Table of contents

1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 36m
5. BONUS: Mathematical Operations and Functions48m
6. Chemical Quantities & Aqueous Reactions3h 53m
7. Gases3h 49m
8. Thermochemistry2h 37m
9. Quantum Mechanics2h 58m
10. Periodic Properties of the Elements3h 9m
11. Bonding & Molecular Structure3h 29m
12. Molecular Shapes & Valence Bond Theory1h 57m
13. Liquids, Solids & Intermolecular Forces2h 23m
14. Solutions3h 1m
15. Chemical Kinetics2h 53m
16. Chemical Equilibrium2h 29m
17. Acid and Base Equilibrium5h 1m
18. Aqueous Equilibrium4h 47m
19. Chemical Thermodynamics1h 50m
20. Electrochemistry2h 42m
21. Nuclear Chemistry2h 36m
22. Organic Chemistry5h 7m
23. Chemistry of the Nonmetals2h 39m
24. Transition Metals and Coordination Compounds3h 16m

16. Chemical Equilibrium

ICE Charts

General Chemistry

16. Chemical Equilibrium

ICE Charts

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Multiple Choice

At a given temperature the gas phase reaction:N₂ (g) + O₂ (g) ⇌ 2 NO (g) has an equilibrium constant of 4.00 x 10 ^-15. What will be the concentration of NO at equilibrium if 2.00 moles of nitrogen and 6.00 moles oxygen are allowed to come to equilibrium in a 2.0 L flask.

1.10 x 10^-7 M

5.48 x 10 ^-7 M

5.48 x 10 ^-8 M

1.10 x 10^-8 M

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Verified step by step guidance

First, write the balanced chemical equation for the reaction: \( \text{N}_2 (g) + \text{O}_2 (g) \rightleftharpoons 2 \text{NO} (g) \).

Calculate the initial concentrations of \( \text{N}_2 \) and \( \text{O}_2 \) using the formula \( \text{Concentration} = \frac{\text{moles}}{\text{volume}} \). For \( \text{N}_2 \), it is \( \frac{2.00 \text{ moles}}{2.0 \text{ L}} = 1.00 \text{ M} \). For \( \text{O}_2 \), it is \( \frac{6.00 \text{ moles}}{2.0 \text{ L}} = 3.00 \text{ M} \).

Set up an ICE (Initial, Change, Equilibrium) table to determine the changes in concentration as the system reaches equilibrium. Let \( x \) be the change in concentration of \( \text{NO} \) formed at equilibrium.

Express the equilibrium concentrations in terms of \( x \): \( [\text{N}_2] = 1.00 - x \), \( [\text{O}_2] = 3.00 - x \), and \( [\text{NO}] = 2x \).

Use the equilibrium constant expression \( K_c = \frac{[\text{NO}]^2}{[\text{N}_2][\text{O}_2]} \) and substitute the equilibrium concentrations into this expression. Set \( K_c = 4.00 \times 10^{-15} \) and solve for \( x \) to find the concentration of \( \text{NO} \) at equilibrium.