Problem 20 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

Nitrogen dioxide gas is heated in a sealed container at \(700 \mathrm{K}\) until the system comes to equilibrium. The nitrogen dioxide dissociates into nitrogen monoxide and oxygen in an endothermic process (Section \(1.9)\) $$2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})$$ The equilibrium constant at \(700 \mathrm{K}\) is \(2.78 \times 10^{-2} \mathrm{moldm}^{-3}\) (a) Write an expression for \(K_{c}\) (b) State how the position of equilibrium would be affected by: (1) an increase in temperature (ii) an increase in the total pressure. (c) At equilibrium at \(700 \mathrm{K}\), the concentration of nitrogen monoxide was found to be \(0.017 \mathrm{moldm}^{-3}\). What was the concentration of nitrogen dioxide in the equilibrium mixture?

Step-by-Step Solution

Answer

(a) \(K_c = \frac{[\text{NO}]^2[\text{O}_2]}{[\text{NO}_2]^2}\); (b) Shift right with increased temperature, left with increased pressure; (c) \([\text{NO}_2] \approx 0.0093 \ \mathrm{moldm}^{-3}\).

Step 1: Writing the expression for Kc

The equilibrium constant expression, \( K_c \), for the reaction \( 2\text{NO}_2(g) \leftrightharpoons 2\text{NO}(g) + \text{O}_2(g) \) is given by the formula: \[ K_c = \frac{[\text{NO}]^2 [\text{O}_2]}{[\text{NO}_2]^2} \] Where \([\text{NO}]\), \([\text{O}_2]\), and \([\text{NO}_2]\) are the equilibrium concentrations of nitrogen monoxide, oxygen, and nitrogen dioxide, respectively.

Step 2: Effect of Temperature on Equilibrium Position

According to Le Chatelier's Principle, if the temperature is increased for an endothermic reaction, the equilibrium position will shift to the right to absorb more heat. This means more reactants will convert to products when the temperature rises.

Step 3: Effect of Pressure on Equilibrium Position

For the given reaction, an increase in total pressure will favor the side with the fewer moles of gas. The reaction \( 2\text{NO}_2 \leftrightharpoons 2\text{NO} + \text{O}_2 \) involves a change from 2 moles of \(\text{NO}_2\) to a total of 3 moles (2 moles of \(\text{NO}\) and 1 mole of \(\text{O}_2\)). Therefore, increasing the pressure will shift the equilibrium to the left, favoring the formation of \(\text{NO}_2\).

Step 4: Calculating [NO2] at Equilibrium

Given that \( K_c = 2.78 \times 10^{-2} \ \mathrm{moldm}^{-3} \) and \([\text{NO}] = 0.017 \ \mathrm{moldm}^{-3}\), use the equilibrium expression: \[ K_c = \frac{[\text{NO}]^2 \cdot [\text{O}_2]}{[\text{NO}_2]^2} \] Substituting the values gives: \[ 2.78 \times 10^{-2} = \frac{(0.017)^2 \cdot [\text{O}_2]}{[\text{NO}_2]^2} \] To find \([\text{NO}_2] \), rearrange and solve: \[ [\text{NO}_2]^2 = \frac{(0.017)^2 \cdot [\text{O}_2]}{2.78 \times 10^{-2}} \] Assuming that the change in \([\text{O}_2]\) due to equilibrium shift is small for this calculation, \([\text{O}_2]\) can be approximated from the equilibrium stoichiometry: \([\text{O}_2] = \frac{1}{2} [\text{NO}] = \frac{0.017}{2} = 0.0085\ \mathrm{moldm}^{-3}\). Then, solve for \([\text{NO}_2]\): \[ [\text{NO}_2]^2 = \frac{(0.017)^2 \cdot 0.0085}{2.78 \times 10^{-2}} \] \[ [\text{NO}_2]^2 \approx \frac{2.89 \times 10^{-4} \cdot 0.0085}{2.78 \times 10^{-2}} \] \[ [\text{NO}_2]^2 \approx 8.58 \times 10^{-5} \] \[ [\text{NO}_2] \approx \sqrt{8.58 \times 10^{-5}} \] \[ [\text{NO}_2] \approx 0.0093 \ \mathrm{moldm}^{-3} \]

Key Concepts

Le Chatelier's Principle

Le Chatelier's Principle helps us understand how a chemical equilibrium will respond to changes in conditions such as temperature, pressure, or concentration. This principle is key in predicting how a reaction at equilibrium will behave if subjected to external changes.
When we talk about temperature changes, Le Chatelier's Principle tells us if we increase the temperature of an endothermic reaction, the equilibrium will shift to create more products. This happens because the system needs to absorb that extra heat, which it does by favoring the reaction that consumes heat. In the case of our nitrogen dioxide example:

• When the temperature increases, more heat is available.
• The reaction shifts right (towards nitrogen monoxide and oxygen formation) to absorb the extra heat.
• This results in more products being formed.

This movement of the equilibrium position ensures that some of this additional heat energy is absorbed, balancing out the change.

Equilibrium Constant

The Equilibrium Constant (\(K_c\)) is a fundamental concept in chemical equilibrium, representing the ratio of product concentrations to reactant concentrations at equilibrium, with each raised to the power of their coefficients in the balanced chemical equation. It is unique for every reaction at a given temperature.
For our reaction of nitrogen dioxide dissociating to nitrogen monoxide and oxygen:

• The expression for\(K_c\) is written as\(K_c = \frac{[\text{NO}]^2 \cdot [\text{O}_2]}{[\text{NO}_2]^2}\)
• The value of\(2.78 \times 10^{-2} \ \mathrm{mol/dm}^3\) indicates the extent of the reaction at equilibrium at\(700 \ \mathrm{K}\)
• If\(K_c\) is large, it means a greater formation of products at equilibrium; if small, more reactants will be present.

This constant depends on the temperature and will change if the temperature changes. However, it remains constant when the concentration or pressure changes at a fixed temperature.

Endothermic Reaction

An endothermic reaction is one that absorbs heat from its surroundings. This is an essential element in understanding chemical reactions that require energy to proceed.
For endothermic reactions like the one involving nitrogen dioxide:

• These reactions absorb heat; energy is needed as a reactant.
• Increasing the temperature supplies additional energy, encouraging the shift towards product formation as heat is absorbed.
• This is different from exothermic reactions which release heat.

Understanding whether a reaction is endothermic or exothermic helps predict how temperature changes will affect reaction behavior. In our nitrogen dioxide example, as it is endothermic, heating the system shifts equilibrium towards more nitrogen monoxide and oxygen formation.