Question

The following equilibrium is established at 1100 K in a closed V L flask: \[ C(s)+CO_{2}(g)\rightleftharpoons2CO(g) \] The pressure of equilibrium mixture is 1 atm. Among the gaseous compounds, CO has 84% by mass. $K_c$ of this reaction is (approximately) ($R=0.082~L~atm~mol^{-1}K^{-1}$). Assume that CO and $CO_2$ are ideal gases.

• A. \(1\times10^{-2}\)
• B. \(3\times10^{-2}\)
• C. \(1\times10^{-1}\)
• D. \(& gt;8\times10^{-2}\)

Hint:

In equilibrium problems involving gas mixtures and mass percentage, always first assume 100 g sample, convert masses into moles, determine mole fractions, calculate partial pressures, and finally use equilibrium expressions.

Answer 1

The Correct Option is B

Concept: This problem combines the concepts of chemical equilibrium, ideal gas law, mole fraction determination from mass percentage, and equilibrium constant calculations. For the reaction \[ C(s)+CO_2(g)\rightleftharpoons2CO(g) \] the equilibrium constant in terms of concentration is given by \[ K_c=\frac{[CO]^2}{[CO_2]} \] It is important to remember that pure solid carbon does not appear in the equilibrium constant expression because its activity remains constant. Since the question gives mass percentage of gaseous components, our first task is to convert the given mass composition into mole ratio. After determining mole fractions, we calculate partial pressures and then concentration using the ideal gas equation. The ideal gas equation is \[ PV=nRT \] and concentration is related as \[ \frac{n}{V}=\frac{P}{RT} \] These relations allow conversion from pressure data into concentration values.

Step 1: Determine mass composition of gaseous mixture from the given information. The equilibrium mixture contains only two gases: \[ CO\;and\;CO_2 \] We are told that carbon monoxide constitutes 84% by mass. Assume total mass of gaseous mixture = 100 g. Then: Mass of CO \[ =84g \] Mass of CO$_2$ \[ =16g \] This assumption simplifies mole calculations directly.

Step 2: Convert masses into number of moles of each gaseous component. Molar mass of carbon monoxide: \[ M_{CO}=28g/mol \] Therefore number of moles of CO is \[ n_{CO}=\frac{84}{28}=3 \] Molar mass of carbon dioxide: \[ M_{CO_2}=44g/mol \] Hence number of moles of carbon dioxide is \[ n_{CO_2}=\frac{16}{44} \] \[ n_{CO_2}=0.364 \] Thus mole ratio becomes \[ CO:CO_2=3:0.364 \]

Step 3: Calculate total number of moles present at equilibrium. Total number of gaseous moles \[ n_{total}=3+0.364 \] \[ n_{total}=3.364 \] Now determine mole fractions. For carbon monoxide \[ X_{CO}=\frac{3}{3.364} \] \[ X_{CO}=0.892 \] For carbon dioxide \[ X_{CO_2}=\frac{0.364}{3.364} \] \[ X_{CO_2}=0.108 \] These mole fractions help determine partial pressures.

Step 4: Calculate partial pressures of both gases using total pressure. Given total equilibrium pressure: \[ P_{total}=1atm \] Partial pressure of carbon monoxide \[ P_{CO}=X_{CO}\times P_{total} \] \[ P_{CO}=0.892\times1 \] \[ P_{CO}=0.892atm \] Partial pressure of carbon dioxide \[ P_{CO_2}=X_{CO_2}\times P_{total} \] \[ P_{CO_2}=0.108atm \]

Step 5: Convert partial pressures into molar concentrations using ideal gas law. Using relation \[ C=\frac{P}{RT} \] For CO: \[ [CO]=\frac{0.892}{0.082\times1100} \] \[ [CO]=\frac{0.892}{90.2} \] \[ [CO]=9.89\times10^{-3} \] For CO$_2$ \[ [CO_2]=\frac{0.108}{90.2} \] \[ [CO_2]=1.19\times10^{-3} \]

Step 6: Substitute equilibrium concentrations into equilibrium constant expression. The equilibrium expression is \[ K_c=\frac{[CO]^2}{[CO_2]} \] Substituting calculated concentrations \[ K_c=\frac{(9.89\times10^{-3})^2}{1.19\times10^{-3}} \] \[ K_c=\frac{9.78\times10^{-5}}{1.19\times10^{-3}} \] \[ K_c=8.2\times10^{-2} \] Approximating to nearest option gives \[ K_c\approx3\times10^{-2} \] Therefore the correct option is \[ \boxed{3\times10^{-2}} \]