Problem 37 - Chemistry The Molecular Science

Question

Carbon dioxide reacts with carbon to give carbon monoxide according to the equation $$ \mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}) $$ At \(700 .{ }^{\circ} \mathrm{C},\) a \(2.0-\mathrm{L}\) sealed flask at equilibrium contains $$ 0.10 \mathrm{~mol} \mathrm{CO}, 0.20 \mathrm{~mol} \mathrm{CO}_{2}, \text { and } 0.40 \mathrm{~mol} \mathrm{C} . \text { Calculate } $$ the equilibrium constant \(K_{\mathrm{P}}\) for this reaction at the specified temperature.

Step-by-Step Solution

Answer

The equilibrium constant \(K_{\mathrm{P}}\) is approximately 2.00 at 700°C.

Step 1: Write the equation for the reaction

The balanced chemical equation is \(\mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})\). Since \(\mathrm{C}\) is a solid, it does not appear in the equilibrium constant expression.

Step 2: Write the expression for the equilibrium constant Kₚ

The equilibrium constant \(K_{\mathrm{P}}\) is given by the expression \(K_{\mathrm{P}} = \frac{(P_{\mathrm{CO}})^2}{P_{\mathrm{CO}_2}}\). Here, \(P_{\mathrm{CO}}\) and \(P_{\mathrm{CO}_2}\) are the partial pressures of \(\mathrm{CO}\) and \(\mathrm{CO}_2\), respectively.

Step 3: Calculate moles to pressures using the ideal gas law

Using the ideal gas law: \(PV = nRT\), we calculate partial pressures. Given \(n_{\mathrm{CO}}=0.10\) mol, \(n_{\mathrm{CO}_2}=0.20\) mol, volume \(V=2.0\) L, and temperature \(T=700^\circ \mathrm{C} = 973\) K, the partial pressures are: - \(P_{\mathrm{CO}} = \frac{n_{\mathrm{CO}}RT}{V} = \frac{0.10 \times 0.0821 \times 973}{2.0}\)- \(P_{\mathrm{CO}_2} = \frac{n_{\mathrm{CO}_2}RT}{V} = \frac{0.20 \times 0.0821 \times 973}{2.0}\)

Step 4: Calculate specific partial pressures

Performing the calculations:- \(P_{\mathrm{CO}} = 3.99\) atm- \(P_{\mathrm{CO}_2} = 7.98\) atm. These values are calculated using the expression \(RT/V\approx 3.99\) using \(R = 0.0821\).

Step 5: Substitute values into the Kₚ expression

Substitute the values calculated into the expression for \(K_{\mathrm{P}}\):\[K_{\mathrm{P}} = \frac{(3.99)^2}{7.98}\]

Step 6: Solve for Kₚ

Calculating the above expression gives \(K_{\mathrm{P}} = \frac{15.92}{7.98} = 1.995\). Therefore, the equilibrium constant \(K_{\mathrm{P}}\) is approximately 2.00 at 700°C.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. The equation is expressed as \( PV = nRT \), where:

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

To solve problems using the Ideal Gas Law, make sure all units are consistent.
Always convert temperatures to Kelvin by adding 273 to the Celsius value. In our exercise, we dealt with values at 700°C, which is converted to 973 K for calculations. Using this law, you can determine any one of the missing variables if the others are known.

Partial Pressure

Partial pressure is a significant concept in chemical reactions involving gases. It refers to the pressure exerted by an individual gas in a mixture of gases. For any gas in a mixture, this can be found using the ideal gas law as \( P = \frac{nRT}{V} \). In equilibrium exercises, understanding partial pressures is crucial as it plays a role in calculating the equilibrium constant, \( K_{\mathrm{P}} \).
Each component gas has a partial pressure contributing to the total pressure. In our exercise,

• Carbon monoxide (CO) had partial pressure calculated using its mole number (0.10 mol)
• Carbon dioxide (CO₂) had partial pressure calculated with its mole number (0.20 mol)

These calculations are essential since partial pressures appear in the expression for \( K_{\mathrm{P}} \), showing how each gas participates in the equilibrium state.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction in a closed system. At this point, the concentrations of reactants and products remain constant but not necessarily equal. The equilibrium constant \( K_{\mathrm{P}} \) specifically applies to systems involving gases, expressed in terms of partial pressures.The equation for our reaction is \( \mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}) \). Since carbon (C) is a solid, it does not appear in the \( K_{\mathrm{P}} \) expression.
The expression is \( K_{\mathrm{P}} = \frac{(P_{\mathrm{CO}})^2}{P_{\mathrm{CO}_2}} \).

• Equilibrium reflects a balance
• Understanding the reaction dynamics is key to chemistry applications

This balance allows us to predict the concentrations of gases at equilibrium when given initial amounts and conditions.

Carbon Dioxide

Carbon dioxide, or CO₂, is a vital component in numerous chemical reactions but is notably significant in the context of gaseous equilibria. In our exercise, it acts as both a reactant and a determinant of the chemical equilibrium involving carbon monoxide.Here is what you should remember about CO₂:

• It contributes to calculations of partial pressure using the ideal gas law.
• It influences the value of the equilibrium constant \( K_{\mathrm{P}} \).
• It participates in reactions that are sensitive to changes in temperature and pressure.

Understanding the role of carbon dioxide in reactions helps chemists anticipate how changes in conditions can shift equilibria, influencing product yields, which is especially important in industrial processes.