equilibrium summary

Question

5.00 moles of CH₂H₅OH were mixed in a container. What is the equilibrium concentration of water under these conditions if K_c = 4.50.

CH₃CO₂H (g) + C₂H₅OH (g) => CH₃CO₂C₂H₅ (g) + H₂O (g)

Answer 1

Am assuming you meant to write that 5.00 moles of each reactant were mixed in a container. Also, not knowing the volume of the container, it is assumed to be 1 liter. Otherwise, we can't answer this question as written. You may want to go back and make sure you copied it correctly. No such thing as CH₂H₅OH.

CH₃CO₂H (g) + C₂H₅OH (g) => CH₃CO₂C₂H₅ (g) + H₂O (g)

5.00......................5.00......................0........................0............Initial

-x..........................-x.........................+x.......................+x..........Change

5-x........................5-x.........................x.........................x...........Equilibrium

Kc = [CH₃CO₂C₂H₅ ] [H₂O] / [CH₃CO₂H] [C₂H₅OH]

4.50 = (x)(x) / (5-x)(5-x) and assuming x is small, we can simplify to ...

4.50 = x² / (5)(5) = x²/25

x² = 112.5

x = 10.6 M = [H₂O]