A gaseous mixture contains \(0.30\) moles \(\mathrm{CO}, 0.10 \mathrm{moles} \mathrm{H}_{2}\), and \(0.03\) moles \(\mathrm{H}_{2} \mathrm{O}\) vapour and an unknown amount of \(\mathrm{CH}_{4}\) per litre. This mixture is at equilibrium at \(1200 \mathrm{~K}\). \(\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) \(K_{\mathrm{C}}=3.9\) What is the concentration of \(\mathrm{CH}_{4}\) in this mixture? The equilibrium constant \(K_{\mathrm{c}}\) equals \(3.92\). (a) \(0.39 \mathrm{M}\) (b) \(0.039 \mathrm{M}\) (c) \(0.78 \mathrm{M}\) (d) \(0.078 \mathrm{M}\)

Short Answer

Expert verified
The concentration of methane (\(\mathrm{CH}_{4}\)) is \(0.039 \mathrm{M}\), which corresponds to option (b).

Step by step solution

01

Identify the Equilibrium Equation

Start by writing down the equilibrium equation: \( \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \).
02

Write the Expression for the Equilibrium Constant

The equilibrium constant expression for the reaction is given by \( K_{\mathrm{C}} = \frac{[\mathrm{CH}_{4}][\mathrm{H}_{2}O]}{[\mathrm{CO}][\mathrm{H}_{2}]^3} \). Note that the brackets represent the concentration of each species in moles per liter (M).
03

Input the Known Values

Substitute the known concentrations into the equilibrium constant expression, leaving \( [\mathrm{CH}_{4}] \) as the unknown: \( 3.92 = \frac{[\mathrm{CH}_{4}](0.03)}{(0.30)(0.10)^3} \).
04

Solve for the Unknown Concentration of CH4

Solve the equation for \( [\mathrm{CH}_{4}] \): \( [\mathrm{CH}_{4}] = \frac{3.92 \times (0.30)(0.10)^3}{0.03} \).
05

Calculate the Concentration of CH4

Perform the arithmetic operation to find \( [\mathrm{CH}_{4}] \).
06

Verify the Answer

Check the calculated concentration value against the given options to select the correct one.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
Understanding chemical equilibrium is essential for grasping the behavior of reactions that can proceed in both forward and reverse directions. In this state, the rate of the forward reaction is equal to the rate of the reverse reaction, resulting in a constant concentration of reactants and products. However, this doesn't imply the quantities of reactants and products are equal; it only indicates that their concentrations do not change over time. In the given exercise, the system reaches equilibrium at 1200 K for the reaction between carbon monoxide (CO) and hydrogen gas (H2) to form methane (CH4) and water vapor (H2O). At equilibrium, there is no net change in the concentrations of these substances, even though the reactions are still occurring on the molecular level.
Equilibrium Constant Expression
The equilibrium constant expression is a mathematical way of representing the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their stoichiometric coefficients. The equilibrium constant, denoted as Kc for reactions in solution or Kp for reactions involving gases and pressures, provides a quantitative measure of the extent of the reaction.

For the reaction in our exercise, the equilibrium constant expression is given by: \[ K_{\mathrm{C}} = \frac{[\mathrm{CH}_{4}][\mathrm{H}_{2}O]}{[\mathrm{CO}][\mathrm{H}_{2}]^3} \].

By knowing the value of Kc and the concentrations of CO, H2, and H2O, we can solve for the unknown concentration of CH4 at equilibrium.
Mole Concept in Chemistry
The mole concept is a foundation in chemistry that relates the mass of a substance to its elemental properties. One mole of any substance contains Avogadro's number (approximately 6.022 x 10^23) of particles, which may be atoms, molecules, ions, or electrons. In the context of the equilibrium problem, the mole concept allows us to relate the given quantities (in moles) to the concentrations of gases (in moles per liter) needed for the equilibrium expression. It is critical to recognize that reaction stoichiometry and the mole concept are intertwined to determine the right proportions of reactants and products involved in the reaction.
Gas Mixtures Equilibrium
When dealing with gas mixtures at equilibrium, it is important to consider the partial pressures of each gas, the total pressure, and volume of the system, as well as the reaction conditions like temperature. These factors collectively affect the position of equilibrium. For reactions in a mixture of gases, the equilibrium constant may also be expressed in terms of partial pressures (Kp). However, for this particular exercise, because we are given moles per liter and Kc, we focus on concentrations in the equilibrium expression. The mole concept plays a crucial role in determining the molar composition of the gas mixture, which helps to calculate the equilibrium constant and solve equilibrium problems involving gas mixtures.

