Question

The equilibrium constant \(K_{c}\) for \(C(s)+\mathrm{CO}_{2}(g) \rightleftharpoons\) 2 \(\mathrm{CO}(g)\) is 1.9 at 1000 \(\mathrm{K}\) and 0.133 at 298 \(\mathrm{K}\) . (a) If excess\(\mathrm{C}\) is allowed to react with 25.0 \(\mathrm{g}\) of \(\mathrm{CO}_{2}\) in a 3.00 -L vessel at \(1000 \mathrm{K},\) how many grams of CO are produced? (b) How many grams of \(\mathrm{C}\) are consumed? (c) If a smaller vessel is used for the reaction, will the yield of CO be greater or smaller? (d) Is the reaction endothermic or exothermic?

Short answer

(a) 15.74 g of CO are produced. (b) 3.373 g of C are consumed. (c) The yield of CO will be smaller if a smaller vessel is used for the reaction. (d) The reaction is endothermic.

Step-by-step solution

Calculate the initial moles of CO2

Since we are given the initial mass of CO2 and the volume of the vessel, we can calculate the moles of CO2 before the reaction takes place: \(n_{CO_2(initial)} = \frac{25.0 \, g}{44.01 \, g/mol} = 0.568 \, mol \)

Relate the changes in moles of CO2, CO, and C

Let the change in the number of moles of CO2, CO, and C be represented by x. At equilibrium: \(n_{CO_2(eq)} = 0.568 - x \, mol\) \(n_{CO(eq)} = 2x \, mol\)

Find moles of CO at equilibrium

We can use the equilibrium constant expression to find the moles of CO at equilibrium for 1000 K: \(K_{c} = 1.9 = \frac{[CO]^2}{[CO_{2}]} = \frac{(2x)^2}{(0.568 - x)}\) Solve for x: \(x \approx 0.281 \, mol\)

Calculate the mass of CO produced and C consumed

Now we can calculate the mass of CO produced and C consumed at equilibrium using their respective molar masses: Mass of CO produced: \(mass_{CO} = n_{CO(eq)} \times \frac{28.01 \, g}{1 \, mol} = 2(0.281) \times 28.01 = 15.74 \, g \) Mass of C consumed: \(mass_{C} = x \times \frac{12.01 \, g}{1 \, mol} = 0.281 \times 12.01 = 3.373 \, g \)

Discuss the effect of vessel size on the yield of CO

Using a smaller vessel will result in an increase in pressure, which will favor the side of the reaction with fewer moles of gas. In this case, since the reaction is producing more moles of CO on the right side, a smaller volume will result in a smaller yield of CO.

Determine if the reaction is endothermic or exothermic

From the given data, we can see that \(K_c\) at 1000 K is greater than \(K_c\) at 298 K: \(K_{c(1000 \, K)} > K_{c(298 \, K)} \) This indicates that the reaction is favored at higher temperatures, which means that the reaction is endothermic, as an increase in temperature favors the endothermic side of the equilibrium. Answers: a) The mass of CO produced at equilibrium at 1000 K is 15.74 g. b) The mass of C consumed at equilibrium at 1000 K is 3.373 g. c) The yield of CO will be smaller if a smaller vessel is used for the reaction. d) The reaction is endothermic.

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \), is a crucial number in chemistry that describes the ratio of concentrations of products to reactants at equilibrium for a given chemical reaction at a certain temperature. For the reaction \( C(s) + \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) \), the equilibrium constant is given as 1.9 at 1000 K. This value indicates how far the reaction proceeds towards the products before reaching equilibrium. A larger \( K_c \) means the reaction favors the formation of products, whereas a smaller \( K_c \) implies that the products are less favored at equilibrium. The equilibrium constant is dimensionless and depends solely on temperature. It does not change with the change in concentration or pressure of the reactants or products.

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept in understanding how equilibrium shifts in a chemical reaction when external conditions are changed. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change.In the context of the given reaction, when the size of the vessel is reduced, the pressure inside increases. According to Le Chatelier's Principle, the system will try to decrease the pressure. Since the reaction \( C(s) + \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) \) has more moles of gas on the product side, increasing pressure shifts the equilibrium towards the reactants, producing less \( \mathrm{CO} \). Hence, using a smaller vessel results in a lower yield of carbon monoxide.

Endothermic Reactions

Endothermic reactions absorb heat from the surroundings, resulting in a net intake of energy. These reactions generally have higher equilibrium constants at increased temperatures. For the chemical equilibrium in the exercise, where \( C(s) + \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) \), the equilibrium constant \( K_c \) is larger at 1000 K than at 298 K. This indicates that the reaction is endothermic, as it is favored by the addition of heat. When the temperature is increased, the reaction absorbs more heat and shifts towards the formation of more products (\( \mathrm{CO} \)), in accordance with Le Chatelier's Principle. Thus, the reaction effectively "likes" the heat being added, which confirms its endothermic nature.

Gas Law

The principles of gas law are essential in predicting how gases behave under various conditions of temperature, pressure, and volume. One of the main aspects of gas law is the ideal gas law, given by the formula \( PV = nRT \), where:

• \( P \) is the pressure of the gas,
• \( V \) is the volume,
• \( n \) is the number of moles,
• \( R \) is the ideal gas constant, and
• \( T \) is the temperature in Kelvin.

This equation helps us to understand how changing the volume of a vessel affects the pressure of the gases within it, as seen in the reaction \( C(s) + \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) \). Reducing the volume of the reaction vessel increases the pressure, which, according to the ideal gas law, affects the equilibrium. This interplay is why understanding gas laws is critical for predicting yields and shifts in equilibrium during reactions in closed containers.