Problem 59 - Chemistry The Central Science In Si Units

Question

Methane, \(\mathrm{CH}_{4}\), reacts with \(\mathrm{I}_{2}\) according to the reaction \(\mathrm{CH}_{4}(g)+\mathrm{I}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{I}(g)+\mathrm{HI}(g) .\) At \(600 \mathrm{~K}, K_{p}\) for this reaction is \(1.95 \times 10^{-4}\). A reaction was set up at 600 \(\mathrm{K}\) with initial partial pressures of methane of \(13.3 \mathrm{kPa}\) and of \(6.67 \mathrm{kPa}\) for \(\mathrm{I}_{2}\). Calculate the pressures, in kPa, of all reactants and products at equilibrium.

Step-by-Step Solution

Answer

The equilibrium pressures are \(13.17\) kPa for \(\mathrm{CH}_4\), \(6.54\) kPa for \(\mathrm{I}_2\), and \(0.1316\) kPa each for \(\mathrm{CH}_3\mathrm{I}\) and \(\mathrm{HI}\).

Step 1: Write the Equilibrium Expression

For the given reaction \( \mathrm{CH}_{4}(g)+\mathrm{I}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{I}(g)+\mathrm{HI}(g) \), the equilibrium expression for \( K_p \) is: \[ K_p = \frac{P_{\mathrm{CH_3I}} \times P_{\mathrm{HI}}}{P_{\mathrm{CH_4}} \times P_{\mathrm{I_2}}} \] where \( P_{\mathrm{species}} \) represents the partial pressures of the respective species.

Step 2: Define Changes in Pressure

Let the change in pressure for the formation of \( \mathrm{CH}_3\mathrm{I} \) and \( \mathrm{HI} \) be \( x \) kPa. Hence, the pressures at equilibrium will be: \[ P_{\mathrm{CH_4,eq}} = 13.3 - x \]\[ P_{\mathrm{I_2,eq}} = 6.67 - x \]\[ P_{\mathrm{CH_3I,eq}} = x \]\[ P_{\mathrm{HI,eq}} = x \]

Step 3: Substitute into the Equilibrium Expression

Substitute the equilibrium pressures into the \( K_p \) expression: \[ K_p = \frac{x \cdot x}{(13.3 - x)(6.67 - x)} = 1.95 \times 10^{-4} \] This leads to the equation: \[ \frac{x^2}{(13.3 - x)(6.67 - x)} = 1.95 \times 10^{-4} \]

Step 4: Solve the Equation for x

Expand and rearrange the equation to solve for \( x \): First, approximate that \( x \) is much smaller than the initial pressures, allowing \( 13.3 - x \approx 13.3 \) and \( 6.67 - x \approx 6.67 \).Thus, \[ x^2 \approx 1.95 \times 10^{-4} \times 13.3 \times 6.67 \]Calculate this approximate value as: \[ x^2 \approx 0.01735 \]Taking the square root gives \[ x \approx 0.1316 \text{ kPa} \]

Step 5: Calculate Equilibrium Pressures

Using \( x = 0.1316 \) kPa:\( P_{\mathrm{CH_4,eq}} = 13.3 - 0.1316 = 13.17 \text{ kPa} \)\( P_{\mathrm{I_2,eq}} = 6.67 - 0.1316 = 6.54 \text{ kPa} \)\( P_{\mathrm{CH_3I,eq}} = 0.1316 \text{ kPa} \)\( P_{\mathrm{HI,eq}} = 0.1316 \text{ kPa} \)

Key Concepts

Equilibrium Constant

The equilibrium constant, typically represented as \( K_p \) for reactions involving gases, is crucial for understanding how far a reaction will proceed before reaching equilibrium. It defines the ratio of the product of the partial pressures of the products to the reactants at equilibrium.

Key Points:

• For the reaction \( \mathrm{CH}_{4}(g) + \mathrm{I}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{I}(g) + \mathrm{HI}(g) \), the equilibrium constant expression is:
  \[ K_p = \frac{P_{\mathrm{CH_3I}} \times P_{\mathrm{HI}}}{P_{\mathrm{CH_4}} \times P_{\mathrm{I_2}}} \]
• An important characteristic is that \( K_p \) is temperature-dependent. For instance, if you change the temperature, \( K_p \) changes as well.
• The value of \( K_p \) provides insight into the position of equilibrium:
  • If \( K_p \) is large (>1), products are favored.
  • If \( K_p \) is small (<1), like in the given reaction with \( 1.95 \times 10^{-4} \), reactants are favored.

By rearranging the equilibrium expression, you can predict how changes to the system, such as removing a product or changing pressure, will affect equilibrium.

Partial Pressure

In chemical reactions involving gases, partial pressure plays a significant role and refers to the pressure each gas exerts in a mixture independently. Each component's partial pressure influences the reaction's behavior as per the equilibrium constant expression.

Understanding Partial Pressure:

• It is denoted as \( P_{\text{species}} \) and is calculated based on the mole fraction and total pressure of the gas mixture.
• In the given reaction, we deal with four gases: \( \mathrm{CH_4} \), \( \mathrm{I_2} \), \( \mathrm{CH_3I} \), and \( \mathrm{HI} \). Their initial and equilibrium partial pressures are needed for computation.
  • Initially, \( P_{\mathrm{CH_4}} = 13.3 \) kPa and \( P_{\mathrm{I_2}} = 6.67 \) kPa.
  • Equilibrium partial pressures for products start from zero and change based on \( x \), the shift in pressure due to reaction.
• Partial pressures are affected by factors like volume and temperature changes, which are crucial for calculating equilibrium states.

Understanding these allows you to solve for equilibrium conditions using the given \( K_p \) value and initial pressures.

Le Chatelier's Principle

Le Chatelier's Principle provides insight into how a system at equilibrium responds to external changes. It is a guiding rule for predicting the behavior of a system when its conditions are altered, such as through changes in pressure, temperature, or concentration.

Applying the Principle:

• States that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change.
  • If pressure is increased by decreasing volume, the system shifts towards the side with fewer gas moles.
  • If a reactant or product's concentration increases, the equilibrium shifts in the direction that consumes the added species.
  • Increasing temperature for endothermic reactions shifts equilibrium towards products.
• In the context of the reaction \( \mathrm{CH}_{4}(g) + \mathrm{I}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{I}(g) + \mathrm{HI}(g) \), reducing pressure would shift the equilibrium towards the side with more gas moles, i.e., the reactants.

Using Le Chatelier's Principle, we can make predictions about equilibrium shifts and adapt the initial conditions to favor the formation of desired reactants or products.