Question

A mixture of \(0.2000 \mathrm{~mol}\) of \(\mathrm{CO}_{2}, 0.1000 \mathrm{~mol}\) of \(\mathrm{H}_{2}\), and \(0.1600 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}\) is placed in a 2.000-L vessel. The following equilibrium is established at \(500 \mathrm{~K}\) : $$ \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ (a) Calculate the initial partial pressures of \(\mathrm{CO}_{2}, \mathrm{H}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} .\) (b) At equilibrium \(P_{\mathrm{H}_{2} \mathrm{O}}=3.51 \mathrm{~atm}\). Calculate the equilibrium partial pressures of \(\mathrm{CO}_{2}, \mathrm{H}_{2}\), and \(\mathrm{CO}\). (c) Calculate \(K_{n}\) for the reaction.

Short answer

(a) The initial partial pressures of CO₂, H₂, and H₂O are calculated to be \(P_{CO_2} = 1.209 atm\), \(P_{H_2} = 0.6046 atm\), and \(P_{H_2O} = 0.9674 atm\). (b) The equilibrium partial pressures of CO₂, H₂, and CO are calculated to be \(P_{CO_2} = 0.7408 atm\), \(P_{H_2} = 0.1364 atm\) and \(P_{CO} = 2.542 atm\). (c) The equilibrium constant, \(K_p = 26.86\).

Step-by-step solution

Calculate initial mole fractions

To calculate the mole fractions of CO₂, H₂, and H₂O, we need to divide their individual molar amounts by the total molar amount. \[x_{CO_2} = \frac{0.2000\,\text{mol}}{0.2000\, \text{mol} + 0.1000\, \text{mol} + 0.1600\, \mathrm{mol}}\] \[x_{H_2} = \frac{0.1000\,\text{mol}}{0.2000\, \text{mol} + 0.1000\, \text{mol} + 0.1600\, \mathrm{mol}}\] \[x_{H_2O} = \frac{0.1600\,\text{mol}}{0.2000\, \text{mol} + 0.1000\, \text{mol} + 0.1600\, \mathrm{mol}}\]

Calculate the total pressure

Using the ideal gas law, we can calculate the total pressure. \[PV = nRT\] \[P = \frac{nRT}{V}\] Plugging in the total number of moles, gas constant (R = 0.0821 L atm / mol K), temperature (500 K), and volume (2.000 L), we get: \[P = \frac{(0.2000 + 0.1000 + 0.1600) \cdot 0.0821 \cdot 500}{2.000}\]

Calculate the initial partial pressures

Now, we can calculate the initial partial pressures by multiplying the mole fractions calculated in step 1 by the total pressure calculated in step 2. \[P_{CO_2} = x_{CO_2} \cdot P\] \[P_{H_2} = x_{H_2} \cdot P\] \[P_{H_2O} = x_{H_2O} \cdot P\] (b) Calculate the equilibrium partial pressures of CO₂, H₂, and CO.

Set up the change table

Since we know the equilibrium partial pressure of H₂O, we can set up a table to see the change in partial pressures for all the species as the reaction progresses to equilibrium. Initial: P(CO₂) , P(H₂) , P(H₂O) - initial, 0 Change: -x , -x , +x , +x Equilibrium: P(CO₂) - x , P(H₂) - x , 3.51 atm , x

Calculate the change

Since we know the equilibrium partial pressure of H₂O is 3.51 atm, we can calculate the change (x) that occurred. \[x = 3.51 - P_{H_2O(initial)}\]

Calculate equilibrium partial pressures

Now, we can calculate the equilibrium partial pressures of CO₂, H₂, and CO by using the values from the change table. \[P_{CO_2}(equilibrium) = P_{CO_2}(initial) - x\] \[P_{H_2}(equilibrium) = P_{H_2}(initial) - x\] \[P_{CO}(equilibrium) = x\] (c) Calculate Kₚ for the reaction.

Write the expression for Kₚ

The equilibrium constant, Kₚ, for this reaction is defined as: \[Kₚ = \frac{P_{CO} \cdot P_{H₂O}}{P_{CO₂} \cdot P_{H₂}}\]

Substitute the equilibrium partial pressures

Now, substitute the equilibrium partial pressures we found in part (b) into the Kₚ expression. \[Kₚ = \frac{P_{CO}(equilibrium) \cdot 3.51}{P_{CO₂}(equilibrium) \cdot P_{H₂}(equilibrium)}\] Now we have calculated the initial partial pressures in part (a), the equilibrium partial pressures in part (b), and the value of the equilibrium constant, Kₚ, in part (c).

Key concepts

Partial Pressure

Partial pressure refers to the pressure exerted by a specific gas in a mixture of gases. It is an important concept when dealing with gas mixtures in chemical reactions, particularly those at equilibrium. You can think of partial pressure as the contribution of one specific component to the overall pressure of a gas mixture. To calculate the partial pressure of a gas, we need to know the mole fraction of the gas in the mixture and the total pressure of the mixture. The mole fraction is the ratio of the number of moles of a particular gas to the total number of moles of all gases in the mixture:

• Mole fraction of a gas: \(x_i = \frac{n_i}{n_{total}}\)
• Partial pressure of gas: \(P_i = x_i \cdot P_{total}\)

These calculations are crucial when determining the individual pressures and contributions of each gas within a reaction, such as the example involving \(\mathrm{CO_2}\), \(\mathrm{H_2}\), and \(\mathrm{H_2O}\) at equilibrium.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and moles of a gas. It is represented by the equation: \[ PV = nRT \] where:

• \(P\) is the pressure of the gas,
• \(V\) is the volume the gas occupies,
• \(n\) is the number of moles of gas,
• \(R\) is the ideal gas constant (usually 0.0821 L atm / mol K),
• \(T\) is the temperature in Kelvin.

This equation is particularly useful for predicting the behavior of gases under certain conditions, especially in equilibrium calculations. By manipulating the equation, one can find the total pressure of gases in a container, which is needed to calculate the initial and varying pressures of individual components as they react and reach equilibrium.

Equilibrium Constant

The equilibrium constant, often denoted as \(K_p\) or \(K_c\), serves as a quantitative measure of the position of equilibrium in a chemical reaction. It reflects the ratio of the concentrations or pressures of the products to the reactants at equilibrium. For gas-phase reactions, the equilibrium constant with respect to partial pressures is referred to as \(K_p\) and can be expressed using the equation:\[ K_p = \frac{(P_{products})}{(P_{reactants})} \]It is essential because:

• It allows comparison between reactions based on how far they proceed towards products.
• It helps predict whether the reaction will favor reactants or products at equilibrium given the initial conditions.
• It lets chemists understand and manipulate reaction conditions to achieve desired outcomes, such as selecting temperature and pressure for industrial processes.

Reaction Stoichiometry

Reaction stoichiometry deals with the quantitative relationships between reactants and products in a chemical reaction. It is integral to equilibrium calculations, as it provides the ratios needed to determine how much of each substance is involved as a reaction progresses:

• It tells us the proportions of each reactant that are needed to form a specific amount of product.
• By using stoichiometry, we can set up an ICE (Initial, Change, Equilibrium) table to track the transformations of reactants into products and calculate how much of each compound remains at equilibrium.
• Stoichiometry is also essential in deriving equations to calculate equilibrium partial pressures based on known or assumed changes during the reaction.

Understanding stoichiometry allows us to properly allocate resources and anticipate the outcomes of chemical processes, particularly in balancing chemical equations and predicting the quantities of products formed.