Question

Consider the heterogeneous equilibrium process: $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ At \(700^{\circ} \mathrm{C}\), the total pressure of the system is found to be 4.50 atm. If the equilibrium constant \(K_{P}\) is 1.52 , calculate the equilibrium partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\)

Short answer

After solving the quadratic equation, it is found that \(x\) has one acceptable value. This value of \(x\) is substituted back into the partial pressure expressions to get the equilibrium partial pressures for \(\mathrm{CO_{2}}\) and \(\mathrm{CO}\).

Step-by-step solution

Identify Relevant Equations

As the problem is dealing with the concept of equilibrium, the expression for equilibrium constant \(K_{P}\) is crucial. For the given reaction:\[\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)\]the equilibrium constant \(K_{P}\) can be written as\[K_{P} = \frac{{(\mathrm{P}_{\mathrm{CO}})^{2}}}{{P_{\mathrm{CO}_{2}}}}\]where \(P_{\mathrm{CO}_{2}}\) and \(P_{\mathrm{CO}}\) are the equilibrium partial pressures of \(CO_{2}\) and \(CO\) respectively.

Set up Equations with Unknowns

Considering that the total pressure of the system is 4.5 atm, we can set:\[P_{\mathrm{CO}} = 2x\P_{\mathrm{CO_{2}}} = 4.5 - 2x\]Where \(x\) is the partial pressure of \(\mathrm{CO}\), at equilibrium.

Apply the Equilibrium Constant

Substitute the equilibrium partial pressures of \(CO_{2}\) and \(CO\) in terms of \(x\) into the \(K_{P}\) equation:\[K_{P} = \frac{{(2x)^{2}}}{{4.5 - 2x}}\]Substitute the given value of \(K_{P}\) = 1.52 into the equation and solve for \(x\).

Solve for \(x\)

In order to solve for \(x\), it is more convenient to simplify the equation from the previous step. This will result in a quadratic equation. Solve this quadratic equation.

Calculate Partial Pressures

Substitute the value of \(x\) back into the equations for \(P_{\mathrm{CO_{2}}}\) and \(P_{\mathrm{CO}}\) to get the equilibrium partial pressures for each.

Key concepts

Chemical Equilibrium

In chemical reactions, chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations (or partial pressures in the case of gases) of reactants and products remain constant over time.

For the reaction given in the exercise, solid carbon reacts with carbon dioxide gas to form carbon monoxide gas. Since carbon is a solid, its concentration remains constant and does not appear in our calculations, which focus on the gases involved. When the system reaches equilibrium, the production of carbon monoxide is counteracted by its conversion back to carbon dioxide and solid carbon at an equal rate, leading to a constant state where none of the amounts of products and reactants change anymore.

Equilibrium Constant

The equilibrium constant (K) is a number that expresses the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their stoichiometric coefficients. For reactions involving gases, this constant is often given in terms of partial pressures and thus referred to as the equilibrium constant for pressure, or K_P.

In the exercise, the equilibrium constant for the reaction \(\mathrm{C}(s) + \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)\) is defined as \(K_{P} = \frac{{(\mathrm{P}_{\mathrm{CO}})^{2}}}{{P_{\mathrm{CO}_{2}}}}\). This expression implies that at equilibrium, the squared partial pressure of carbon monoxide must be proportional to the partial pressure of carbon dioxide by the constant factor K_P, which is given as 1.52. It's crucial when solving such problems to set up the equilibrium expression correctly and understand how to manipulate and solve for the unknown quantities.

Le Chatelier's Principle

Le Chatelier's Principle predicts how an equilibrium system will respond to changes in concentration, pressure, or temperature. According to this principle, if an external stress is applied to a system at equilibrium, the system will adjust in such a way as to partially counteract the change and establish a new equilibrium.

For instance, increasing the pressure of the system will often shift the equilibrium towards the side with fewer gas molecules, as seen in the provided reaction where pressure favors the production of carbon monoxide, reducing the number of gas molecules from one mole of carbon dioxide to two moles of carbon monoxide. Understanding this principle can aid students in anticipating the direction of the shift in equilibrium in response to changes, which is a critical concept in practical chemistry.