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: In air at 25°C and 1.00 atm, the N2 concentration is 0.01 M and the O2 concentration is 0.033 M. The reactionN2(g) + O2(g) ⇌ 2N0(g) has Kc = 4.8 x 10-31 at 25°C.Taking the N2 and O2 concentrations giv

Problem

In air at 25°C and 1.00 atm, the N2 concentration is 0.01 M and the O2 concentration is 0.033 M. The reaction

N2(g) + O2(g) ⇌ 2N0(g) 

has Kc = 4.8 x 10-31 at 25°C.

Taking the N2 and O2 concentrations given above as initial values, calculate the equilibrium NO concentration that should exist in our atmosphere from this reaction at 36°C, Kc = 6.4E-30. 

Make simplifying assumptions in your calculations. (Express your answer in scientific notation.)