Ch.14 - Chemical EquilibriumWorksheetSee all chapters
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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: For the gas phase reaction N2 + 3 H2 ⇌ 2 NH3 ΔH° = -92 kJ for the forward reaction. In order to decrease the yield of NH3, the reaction should be runa. at high P, low T.b. at low P, high T.c. at high

Problem

For the gas phase reaction N2 + 3 H2 ⇌ 2 NH3 ΔH° = -92 kJ for the forward reaction. In order to decrease the yield of NH3, the reaction should be run

a. at high P, low T.
b. at low P, high T.
c. at high P, high T.
d. at low P, low T.

Solution

We’re being asked to identify the condition in order to decrease the yield of NH3 for the gas-phase reaction:

N2 + 3 H2 ⇌ 2 NH3          ΔH° = -92 kJ



According to Le Chatelier’s Principle, if a system (chemical reaction) is at equilibrium and we disturb it, then the system will readjust to maintain its equilibrium state.



Since the reaction is exothermic, ΔH = (–), heat is given off as a product:

N2(g) + 3 H2(g) ⇌ 2 NH3(g) + heat


Now, let’s look at the effect of changing pressure and temperature on the reaction.


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