Table of contents

1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 38m
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

Equilibrium Constant K

General Chemistry

16. Chemical Equilibrium

Equilibrium Constant K

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Multiple Choice

At T = 700 K, the equilibrium constant for the reaction CCl4(g) ⇌ C(s) + 2 Cl2(g) is K_p = 0.76. A flask is charged with 3.0 atm of CCl4(g) at 700 K, which is then allowed to reach equilibrium. What is the partial pressure of Cl2(g) in the flask?

1.52 atm

3.04 atm

0.76 atm

2.28 atm

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Verified step by step guidance

Identify the balanced chemical equation: CCl4(g) ⇌ C(s) + 2 Cl2(g). Note that C(s) is a solid and does not appear in the expression for the equilibrium constant K_p.

Write the expression for the equilibrium constant K_p in terms of partial pressures: K_p = (P_{Cl2}^2) / (P_{CCl4}).

Let x be the change in pressure of CCl4 that dissociates at equilibrium. Therefore, the change in pressure for Cl2 will be 2x, since 2 moles of Cl2 are produced for every mole of CCl4 that dissociates.

Set up the equilibrium expressions: P_{CCl4} = 3.0 - x and P_{Cl2} = 2x.

Substitute these expressions into the K_p expression: 0.76 = (2x)^2 / (3.0 - x). Solve this equation for x to find the equilibrium partial pressure of Cl2, which is 2x.