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 \( \mathrm{B} \) is 0.8 mol/L, which does not match the provided options.

Step-by-step solution

Set Up the Initial Conditions

Initially, you have 1 mole of \( \mathrm{A(g)} \) in a 1-litre vessel, so its concentration is 1 mol/L. \( \mathrm{B(g)} \) starts with a concentration of 0 mol/L because none of it is present initially.

Define the Change at Equilibrium

As the reaction reaches equilibrium, let \( x \) be the amount of \( \mathrm{A(g)} \) that reacts to form \( \mathrm{B(g)} \). Therefore, the concentration of \( \mathrm{B(g)} \) at equilibrium will be \( x \), and the concentration of \( \mathrm{A(g)} \) will be \( 1 - x \).

Write the Expression for the Equilibrium Constant

The equilibrium constant \( K_c \) for the reaction \( \mathrm{A(g)} \rightleftharpoons \mathrm{B(g)} \) is given by:\[ K_c = \frac{[B]}{[A]} \]We know \( K_c = 4 \), \([B] = x\), and \([A] = 1 - x\).

Solve for x

Substituting into the equilibrium constant expression, we have:\[ 4 = \frac{x}{1-x} \]Solving for \( x \),\[ 4(1 - x) = x \]\[ 4 - 4x = x \]\[ 4 = 5x \]\[ x = \frac{4}{5} \]\[ x = 0.8 \]

Check the Options

The concentration of \( B \) at equilibrium is 0.8 mol/L. Let's verify the available options:(a) 0.2(b) 0.6(c) 0.75(d) 0.1None of these match 0.8 mol/L. The given answers seem incorrect based on our calculation.

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \), is a crucial concept in chemical equilibrium that helps us understand how a reversible reaction behaves at a given temperature. It is essentially a ratio that reflects the concentrations of products to reactants when the system has reached equilibrium. For the reaction \( \mathrm{A(g)} \rightleftharpoons \mathrm{B(g)} \), \( K_c \) is expressed as: \[ K_c = \frac{[\mathrm{B}]}{[\mathrm{A}]} \]This formula shows that the constant is a comparison between the concentration of \( \mathrm{B} \) to that of \( \mathrm{A} \). If \( K_c \) is greater than 1, it indicates that, at equilibrium, the reaction favors the formation of products. If it’s less than 1, reactants are favored. Here, having a \( K_c \) value of 4 suggests a significant formation of product \( \mathrm{B} \) at equilibrium.Understanding equilibrium constants helps chemists predict the composition of a reaction mixture at various stages and consider how changes to the system might affect equilibrium. It's a valuable tool in designing reactions and optimizing conditions for desired outcomes.

Concentration Calculations

To fully grasp how to determine concentrations at equilibrium, one must understand how to calculate initial concentrations and changes during the reaction. Initially, in our problem, there is 1 mole of \( \mathrm{A(g)} \) in a 1-liter container, giving it a concentration of 1 mol/L.When equilibrium is attained, a certain amount of \( \mathrm{A(g)} \) has been converted into \( \mathrm{B(g)} \). If we let \( x \) represent the concentration of \( \mathrm{A(g)} \) that has changed, then the concentration of \( \mathrm{B(g)} \) at equilibrium will also be \( x \), while \( \mathrm{A(g)} \) will be \( 1-x \). These calculations provide a snapshot of the reaction’s current state.In practice, this means setting up a simple equation where we balance out what changes occur in reactants and products, using known values like \( K_c \). This helps illuminate how the system has shifted from its initial state to equilibrium and is useful for predicting concentrations in similar chemical systems.

Reaction Stoichiometry

Reaction stoichiometry involves using coefficients from the balanced chemical equation to understand how much reactants turn into products. For the equilibrium reaction \( \mathrm{A(g)} \rightleftharpoons \mathrm{B(g)} \), there is a 1:1 stoichiometric relationship. This implies that for every mole of \( \mathrm{A} \) that reacts, one mole of \( \mathrm{B} \) is produced.This proportionality is crucial for setting up calculations for changes in concentration, as it simplifies understanding of how the system progresses to equilibrium. Knowing that \( 1 \) mole of \( \mathrm{A} \) creates \( 1 \) mole of \( \mathrm{B} \) when equilibrium is established means that our changes in concentration can be expressed directly and simply as \( x \) and \( 1-x \).Using stoichiometry, we are able to link the microscopic steps that molecules undergo during reactions to the macroscopic changes we observe, such as shifts in concentration. It’s not just fundamental for calculating outcomes, but also a backbone for predicting how changing conditions might influence a reaction sequence in industrial or laboratory settings.