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Most popular questions from this chapter

\(\mathrm{XeF}_{6}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{XeOF}_{4}+2 \mathrm{HF}\) equilibrium constant \(=K_{1}\). \(\mathrm{XeO}_{4}+\mathrm{XeF}_{6} \rightleftharpoons \mathrm{XeOF}_{4}+\mathrm{XeO}_{3} \mathrm{~F}_{2}\) equilibrium constant \(=K_{2} .\) Then equilibrium constant for the following reaction will be: \(\mathrm{XeO}_{4}+2 \mathrm{HF} \rightleftharpoons \mathrm{XeO}_{3} \mathrm{~F}_{2}+\mathrm{H}_{2} \mathrm{O}\) (a) \(\frac{K_{1}}{K_{2}}\) (b) \(K_{1}+K_{2}\) (c) \(\frac{K_{2}}{K_{1}}\) (d) \(K_{2}-K_{1}\)

For a gaseous equilibrium: \(2 \mathrm{~A}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g}), K_{\mathrm{p}}\) has a value \(1.8\) at \(700 \mathrm{~K}\). The value of \(K_{c}\) for the equilibrium: \(2 \mathrm{~B}(\mathrm{~g})\) \(+\mathrm{C}(\mathrm{g}) \rightleftharpoons 2 \mathrm{~A}(\mathrm{~g})\) at that temperature is about (a) \(0.031\) (b) 32 (c) \(57.4\) (d) \(103.3\)

The equilibrium: \(\mathrm{SOCl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})\) \(+\mathrm{Cl}_{2}(\mathrm{~g})\) is attained at \(25^{\circ} \mathrm{C}\) in a closed container and helium gas is introduced. Which of the following statements is correct? (a) concentration of \(\mathrm{SO}_{2}\) is increased (b) more \(\mathrm{Cl}_{2}\) is formed (c) concentrations of all change (d) concentrations will not change

In a closed container maintained at 1 atm pressure and \(25^{\circ} \mathrm{C}, 2\) moles of \(\mathrm{SO}_{2}(\mathrm{~g})\) and 1 mole of \(\mathrm{O}_{2}(\mathrm{~g})\) were allowed to react to form \(\mathrm{SO}_{3}(\mathrm{~g})\) under the influence of a catalyst. \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) At equilibrium, it was found that \(50 \%\) of \(\mathrm{SO}_{2}(\mathrm{~g})\) was converted to \(\mathrm{SO}_{3}(\mathrm{~g})\). The partial pressure of \(\mathrm{O}_{2}(\mathrm{~g})\) at equilibrium will be (a) \(0.17\) atm (b) \(0.5 \mathrm{~atm}\) (c) \(0.33\) atm (d) \(0.20 \mathrm{~atm}\)

A quantity of \(25 \mathrm{~g}\) sample of \(\mathrm{BaO}_{2}\) is heated to \(954 \mathrm{~K}\) in a closed and rigid evacuated vessel of \(8.21\) L capacity. What percentage of peroxide is converted into oxide? \(2 \mathrm{BaO}_{2}(\mathrm{~s}) \rightleftharpoons 2 \mathrm{BaO}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g})\) \(K_{\mathrm{F}}=0.5 \mathrm{~atm}\) (a) \(20 \%\) (b) \(50 \%\) (c) \(75 \%\) (d) \(80 \%\)

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