Question

At \(986^{\circ} \mathrm{C}\), you have the following equilibrium: \(\mathrm{C} \mathrm{O}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \rightleftarrows \mathrm{C} 0(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) Initially, \(49.3\) mole percent \(\mathrm{CO}_{2}\) is mixed with \(50.7 \mathrm{~mole}\) per cent \(\mathrm{H}_{2}\). At equilibrium, you find \(21.4\) mole percent \(\mathrm{CO}_{2}\), \(22.8\) mole percent \(\mathrm{H}_{2}\), and \(27.9 \mathrm{~mole}\) percent of \(\mathrm{CO}\) and \(\mathrm{H}_{2} \mathrm{O}\). Find \(\mathrm{K}\). If you start with a mole percent ratio of \(60: 40\) \(\mathrm{CO}_{2}\) to \(\mathrm{H}_{2}\), find the equilibrium concentrations of both reactants and products.

Short answer

The equilibrium constant (K) for the given reaction is found to be 2.153. For the initial mole percent ratio of CO2 to H2 as 60:40, the equilibrium concentrations of the species are: - CO2: 38.878 mole percent - H2: 18.878 mole percent - CO: 21.122 mole percent - H2O: 21.122 mole percent

Step-by-step solution

Write down the given information

We have the following initial conditions: - CO2: 49.3 mole percent - H2: 50.7 mole percent At equilibrium, we have: - CO2: 21.4 mole percent - H2: 22.8 mole percent - CO: 27.9 mole percent - H2O: 27.9 mole percent

Calculate the mole fractions of all the species

We need to find the mole fractions of all the species involved in the reaction. To do that, we will first use the equilibrium mole percentages that are given in the problem. The sum of all mole percentages should always equal 100. Therefore, CO2: \(frac{21.4}{100}\) H2: \(frac{22.8}{100}\) CO: \(frac{27.9}{100}\) H2O: \(frac{27.9}{100}\)

Determining the change in concentrations

Notice that the mole fractions of CO and H2O are equal and the mole fractions in CO2 and H2 are equal. Therefore, we can determine the change in concentrations during the reaction as follows: Change in CO2 concentration: 49.3 - 21.4 = 27.9 mole percent Change in H2 concentration: 50.7 - 22.8 = 27.9 mole percent Change in CO concentration: 27.9 mole percent Change in H2O concentration: 27.9 mole percent

Calculate the equilibrium constant (K)

Now we can calculate the equilibrium constant (K) for the given reaction using the equilibrium mole fractions: \[K = \frac{[CO][H_2O]}{[CO_2][H_2]}\] \[K = \frac{\frac{27.9}{100} \times \frac{27.9}{100}}{\frac{21.4}{100} \times \frac{22.8}{100}} \] \[K = 2.153\]

Determine equilibrium concentrations for a new initial ratio

For initial mole percent ratio of CO2 to H2 as 60:40, we need to find the equilibrium concentrations of all the species. To do that, we'll use the balanced equation and the K value calculated above. Initial conditions: - CO2: 60 mole percent - H2: 40 mole percent Let x be the mole percent change of both CO2 and H2 in the reaction to achieve equilibrium: Equilibrium concentrations: - CO2: 60 - x mole percent - H2: 40 - x mole percent - CO: x mole percent - H2O: x mole percent Now we use the K value and the equilibrium concentrations to find x. \[K = \frac{[CO][H_2O]}{[CO_2][H_2]} = \frac{x^2}{(60-x)(40-x)}\] Solving for x using the calculated K value, we get: \[2.153 = \frac{x^2}{(60-x)(40-x)}\] By solving this equation, we find that \(x = 21.122\). Now, we can find the equilibrium concentrations for all species: - CO2: 60 - 21.122 = 38.878 mole percent - H2: 40 - 21.122 = 18.878 mole percent - CO: 21.122 mole percent - H2O: 21.122 mole percent

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as K, is a numerical value that relates the concentrations of reactants and products in a chemical equilibrium. It is a significant value in chemistry as it provides a quantitative measure of the extent of the reaction. When a reaction reaches equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant.

For the reaction at hand, \[ CO_2(g) + H_2(g) \rightleftarrows CO(g) + H_2O(g) \]
The equilibrium constant expression is given by the formula \[ K = \frac{[CO][H_2O]}{[CO_2][H_2]} \]
where the square brackets denote the concentration of each species at equilibrium. It's crucial to note that only gases and dissolved species are included in the expression and that the concentrations must be in equilibrium.

To determine the equilibrium constant from the exercise, you first calculate the mole fractions of reactants and products at equilibrium and then plug those numbers into the equilibrium constant expression. The change in reactants and products gives insight into the extent of the reaction, and through the proper mathematical manipulation, as shown in the solution steps, you arrive at the equilibrium constant K.

Mole Percent

Mole percent is one way to express the concentration of a component in a mixture. This unit of measure specifies what percentage of the total moles of all substances in a mixture is made up by a specific substance. Mole percent is especially handy when dealing with gases, as it allows for an easy conversion to partial pressures, which are in the same ratio as mole percents when total pressure is constant.

In our exercise, the equilibrium of the reaction mixture is described using mole percent. Initially, we have 49.3 mole percent of CO2 mixed with 50.7 mole percent of H2. At equilibrium, these percentages change as the reaction occurs. By using the initial and equilibrium mole percents, you can calculate how much of each reactant has been converted to product. For example, if you start with 60 mole percent of CO2 and, upon reaching equilibrium, the mole percent drops to 38.878, you know that the difference has been converted into CO and H2O.

Equilibrium Concentrations

Understanding equilibrium concentrations is crucial when analyzing any chemical equilibrium situation. These concentrations are the amounts of reactants and products present when a chemical system is at equilibrium. In the context of a reaction, it refers to the stabilized concentrations of the reacting species after the rate of the forward reaction equals the rate of the reverse reaction.

In relation to the exercise presented, after finding the equilibrium constant K, the task is to calculate the equilibrium concentrations of the reactants and products from an initial condition that is different from the one the K value was determined. This demonstrates the concept that the equilibrium constant K remains the same for a given reaction at a specific temperature, irrespective of the starting concentrations of reactants and products.

To find the new equilibrium concentrations, you assume a change in the starting mole percent of reactants, which will consequently lead to a change in the product's mole percent. This shift in concentration is denoted by 'x', which is then used in the equilibrium constant expression to solve for the new equilibrium concentrations. As always in chemical equilibrium problems, ensuring that the mole percent sums to 100 and that stoichiometry is taken into account is essential in these calculations.