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: An equilibrium is established according to the following equationHg2 2+(aq) + NO3 −(aq) + 3H+(aq) ⇌ 2Hg2+(aq) + HNO2(aq) + H2 O(l)    Kc = 4.6What will happen in a solution that is 0.20 M each in Hg 2

Problem

An equilibrium is established according to the following equation
Hg2 2+(aq) + NO(aq) + 3H+(aq) ⇌ 2Hg2+(aq) + HNO2(aq) + H2 O(l)    K= 4.6
What will happen in a solution that is 0.20 M each in Hg 2+, NO, H+, Hg2+, and HNO2?

(a) Hg2+ will be oxidized and NO reduced.
(b) Hg2 2+ will be reduced and NO3 oxidized.
(c) Hg2+ will be oxidized and HNO2 reduced.
(d) Hg2+ will be reduced and HNO2 oxidized.
(e) There will be no change because all reactants and products have an activity of 1.