Chemistry HL
Chemistry HL
6
Chapters
243
Notes
Chapter 1 - Models Of The Particulate Nature Of Matter
Chapter 1 - Models Of The Particulate Nature Of Matter
Chapter 2 - Models Of Bonding & Structure
Chapter 2 - Models Of Bonding & Structure
Chapter 3 - Classification Of Matter
Chapter 3 - Classification Of Matter
Chapter 4 - What Drives Chemical Reactions?
Chapter 4 - What Drives Chemical Reactions?
Chapter 5 - How Much, How Fast & How Far?
Chapter 5 - How Much, How Fast & How Far?
Chapter 6 - What Are The Mechanisms Of Chemical Change?
Chapter 6 - What Are The Mechanisms Of Chemical Change?
IB Resources
Chapter 5 - How Much, How Fast & How Far?
Chemistry HL
Chemistry HL

Chapter 5 - How Much, How Fast & How Far?

Unlocking Gibbs Energy: Equilibrium Secrets!

Word Count Emoji
397 words
Reading Time Emoji
2 mins read
Updated at Emoji
Last edited on 5th Nov 2024

Table of content

Hello young chemists! 🧪 Today, let’s have a blast learning about Gibbs energy and equilibrium. We’re going to keep it fun, lively, and simple! 🎉

Key Concepts 🌟

  • Equilibrium Constant (K)
    • If >1K>1: Forward reaction is spontaneous 🚀
    • If <1K<1: Reverse reaction is favoured 🔄
    • If =1K=1: Equilibrium is established—no reaction is favoured! 🤷
  • Gibbs Energy (ΔG)
    • Negative ΔG: Forward reaction is spontaneous 🚀
    • Positive ΔG: Reverse reaction is favoured 🔄
  • Standard Gibbs Energy (Δ�∘ΔG∘)
    • Refers to changes from reactants to products under standard conditions.
    • Standard Conditions: p = 100kPa for gases and c=1moldm−3 for aqueous species at 298K.

Real-World Example: Dinitrogen Dioxide 🔍

Dinitrogen dioxide, N2​O2​, is so unstable, it’s like a house of cards 🏠💨! Figuring out its Gibbs energy change of formation, ∘ΔGf∘​, directly is a no-go. But don’t worry, we can use the equilibrium constant, K, and a bit of chemistry magic to find it out!

👩‍🔬 Worked Example

  • Given: 2NO(g) ⇌ N2​O2​(g) with  = 1.39 ×10−5 at 298K.
  • To Find Δ∘
    • Use: ΔG∘= − RTlnK
    • Substitute values and remember, R = 8.31JK−1mol−1
    • Solution: ΔG∘ = 27.7kJmol−1
  • To Find ∘ΔGf∘​ forN2​O2​(g)
    • Use the given ∘ΔGf∘​ for NO(g) which is 87.6 kJmol−1.
    • Solution: ∘(NO)=202.9 kJmol−1ΔGf∘​(NO)=202.9kJmol−1

Mathematical Magic with GDC 📐

  • A Graphic Display Calculator (GDC) can help you with solving equations, especially when finding equilibrium constants.
  • Example: If you have a reaction with  ΔG∘ = − 0.82kJmol−1 at 298K, use your GDC to find K by solving  ΔG∘ = − RTlnK.

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IB Resources
Chapter 5 - How Much, How Fast & How Far?
Chemistry HL
Chemistry HL

Chapter 5 - How Much, How Fast & How Far?

Unlocking Gibbs Energy: Equilibrium Secrets!

Word Count Emoji
397 words
Reading Time Emoji
2 mins read
Updated at Emoji
Last edited on 5th Nov 2024

Table of content

Hello young chemists! 🧪 Today, let’s have a blast learning about Gibbs energy and equilibrium. We’re going to keep it fun, lively, and simple! 🎉

Key Concepts 🌟

  • Equilibrium Constant (K)
    • If >1K>1: Forward reaction is spontaneous 🚀
    • If <1K<1: Reverse reaction is favoured 🔄
    • If =1K=1: Equilibrium is established—no reaction is favoured! 🤷
  • Gibbs Energy (ΔG)
    • Negative ΔG: Forward reaction is spontaneous 🚀
    • Positive ΔG: Reverse reaction is favoured 🔄
  • Standard Gibbs Energy (Δ�∘ΔG∘)
    • Refers to changes from reactants to products under standard conditions.
    • Standard Conditions: p = 100kPa for gases and c=1moldm−3 for aqueous species at 298K.

Real-World Example: Dinitrogen Dioxide 🔍

Dinitrogen dioxide, N2​O2​, is so unstable, it’s like a house of cards 🏠💨! Figuring out its Gibbs energy change of formation, ∘ΔGf∘​, directly is a no-go. But don’t worry, we can use the equilibrium constant, K, and a bit of chemistry magic to find it out!

👩‍🔬 Worked Example

  • Given: 2NO(g) ⇌ N2​O2​(g) with  = 1.39 ×10−5 at 298K.
  • To Find Δ∘
    • Use: ΔG∘= − RTlnK
    • Substitute values and remember, R = 8.31JK−1mol−1
    • Solution: ΔG∘ = 27.7kJmol−1
  • To Find ∘ΔGf∘​ forN2​O2​(g)
    • Use the given ∘ΔGf∘​ for NO(g) which is 87.6 kJmol−1.
    • Solution: ∘(NO)=202.9 kJmol−1ΔGf∘​(NO)=202.9kJmol−1

Mathematical Magic with GDC 📐

  • A Graphic Display Calculator (GDC) can help you with solving equations, especially when finding equilibrium constants.
  • Example: If you have a reaction with  ΔG∘ = − 0.82kJmol−1 at 298K, use your GDC to find K by solving  ΔG∘ = − RTlnK.

Unlock the Full Content! File Is Locked Emoji

Dive deeper and gain exclusive access to premium files of Chemistry HL. Subscribe now and get closer to that 45 🌟

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