Question
Asked By ObsidianSkies99 at
Answered By Expert
Shawn
Expert · 3.7k answers · 3k people helped
Step 1/5
Moles of `Fe3+` initially: `2.00×10−3mol/L×0.005L=1.00×10−5mol`Moles of `SCN` − initially: `2.00×10−3mol/L×0.003L=6.00×10−6mol`Explanation:
Here is the Calculate Initial Moles
Step 2/5
The number of Moles of `FeSCN2+` at equilibrium: `7.0×10−3mol/L×0.01L=7.00×10−5mol`Moles of `Fe3+` and `SCN− ` used up: `7.00×10−7mol`Explanation:
Here we Determine Equilibrium Moles
Step 3/5
Remaining moles of `Fe3+: ` `1.00×10−5−7.00×10−7=9.3×10−6mol`Remaining moles of `SCN−` `6.00×10−6−7.00×10−7=5.3×10−6mol`Explanation:
Here we Calculate Remaining Moles
Step 4/5
Concentration of `Fe3+: 0.0129.3×10−6=7.75×10−4M` Concentration of `SCN−: 0.0125.3×10−6=4.42×10−4M` Concentration of `FeSCN2+: 7.00×10−5/0.012=5.83×10−3M`Explanation:
Here we Determine Equilibrium Concentrations
Step 5/5
`Kc=[Fe3+]⋅[SCN−][FeSCN2+]=` `(7.e75×10−4)⋅(4.42×10−4)5.83×10−3` `≈142.01`Therefore, the value of `Kc≈142.01.`Explanation:
Here we Calculate Equilibrium Constant `(Kc)`Final Answer
Here the Determination of the Equilibrium Constant for a Chemical Reaction
Therefore, `K c ≈142.01.`🧑🏫 More Questions
👉 Interested in exploring further?
Chrome Extension
1. Search answers from our 90+ million questions database.
2. Get instantly AI Solutions powered by most advanced models like GPT-4, Bard, Math GPT, etc.
3. Enjoy one-stop access to millions of textbook solutions.
4. Chat with 50+ AI study mates to get personalized course studies.
5. Ask your questions simply with texts or screenshots everywhere.