Chapter 14: Problem 20
Calculate the equilibrium concentrations of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) at \(25^{\circ} \mathrm{C}\) in a vessel that contains an initial \(\mathrm{N}_{2} \mathrm{O}_{4}\) concentration of \(0.500 \mathrm{M}\). The equilibrium constant \(K_{\mathrm{c}}\) for the reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) is \(4.64 \times 10^{-3}\) at \(25^{\circ} \mathrm{C}\)
Short Answer
Step by step solution
Write the balanced equation
Define initial concentrations
Set up changes in concentration
Express equilibrium concentrations
Substitute into expression for Kc
Solve the equation for x
Solve the quadratic equation using the quadratic formula
Calculate equilibrium concentrations
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equilibrium Concentrations
In the case of the reaction between \(\text{N}_2\text{O}_4\) and \(\text{NO}_2\), the equilibrium concentrations are determined by initially given concentrations and the changes that occur as the system approaches equilibrium. Starting with \(0.500 \, \text{M} \, \text{N}_2\text{O}_4\) and none of the product \(\text{NO}_2\), the concentrations shift until a balance is achieved where \( [\text{N}_2\text{O}_4] = 0.4764 \, \text{M} \) and \( [\text{NO}_2] = 0.0472 \, \text{M} \) at equilibrium.
Understanding equilibrium concentrations helps in predicting how concentrations of substances in a chemical reaction will change when the system reaches equilibrium. It highlights the dynamic nature of chemical equilibrium, as opposed to being a static endpoint.
Equilibrium Constant
For the reaction \(\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)\), the equilibrium constant expression is given by \( K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} \). In this specific exercise, the equilibrium constant was provided as \(4.64 \times 10^{-3}\) at \(25^\circ \text{C}\). This small value of \( K_c \) indicates that at equilibrium, the concentration of the reactants (\(\text{N}_2\text{O}_4\)) is significantly higher than that of the products (\(\text{NO}_2\)).
An equilibrium constant gives insight into the directionality of the reaction and aids in calculating equilibrium concentrations from initial concentrations and known reaction conditions.
ICE Table
- Initial: Begin by writing down the initial concentrations of the reactants and products. In this exercise, we had \( [\text{N}_2\text{O}_4] = 0.500 \, \text{M} \) and \( [\text{NO}_2] = 0 \, \text{M} \).
- Change: Next, determine the change in concentration as the system approaches equilibrium. The change will depend on the stoichiometry of the equation, which in this case means letting the concentration of \(\text{N}_2\text{O}_4\) decrease by \(-x\), and \(\text{NO}_2\) increase by \(+2x\).
- Equilibrium: Finally, express the concentrations at equilibrium based on the initial and change values. Substituting the appropriate values gives \(\text{N}_2\text{O}_4\) \(0.500-x\) and \(\text{NO}_2\) \2x\.
The ICE table is a powerful visualization and calculation strategy for tracking concentration changes and solving equilibrium problems.
Quadratic Equation in Chemistry
The quadratic equation follows the standard form \ax^2 + bx + c = 0\. In our example, substituting known values into the equilibrium expression led to the equation \(4x^2 + 4.64 \times 10^{-3}x - 2.32 \times 10^{-3} = 0\). Solving this required using the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Inserting the coefficients \(a = 4\), \(b = 4.64 \times 10^{-3}\), and \(c = -2.32 \times 10^{-3}\) into the formula yielded a potential value for \(x\): \(x \approx 0.0236\). Only the positive root is viable here, as a concentration cannot be negative.
The quadratic approach helps to precisely determine concentrations at equilibrium and is fundamental in solving many chemical equilibrium puzzles effectively.