Determine the vapor pressure (atm) of rubbing alcohol (isopropanol) at 20.0 °C. The normal boiling point of isopropanol is 82.3
1 answer:
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Explanation:
The initial concentrations for a mixture :
Acetic acid at equilibrium = 0.15 M
Ethanol at equilibrium = 0.15 M
Ethyl acetate at equilibrium = 0.40 M
Water at equilibrium = 0.40 M
Initially:
0.15 M 0.15 M 0.40 M 0.40 M
At equilibrium
(0.15-x)M (0.15-x) M (0.40+x) M (0.40+x) M
The equilibrium constant is given by expression
Solving for x:
x = 0.0333
The equilibrium concentrations for a mixture :
Acetic acid at equilibrium = (0.15-x)M = (0.15-0.033) M = 0.117 M
Ethanol at equilibrium = (0.15-x)M = (0.15-0.033) M = 0.117 M
Ethyl acetate at equilibrium = (0.40+x)M = (0.40+0.033) M = 0.433 M
Water at equilibrium = (0.40+x)M = (0.40+0.033) M = 0.433 M