Problem 71 - Chemistry The Central Science

Question

A mixture of \(\mathrm{CH}_{4}\) and \(\mathrm{H}_{2} \mathrm{O}\) is passed over a nickel catalyst at \(1000 \mathrm{~K}\). The emerging gas is collected in a \(5.00-\mathrm{L}\) flask and is found to contain \(8.62 \mathrm{~g}\) of \(\mathrm{CO}, 2.60 \mathrm{~g}\) of \(\mathrm{H}_{2}, 43.0 \mathrm{~g}\) of \(\mathrm{CH}_{4},\) and \(48.4 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) Assuming that equilibrium has been reached, calculate \(K_{c}\) and \(K_{p}\) for the reaction.

Step-by-Step Solution

Answer

The equilibrium constants for the reaction \(CO + 3H_2 \rightleftharpoons CH_4 + H_2O\) at 1000 K are \(K_c = 3.90\) and \(K_p = 5.07\).

Step 1: Write the balanced chemical equation

The balanced chemical equation is: \(CO + 3H_2 \rightleftharpoons CH_4 + H_2O\)

Step 2: Calculate moles of each species at equilibrium

To find moles, we will use the formula: moles = mass / molar mass For CO: Moles of CO = \(\frac{8.62\, g}{28.01\, g/mol} = 0.3077\, mol\) For H2: Moles of H2 = \(\frac{2.60\, g}{2.02\, g/mol} = 1.287\, mol\) For CH4: Moles of CH4 = \(\frac{43.0\, g}{16.04\, g/mol} = 2.68\, mol\) For H2O: Moles of H2O = \(\frac{48.4\, g}{18.02\, g/mol} = 2.685\, mol\)

Step 3: Convert moles to concentrations

To find the concentrations, we will use the formula: concentration = moles / volume Volume = \(5.00\, L\) For CO: Concentration of CO = \(\frac{0.3077\, mol}{5.00\, L} = 0.06154\, M\) For H2: Concentration of H2 = \(\frac{1.287\, mol}{5.00\, L} = 0.2574\, M\) For CH4: Concentration of CH4 = \(\frac{2.68\, mol}{5.00\, L} = 0.536\, M\) For H2O: Concentration of H2O = \(\frac{2.685\, mol}{5.00\, L} = 0.537\, M\)

Step 4: Calculate the equilibrium constant Kc

Now plug the concentrations into the equilibrium expression for Kc: \(K_c = \frac{[CH_4][H_2O]}{[CO][H_2]^3}\) \(K_c = \frac{(0.536\, M)(0.537\, M)}{(0.06154\, M)(0.2574\, M)^3} = 3.90\)

Step 5: Calculate the equilibrium constant Kp

To find Kp from Kc, use the relationship: \(Kp = Kc \times (RT)^{\Delta n}\) Here, R is the ideal gas constant (0.08206 L atm/mol K), T is the temperature in Kelvin (1000 K), and Δn is the change in the number of moles of gas in the balanced equation. \(\Delta n = (1 + 1) - (1 + 3) = -2\) Now, plug in the values to find Kp: \(K_p = (3.90) \times (0.08206\, L\, atm/mol\, K \times 1000\, K)^{-2} = 5.07\) So, the equilibrium constants for the reaction are \(K_c = 3.90\) and \(K_p = 5.07\).

Key Concepts

Equilibrium Constant Derivation

Understanding chemical equilibrium is essential for many aspects of chemistry, and at the heart of this concept is the equilibrium constant. The equilibrium constant, denoted as K, is a measure of the extent of a reaction at equilibrium. It is determined by the relative concentrations of the reactants and products involved in the reaction.

Deriving the equilibrium constant involves writing an expression based on the balanced chemical equation. For a general reaction where aA + bB ↔ cC + dD, the equilibrium constant expression is:

\(K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}\)

Here, [A], [B], [C], and [D] represent the molar concentrations of the reactants and products, while a, b, c, and d are their respective stoichiometric coefficients. In the case of a gas-phase reaction, the equilibrium constant can also be expressed in terms of partial pressures, leading to the equilibrium constant for pressure, Kp. The relationship between Kc and Kp is affected by the change in the number of moles of gas, symbolized by Δn, through the equation:

\(K_p = K_c \times (RT)^{\Delta n}\)

This equation links the dimensionless concentration-based constant to the pressure-based constant and includes the ideal gas constant R and the temperature T in Kelvin.

Mole-to-Mass Conversion

To grasp chemical reactions quantitatively, converting between moles and mass is a crucial skill. To do this, we apply the formula: number of moles = mass (in grams) / molar mass (in g/mol).

For example, if we know a compound's mass and its molar mass, we can calculate the number of moles present using this formula. As seen in the exercise solution, the mass of CO is divided by the molar mass of CO to obtain the moles of CO. This is a fundamental step in determining the concentrations required for the equilibrium constant expression. Let's recap with an example from the text:

Moles of CO = \(\frac{8.62\, g}{28.01\, g/mol} = 0.3077\, mol\)

Mole-to-Concentration Conversion

After determining the number of moles of each reactant and product in a reaction at equilibrium, we then convert these to concentrations. The concentration is calculated by dividing the number of moles by the volume of the reaction mixture. The formula is concentration = moles / volume.

In a reaction vessel of a known volume, this conversion allows us to find the molarity (M), which is the moles of solute per liter of solution. Molarity is the standard unit for concentration in equilibrium expressions. For instance, if we have 0.3077 moles of CO in a 5.00 L flask, the concentration of CO is:

Concentration of CO = \(\frac{0.3077\, mol}{5.00\, L} = 0.06154\, M\)

Le Châtelier's Principle

Le Châtelier's principle is a cornerstone of chemical equilibrium that helps us predict how a system at equilibrium responds to external changes. According to this principle, if a dynamic equilibrium is disturbed by changing the conditions, such as concentration, temperature, or pressure, the position of equilibrium moves to counteract the change.

For example, if the concentration of a reactant is increased, the system responds by consuming the additional reactant, shifting the equilibrium toward the products. Conversely, if the temperature is increased for an exothermic reaction, the equilibrium shifts toward the reactants to absorb excess thermal energy. Le Châtelier's principle allows chemists to optimize reactions ensuring maximum yield under varying conditions. This principle also implies that changing the conditions can affect the value of K, as equilibrium constants are dependent on temperature.