Question

The following equilibria were measured at \(823 \mathrm{~K}\) : $$\begin{array}{l} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=67 \\ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=0.14 \end{array}$$ (a) Use these equilibria to calculate the equilibrium constant, \(K_{c},\) for the reaction \(\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)\) \(+\mathrm{CO}_{2}(g)\) at \(823 \mathrm{~K}\). (b) Based on your answer to part (a), would you say that carbon monoxide is a stronger or weaker reducing agent than \(\mathrm{H}_{2}\) at \(T=823 \mathrm{~K} ?(\mathbf{c})\) If you were to place \(5.00 \mathrm{~g}\) of \(\mathrm{CoO}(s)\) in a sealed tube with a volume of \(250 \mathrm{~mL}\) that contains \(\mathrm{CO}(g)\) at a pressure of \(101.3 \mathrm{kPa}\) and a temperature of \(298 \mathrm{~K},\) what is the concentration of the CO gas? Assume there is no reaction at this temperature and that the CO behaves as an ideal gas (you can neglect the volume of the solid). (d) If the reaction vessel from part (c) is heated to \(823 \mathrm{~K}\) and allowed to come to equilibrium, how much \(\operatorname{CoO}(s)\) remains?

Short answer

In summary, the equilibrium constant \(K_{c}\) for the reaction $\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g)$ at \(823 \mathrm{K}\) is 9.38. Carbon monoxide (CO) is a weaker reducing agent than hydrogen gas (\(\mathrm{H}_{2}\)) at \(T=823 \mathrm{K}\). The concentration of CO gas at \(298 \mathrm{K}\) is 0.0408 M. Finally, there are 1.99 g of CoO remaining after the reaction reaches equilibrium at \(823 \mathrm{K}\).

Step-by-step solution

Calculate the equilibrium constant for the desired reaction

To find the equilibrium constant for the reaction $\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g)$, we can manipulate and combine the given equilibria: (1) $\mathrm{CoO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c1}=67$ (2) $\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c2}=0.14$ We can multiply the two equilibrium equations: (1) * (2), to get the desired reaction: (3) $\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g)$ The equilibrium constant for the combined reaction (3) is obtained by multiplying the equilibrium constants of the original reactions: \(K_{c3} = K_{c1} \cdot K_{c2} = 67 \cdot 0.14\) Now, we can calculate \(K_{c3}\): \(K_{c3} = 9.38\) Thus, the equilibrium constant for the reaction, \(K_{c3}\), is 9.38 at \(823 \mathrm{~K}\).

Determine the strength of the reducing agents

We would say that CO (carbon monoxide) is a stronger reducing agent if \(K_{c3}\) is higher than \(K_{c1}\) (67). We calculated \(K_{c3}\) to be 9.38, which is lower than 67. Therefore, CO is a weaker reducing agent than \(\mathrm{H}_{2}\) at \(T=823 \mathrm{~K}\).

Calculate the concentration of CO gas at \(298 \mathrm{~K}\)

We can use the ideal gas law to determine the concentration of CO gas. The ideal gas law equation is: \(PV = nRT\) Where: \(P\) = pressure (101.3 kPa) \(V\) = volume (250 mL = 0.250 L) \(n\) = moles of CO \(R\) = gas constant (8.314 J/(mol·K) = 0.0821 L·atm/(mol·K)) \(T\) = temperature (298 K) First, we need to convert the pressure to atmospheres: \(P_{atm} = 101.3 \mathrm{kPa} \times \frac{1 \mathrm{atm}}{101.325 \mathrm{kPa}} = 1 \mathrm{atm}\) Now, we can solve for the number of moles of CO: \(n = \frac{PV}{RT} = \frac{(1 \mathrm{atm})(0.250 \mathrm{L})}{(0.0821 \mathrm{L\cdot atm/mol\cdot K})(298 \mathrm{K})} = 0.0102 \mathrm{mol}\) Next, we can calculate the concentration of CO gas: \(\text{Concentration of CO} = \frac{n}{V} = \frac{0.0102 \mathrm{mol}}{0.250 \mathrm{L}} = 0.0408 \mathrm{M}\)

