Problem 19 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

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) \(0.078 \mathrm{M}\)

Step-by-Step Solution

Answer

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

Step 1: Identify the Equilibrium Equation

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

Step 2: 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).

Step 3: Input the Known Values

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

Step 4: Solve for the Unknown Concentration of CH4

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

Step 5: Calculate the Concentration of CH4

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

Step 6: Verify the Answer

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

Key Concepts

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.