WJEC A-Level Chemistry 1410QS
Formulae and Equations
Basic Ideas About Atoms
Chemical Calculations
Bonding
Solid Structures
The Periodic Table
Simple Equilibria and Acid-Base Reactions
Thermochemistry
Rates of Reaction
The Wider Impact of Chemistry
Organic Compounds
Hydrocarbons
Halogenoalkanes
Alcohols and Carboxylic Acids
Instrumental Analysis
Redox and Standard Electrode Potential
Redox Reactions
Chemistry of the P-Block
Chemistry of the D-Block Transition Metals
Chemical Kinetics
Enthalpy Changes for Solids and Solutions
Entropy and Feasibility of Reactions
Equilibrium Constants
Acid-Base Equilibria
Stereoisomerism
Aromaticity
Alcohols and Phenols
Aldehydes and Ketones
Carboxylic Acids and Their Derivatives
Amines
Amino Acids, Peptides and Proteins
Organic Synthesis and Analysis
Practical Skills and Techniques
Planning, Analysis and Evaluation
Core Experiments and Their Chemistry
Mathematical Skills in Chemistry
Equilibrium Constants
Partial pressures and Kp calculations
- 3 things to know
- A2 Unit 3
Overview
Recall questions on Partial pressures and Kp calculations.
What you need to know
Answer each one before you reveal it.
Why must the total number of moles used to find mole fractions include every gas present, including any that takes no part in the reaction?
In an equilibrium mixture the partial pressures are 40 kPa of NO₂ and 20 kPa of N₂O₄. Calculate Kp for N₂O₄(g) ⇌ 2NO₂(g).
Questions from this subtopic
Each one with the model answer underneath.
- 1
Set out the steps for calculating Kp from the numbers of moles at equilibrium and the total pressure.
Model answer
First add the moles of all the gases to obtain the total. Then divide the moles of each gas by that total to obtain its mole fraction. Multiply each mole fraction by the total pressure to obtain that gas's partial pressure. Finally substitute these partial pressures into the Kp expression and derive the units.
- 2
An equilibrium mixture contains 2.0 mol N₂, 6.0 mol H₂ and 2.0 mol NH₃ at a total pressure of 100 atm. Calculate Kp for N₂(g) + 3H₂(g) ⇌ 2NH₃(g).
Model answer
The total is 10.0 mol, so the mole fractions are 0.20, 0.60 and 0.20, giving partial pressures of 20 atm for N₂, 60 atm for H₂ and 20 atm for NH₃. Then Kp = (20)² / (20 × (60)³) = 400 / (20 × 216000) = 400 / (4.32 × 10⁶) = 9.26 × 10⁻⁵ atm⁻².
- 3
An equilibrium mixture contains 0.50 mol SO₂, 0.25 mol O₂ and 1.50 mol SO₃ at a total pressure of 4.50 atm. Calculate Kp for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g).
Model answer
The total is 2.25 mol, so the partial pressures are (0.50/2.25) × 4.50 = 1.00 atm for SO₂, (0.25/2.25) × 4.50 = 0.50 atm for O₂ and (1.50/2.25) × 4.50 = 3.00 atm for SO₃, which sum correctly to 4.50 atm. Then Kp = (3.00)² / ((1.00)² x 0.50) = 9.00/0.50 = 18.0 atm⁻¹.
- 4
An equilibrium mixture contains 0.20 mol H₂, 0.20 mol I₂ and 1.60 mol HI at a total pressure of 10.0 atm. Calculate Kp for H₂(g) + I₂(g) ⇌ 2HI(g).
Model answer
The total is 2.00 mol, so the partial pressures are 1.00 atm for H₂, 1.00 atm for I₂ and 8.00 atm for HI. Then Kp = (8.00)² / (1.00 × 1.00) = 64.0, and since there are two moles of gas on each side the units cancel, so Kp has no units.
- 5
One mole of dinitrogen tetraoxide is heated until 40% has dissociated, at a total pressure of 1.00 atm. Calculate Kp for N₂O₄(g) ⇌ 2NO₂(g).
Model answer
Dissociation of 0.40 mol leaves 0.60 mol of N₂O₄ and forms 0.80 mol of NO₂, a total of 1.40 mol. The partial pressures are (0.60/1.40) × 1.00 = 0.4286 atm and (0.80/1.40) × 1.00 = 0.5714 atm. Then Kp = (0.5714)² / 0.4286 = 0.3265/0.4286 = 0.762 atm.
- 6
One mole of phosphorus pentachloride is heated until 25% has dissociated, at a total pressure of 2.00 atm. Calculate Kp for PCl₅(g) ⇌ PCl₃(g) + Cl₂(g).
Model answer
Dissociation of 0.25 mol leaves 0.75 mol of PCl₅ and forms 0.25 mol each of PCl₃ and Cl₂, a total of 1.25 mol. The partial pressures are (0.75/1.25) × 2.00 = 1.20 atm for PCl₅ and (0.25/1.25) × 2.00 = 0.40 atm for each of the products. Then Kp = (0.40 × 0.40)/1.20 = 0.16/1.20 = 0.133 atm.
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