Problem 66 - Chemistry: Principles and Reactions

Question

Consider the equilibrium $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)$$ When this system is at equilibrium at \(700^{\circ} \mathrm{C}\) in a \(2.0\) - \(\mathrm{L}\) container, \(0.10 \mathrm{~mol}\) \(\mathrm{CO}, 0.20 \mathrm{~mol} \mathrm{CO}_{2}\), and \(0.40 \mathrm{~mol} \mathrm{C}\) are present. When the system is cooled to \(600^{\circ} \mathrm{C}\), an additional \(0.040 \mathrm{~mol} \mathrm{C}(s)\) forms. Calculate \(K\) at \(700^{\circ} \mathrm{C}\) and again at \(600^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The equilibrium constant K is 0.025 at 700°C and 0.00125 at 600°C.

Step 1: Write the balanced chemical equation and the equilibrium expression

We are given the balanced chemical equation: $$\mathrm{C}(s) + \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)$$ The equilibrium expression for this reaction is: $$K = \frac{[\mathrm{CO}]^{2}}{[\mathrm{CO}_2 ]}$$

Step 2: Calculate the initial concentrations

We are given the initial moles of the substances in a 2.0 L container. We can calculate the initial concentrations: $$[\mathrm{CO}] = \frac{0.10 \mathrm{~mol}}{2 \mathrm{~L}} = 0.050 \mathrm{~M}$$ $$[\mathrm{CO}_{2}] = \frac{0.20 \mathrm{~mol}}{2 \mathrm{~L}} = 0.10 \mathrm{~M}$$

Step 3: Calculate K at \(700^{\circ} \mathrm{C}\)

We can now substitute the initial concentrations into the equilibrium expression and solve for K: $$K = \frac{[\mathrm{CO}]^{2}}{[\mathrm{CO}_2]} = \frac{(0.050 ^\mathrm{M})^{2}}{0.10 ^\mathrm{M}} = 0.025$$

Step 4: Calculate the new concentrations at \(600^{\circ} \mathrm{C}\)

When the system is cooled to \(600^{\circ} \mathrm{C}\), an additional \(0.040 \mathrm{~mol} \mathrm{C}(s)\) forms. This change will affect the concentrations of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) as well: Initial moles of \(\mathrm{CO}_{2}\): \(0.20-0.040=0.16\ \mathrm{mol}\) $$[\mathrm{CO}_{2}] = \frac{0.16 \mathrm{~mol}}{2 \mathrm{~L}} = 0.080 \mathrm{~M}$$ Initial moles of \(\mathrm{CO}\): \(0.10-2\times0.040=0.020\ \mathrm{mol}\) $$[\mathrm{CO}] = \frac{0.020 \mathrm{~mol}}{2 \mathrm{~L}} = 0.010 \mathrm{~M}$$

Step 5: Calculate K at \(600^{\circ} \mathrm{C}\)

Using the new concentrations, we can now calculate K for the reaction at \(600^{\circ} \mathrm{C}\) : $$K = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2 ]} = \frac{(0.010 \mathrm{~M})^2}{0.080 \mathrm{~M}} = 0.00125$$ Therefore, the equilibrium constant K is 0.025 at \(700^{\circ} \mathrm{C}\) and 0.00125 at \(600^{\circ} \mathrm{C}\).

Key Concepts

Balanced Chemical Equation

A balanced chemical equation is crucial in showing the correct proportions of reactants and products in a chemical reaction. In this exercise, the balanced chemical equation is \( \mathrm{C}(s) + \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) \). This equation tells us that one mole of solid carbon (C) reacts with one mole of carbon dioxide (\( \mathrm{CO}_2 \)) gas to produce two moles of carbon monoxide (CO) gas.
Balanced equations are critical as they adhere to the law of conservation of mass, ensuring that the number of atoms of each element is the same on both sides of the equation. This balanced equation allows us to precisely calculate reactant consumption and product formation, ensuring accurate concentration and equilibrium calculations.

Equilibrium Expression

The equilibrium expression gives us a mathematical representation of the concentrations of products and reactants at equilibrium. For the given reaction, the equilibrium expression is written as:
\[ K = \frac{[\mathrm{CO}]^{2}}{[\mathrm{CO}_2]} \]
In this expression, \( K \) is the equilibrium constant, \( [\mathrm{CO}] \) represents the concentration of carbon monoxide, and \( [\mathrm{CO}_2] \) is the concentration of carbon dioxide. The exponents in the equilibrium expression match the coefficients from the balanced chemical equation. For this reaction, the coefficient for CO is 2, therefore the concentration of CO in the expression is squared.
This equilibrium expression helps us understand how the concentrations of reactants and products relate to the equilibrium constant \( K \), which at a fixed temperature, indicates the position of equilibrium.

Concentration Calculation

Calculating concentrations accurately is key for determining the equilibrium constant. In this scenario, we start by calculating the initial concentrations using the given moles and volume of the reaction container.
The concentrations are calculated using the formula:
\[ [\text{Substance}] = \frac{\text{Moles}}{\text{Volume in Liters}} \]
For \( \mathrm{CO} \), with 0.10 moles in a 2.0 L container, the concentration is 0.050 M.
For \( \mathrm{CO}_{2} \), with 0.20 moles, the concentration becomes 0.10 M.
These initial concentrations are used in the equilibrium expression to calculate the equilibrium constant at 700°C, resulting in \( K = 0.025 \).
Then, when the system is cooled to 600°C, concentrations are recalculated, considering changes due to additional solid carbon formation, and the new \( K \) is found to be 0.00125.

Temperature Effect on Equilibrium

Temperature significantly affects the position of equilibrium and the value of the equilibrium constant \( K \). In the given chemical reaction, when the system is at 700°C, the equilibrium constant is calculated to be 0.025. However, when the temperature drops to 600°C, \( K \) changes to 0.00125.
Such changes occur because equilibrium constants are temperature-dependent. **Le Chatelier's Principle** helps explain why: if a system at equilibrium is subjected to a change in temperature, the system will adjust itself to counteract that change.
If the reaction is exothermic, reducing the temperature will increase \( K \), favoring the formation of products. Conversely, if it's endothermic, reducing temperature decreases \( K \), favoring reactants. Accurately predicting the effect of temperature on equilibrium constants is crucial, especially in industrial processes where conditions are tightly controlled to maximize product yield.