Calculate the amount of CoO remaining

We first need to determine how many moles of CoO there are initially. The molar mass of CoO is: \(\mathrm{CoO}:~ 58.93\mathrm{g/mol}(\mathrm{Co}) + 16.00\mathrm{g/mol}(\mathrm{O}) = 74.93\mathrm{g/mol}(\mathrm{CoO}\) The initial moles of CoO is \(\frac{5.00\mathrm{g}(\mathrm{CoO})}{74.93\mathrm{g/mol}(\mathrm{CoO}\) Since the given information states that the reaction CoO(s)+CO(g)⟷Co(s)+CO2(g) is at equilibrium, the amount of CoO(s) remaining can be calculated as: \(x=\left[\frac{5.00\mathrm{g}(\mathrm{CoO})}{74.93\mathrm{g/mol}(\mathrm{CoO}\right] \times \frac{1}{\sqrt{K_{c3}}} = 0.0266\mathrm{mol}\) Now, we can calculate the mass of the remaining CoO: \(mass_{CoO}=\left(0.0266\mathrm{mol}\right) \times \left(74.93\mathrm{g/mol}\right)= 1.99\mathrm{g}\) Thus, there are 1.99 g of CoO remaining after the reaction reaches equilibrium.

Key concepts

Equilibrium Constant

The equilibrium constant, represented as \(K_c\), is a numerical value that indicates the ratio of the concentrations of products to reactants at equilibrium for a reversible chemical reaction. In essence, it helps us understand how much of the product is formed at equilibrium under a given temperature.
The concept of equilibrium constant is vital because it tells us the direction in which a reaction tends to proceed. When \(K_c\) is greater than 1, the reaction favors the formation of products at equilibrium. Conversely, if \(K_c\) is less than 1, the formation of reactants is favored.

• The calculation involves considering the balanced chemical reaction where the concentration of products raised to their stoichiometric coefficients is divided by the concentration of reactants similarly raised.
• It only includes species in the gaseous and aqueous states, excluding solids and liquids.

This principle was applied in the exercise by using two given equilibria to calculate the \(K_c\) for a desired reaction. By manipulating and multiplying these equations, we derived the equilibrium constant for the complex equation.

Ideal Gas Law

The ideal gas law is a mathematical expression, \(PV = nRT\), that approximates the behavior of ideal gases. Here, \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.
This law enables us to relate these variables and predict how gases will behave under different conditions. It's key in calculating properties like the concentration of a gas when other variables are known.

• The exercise utilized the ideal gas law to determine the concentration of carbon monoxide gas at a lower temperature, assuming no reaction occurred. This was achieved by rearranging the equation to solve for \(n\), the number of moles.
• The pressure conversion from kPa to atm was necessary because the gas constant \(R\) was expressed in terms of L·atm/mol·K.
• Ultimately, this helped in calculating the moles and, hence, the concentration of CO under ideal conditions.

Understanding this law offers substantial insight into calculations involving gases, especially in ideal equilibrium conditions.

Reducing Agents

In chemical reactions, reducing agents play an important role. They are substances that donate electrons to another substance, thereby reducing it. This process is essential in many chemical equilibria where a reducing agent donates electrons to transform reactants to products, often resulting in a change in oxidation state of other materials.
The exercise compares two reducing agents: hydrogen (\(H_2\)) and carbon monoxide (CO). By analyzing the equilibrium constant calculated for the reaction involving CO, it becomes evident that CO is a weaker reducing agent than \(H_2\) at the given temperature.

• The strength of a reducing agent is often assessed by examining its ability to donate electrons effectively. A high equilibrium constant in a reduction reaction often suggests a more effective reducing agent.
• The exercise demonstrated that since \(K_{c3} = 9.38\) is less than \(67\), CO is not as strong as \(H_2\) in reducing CoO to Co under the specified conditions.

Recognizing the capabilities of different reducing agents is crucial in predicting and controlling the outcome of chemical reactions, particularly those involving metal oxides like CoO.