Asked by JP
Consider the evaporation of methanol at 25.0 c: CH3OH(l)-- CH3OH(g)
I figured out that Delta G is 4.3 kJ.
Find the Delta G at 25 C under the following conditions:
Pressure of CH3OH= 157mmHg, 103mmHg, 14mmHg
thanks! please explain
I figured out that Delta G is 4.3 kJ.
Find the Delta G at 25 C under the following conditions:
Pressure of CH3OH= 157mmHg, 103mmHg, 14mmHg
thanks! please explain
Answers
Answered by
Anonymous
Thank you for finding the delta G for me so here is how to solve this problem,
Formula: Delta G(rxn)= DeltaG(rxn)^degree + RTlnQ
1 atm = 760 mmHg
4300J + 8.314*298K*ln(157/760)=
tyvm again.
Formula: Delta G(rxn)= DeltaG(rxn)^degree + RTlnQ
1 atm = 760 mmHg
4300J + 8.314*298K*ln(157/760)=
tyvm again.
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