Problem 63 - Chemistry The Molecular Science

Question

The value of \(K_{\mathrm{c}}\) is \(3.7 \times 10^{-23}\) at \(25^{\circ} \mathrm{C}\) for $$ \mathrm{C}(\text { graphite })+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}) $$ Describe what will happen if \(3.5 \mathrm{~mol} \mathrm{CO}\) and \(3.5 \mathrm{~mol}\) \(\mathrm{CO}_{2}\) are mixed in a 1.5-L sealed graphite container with a suitable catalyst so that the reaction rate is rapid at this temperature.

Step-by-Step Solution

Answer

The reaction will proceed in the reverse direction, converting CO to CO2, until equilibrium is reached.

Step 1: Write the Expression for Equilibrium Constant

The equilibrium constant expression for the reaction \( \mathrm{C(\text{graphite})} + \mathrm{CO}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{CO}(\mathrm{g}) \) is \( K_c = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2]} \). This equation represents the ratio of the concentration of products and reactants at equilibrium.

Step 2: Calculate the Initial Concentrations

To find the initial concentrations, divide the number of moles by the volume of the container. For \( \mathrm{CO} \), \([\mathrm{CO}] = \frac{3.5}{1.5} = 2.33\, \mathrm{mol/L} \). For \( \mathrm{CO}_2 \), \([\mathrm{CO}_2] = \frac{3.5}{1.5} = 2.33\, \mathrm{mol/L} \).

Step 3: Set Up the Reaction Quotient

The reaction quotient, \( Q \), is calculated using the same expression as for \( K_c \): \( Q = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2]} = \frac{(2.33)^2}{2.33} = 2.33 \).

Step 4: Compare Q to K_c

Compare \( Q \) and \( K_c \). Here, \( Q = 2.33 \) and \( K_c = 3.7 \times 10^{-23} \). Since \( Q \gg K_c \), the system is not at equilibrium and the concentration of \( \mathrm{CO} \) is much higher than if at equilibrium.

Step 5: Predict the Direction of the Reaction

Because \( Q > K_c \), the reaction will proceed in the reverse direction to reach equilibrium. This means \( \mathrm{CO} \) will convert back into \( \mathrm{CO}_2 \) until the concentrations adjust and \( Q = K_c \).

Step 6: Conclude the Adjustment Process

As the reaction approaches equilibrium, the concentration of \( \mathrm{CO} \) will decrease and the concentration of \( \mathrm{CO}_2 \) will increase. The process will continue until \( Q = K_c \).

Key Concepts

Equilibrium Constant

In chemical reactions, the equilibrium constant, denoted as \(K_c\), helps us understand the balance between reactants and products at equilibrium.
It's specific to a particular reaction at a certain temperature. The value of \(K_c\) tells us whether the products or reactants are favored in the equilibrium state. A high \(K_c\) indicates a higher concentration of products, while a low \(K_c\) suggests products convert back to reactants. For the given reaction, \( \mathrm{C(graphite) + CO_2(g) \rightleftharpoons 2CO(g)} \), the equilibrium constant expression is:

• \(K_c = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2]}\)

Here, \(K_c = 3.7 \times 10^{-23}\) at 25°C, indicating that there are far more reactants than products when the system is in equilibrium. This extremely small value signifies that, under these conditions, almost none of the reactants are transformed into products.

Reaction Quotient

The reaction quotient, denoted as \(Q\), is a measure we use to compare the current state of a reaction to its equilibrium state.
Like \(K_c\), its calculation involves the concentrations of products and reactants but can be assessed whether the system is at equilibrium or not. By substituting the initial concentrations into the equilibrium expression:

• \(Q = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2]} = \frac{(2.33)^2}{2.33} = 2.33\)

We compare \(Q\) to \(K_c\). If \(Q = K_c\), the reaction is at equilibrium.
If \(Q < K_c\), the reaction will proceed forward, forming more products. If \(Q > K_c\), as in this case, the reaction will proceed in reverse, converting products back into reactants until equilibrium is achieved.

Le Chatelier's Principle

Le Chatelier's Principle helps predict how a system at equilibrium responds to disturbances.
When the equilibrium is disturbed by changes in concentration, temperature, or pressure, the system will adjust to counteract the disturbance and restore a new equilibrium state. In the context of this reaction, given:

• \(Q > K_c\)

According to Le Chatelier, the system will shift to the left, towards the reactants, to reduce the concentration of products (\(\mathrm{CO}\)) and increase that of the reactants (\(\mathrm{CO}_2\)).
This principle demonstrates the dynamic nature of chemical equilibria and their tendency to resist changes to the established balance.