Consider the evaporation of methanol at 25.0 ?C : CH3OH(l)?CH3OH(g) . Why methanol spontaneously evaporates in open...

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Chemistry

Consider the evaporation of methanol at 25.0 ?C :
CH3OH(l)?CH3OH(g) .
Why methanol spontaneously evaporates in open air at 25.0 ?C :Methanol evaporates at room temperature because there is anequilibrium between the liquid and the gas phases. The vaporpressure is moderate (143 mmHg at 25.0 degrees centigrade), so amoderate amount of methanol can remain in the gas phase, which isconsistent with the free energy values.
A.) Find ?G? at 25.0 ?C .
B.) Find ?G at 25.0 ?C under the following nonstandardconditions:
(i)PCH3OH= 158.0mmHg
C). Find ?G at 25.0 ?C under the following nonstandardconditions:
(ii) PCH3OH= 110.0mmHg
D.) Find ?G at 25.0 ?C under the following nonstandardconditions:
(iii) PCH3OH= 11.0mmHg

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3.8 Ratings (500 Votes)

the given reaction is Ch3OHl Ch3OH g A we know that dGo dGo products dGo reactants so dGo dGo Ch3OHg dGo CH3OHl dGo 1623 1666 dGo 43 kJmol B now See Answer

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