Question
Explain why reactions with large equilibrium constants, such as the formation of rust $\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right),$ may have very slow rates.
Step 1
This can be represented as $K = \frac{k_{1}}{k_{2}}$. Show more…
Show all steps
Your feedback will help us improve your experience
Natalie Almond and 69 other Chemistry 102 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Explain why reactions with large equilibrium constants, such as the formation of rust $\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)$, may have very slow rates.
Explain how it is possible for a reaction to have a large equilibrium constant but small forward and reverse rate constants.
Why is rust (Fe2O3) not soluble in water? Why does it not dissolve with the like-dissolve-like theory? Explain using intermolecular forces.
Transcript
600,000+
Students learning Chemistry with Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD
