write the Expression for Equilibrium Constant

For the equilibrium \( \mathrm{A(g)} \rightleftharpoons \mathrm{B(g)} \), the equilibrium constant \( K_c \) is given by:\[ K_c = \frac{[\mathrm{B}]}{[\mathrm{A}]} = \frac{x}{1-x} \]

Solve the Equilibrium Expression

Given \( K_c = 4 \), substitute this into the expression:\[ 4 = \frac{x}{1-x} \]Cross-multiply to solve for \( x \):\[ 4(1-x) = x \]\[ 4 - 4x = x \]\[ 4 = 5x \]\[ x = \frac{4}{5} = 0.8 \]

Determine the Concentration of B

The concentration of \( \mathrm{B} \) at equilibrium is \( x = 0.8 \).

Key concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K_c \) for reactions involving concentrations, is a key concept in understanding the dynamics of a chemical equilibrium. It quantifies the ratio of concentrations of products to reactants at equilibrium, providing insight into the position of equilibrium. For our reaction \( \mathrm{A(g)} \rightleftharpoons \mathrm{B(g)} \), the equilibrium constant is expressed as:\[K_c = \frac{[\mathrm{B}]}{[\mathrm{A}]}\]where \([\mathrm{B}]\) and \([\mathrm{A}]\) are the concentrations of products and reactants respectively.

• The value of \( K_c \) indicates how far the reaction proceeds before reaching equilibrium. A large \( K_c \) (\( > 1 \)) suggests a product-favored reaction, where equilibrium lies towards the right, while a small \( K_c \) (\( < 1 \)) indicates a reactant-favored equilibrium.
• In our specific example, \( K_c = 4 \), which implies that at equilibrium, the product \( \mathrm{B(g)} \) is present in a greater concentration than the reactant \( \mathrm{A(g)} \).

Understanding \( K_c \) helps predict how changes in conditions such as concentration or temperature will impact the system, an essential factor in optimizing chemical processes in industries.

Concentration Calculations

Concentration calculations play a pivotal role in predicting the quantities of substances present in a system at equilibrium. In our exercise, here's how the concentrations change:
Initially, we start with 1 mole of \( \mathrm{A(g)} \) in one liter, hence the initial concentration of \( \mathrm{A} \) is 1 M.

• As the reaction progresses towards equilibrium, \( \mathrm{A} \) decreases by an amount \( x \) and \( \mathrm{B} \) increases by the same amount \( x \).
• At equilibrium, we write the concentrations as follows: \( [\mathrm{A}] = 1-x \) and \( [\mathrm{B}] = x \).
• Using the equilibrium constant expression, \( K_c = \frac{x}{1-x} \), and knowing \( K_c = 4 \), we solve for \( x \) by rearranging and simplifying the equation to find \( x = 0.8 \).

The calculation of \( x \) indicates that the concentration of \( \mathrm{B(g)} \) at equilibrium is 0.8 mol/L, highlighting the significance of concentration changes in determining system dynamics at equilibrium.

Chemical Reaction Dynamics

Chemical reaction dynamics involves studying how changes in concentration, temperature, or pressure affect the rates and outcomes of reactions. In the context of chemical equilibrium, these dynamics describe the reversible nature of reactions and how they adjust back and forth until equilibrium is achieved.At equilibrium, the rate of the forward reaction (\( \mathrm{A} \rightarrow \mathrm{B} \)) equals the rate of the reverse reaction (\( \mathrm{B} \rightarrow \mathrm{A} \)). This idea is fundamental as it explains why concentrations remain constant at equilibrium rather than equal. The dynamic is always active, with ongoing molecular interactions, despite the apparent static nature when viewed on a macroscopic scale.

• Le Chatelier's Principle plays a critical role here, predicting how a system at equilibrium responds to external changes. For instance, an increase in temperature or pressure could shift the equilibrium position.
• Understanding these dynamics allows chemists to manipulate the conditions to 'push' a reaction towards more product or reactant formation, optimizing yield and efficiency in industrial settings.

By exploring these dynamics thoroughly, students can gain a better appreciation of the delicate balance within chemical reactions and the practicality of controlling them.