Ch.14 - Chemical EquilibriumWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: A Student ran the following reaction in the laboratory at 502 K:PCl3(g) + Cl2(g) ⇌ PCl5(g)When she introduced 0.199 moles of PCI3(g) and 0.235 moles of Cl2(g) into a 1.00 liter container, she found th

Solution: A Student ran the following reaction in the laboratory at 502 K:PCl3(g) + Cl2(g) ⇌ PCl5(g)When she introduced 0.199 moles of PCI3(g) and 0.235 moles of Cl2(g) into a 1.00 liter container, she found th

Problem

A Student ran the following reaction in the laboratory at 502 K:

PCl3(g) + Cl2(g) ⇌ PCl5(g)

When she introduced 0.199 moles of PCI3(g) and 0.235 moles of Cl2(g) into a 1.00 liter container, she found the equilibrium concentration of Cl2(g) to be 6.84 x 10-2 M. Calculate the equilibrium constant, Kc, she obtained for this reaction.