Chapter 16: Problem 16
The equilibrium constant for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}_{2}(\mathrm{g}) $$ at \(25^{\circ} \mathrm{C}\) is \(5.88 \times 10^{-3} .\) Suppose \(15.6 \mathrm{g}\) of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is placed in a \(5.00-\) L. flask at \(25^{\circ}\) C. Calculate the following: (a) the amount of \(\mathrm{NO}_{2}(\mathrm{mol})\) present at equilibrium (b) the percentage of the original \(\mathrm{N}_{2} \mathrm{O}_{4}\) that is dissociated
Short Answer
Step by step solution
Determine Initial Concentrations
Define Equilibrium Expressions
Apply the Equilibrium Constant Expression
Solve for \( x \)
Find Moles of \( \mathrm{NO}_2 \) at Equilibrium
Calculate Percentage of \( \mathrm{N}_2\mathrm{O}_4 \) Dissociation
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equilibrium Constant
- \( K_c = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]} \)
Reaction Quotient
- \( Q_c = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]} \)
Mole Calculation
- \( n = \frac{\text{mass}}{\text{molar mass}} = \frac{15.6 \ \text{g}}{92.02 \ \text{g/mol}} = 0.1696 \ \text{mol} \)
Dissociation Percentage
- \( \text{Dissociation percentage} = \left( \frac{0.00626 \ \text{mol}}{0.1696 \ \text{mol}} \right) \times 100\% \approx 3.69\% \)