posted by Lizichka .
From data below, calculate the total heat (J) needed to convert 0.172 mol of gaseous ethanol (C2H6O) at 351° C and 1 atm to liquid ethanol at 25.0° C and 1 atm.
Boiling point at 1 atm 78.5°C
c gas 1.43 J/g · °C
c liquid 2.45 J/g · °C
ΔH°vap 40.5 kJ/mol
.172 mol = .003734 g
1) q = mc(T2-T1)= (.003734)(1.43)(351-78.5) = 1.455 J
2) q = n*delta Hvap =.172*40500 = 6966J
3) q = mc(T2-T1) = (.003734)(2.45)(78.5-25) = .4894 J
-------------- adding all the q together gives me -6.97e3 J, its negative since im going from g to l. Its also 3 sig figs. The computer tell me my answer is off by 10%. So what am I doing wrong. Someone please tell me or I will become insane and fail my chem test then drop out of college which will lead me to become homeless, all because of this stupid problem!
Is 0.172 mol ethanol really 0.003734 g? Isn't it 0.172mol x 46 g/mol = 7.912 g?
If you still have a problem, post the boiling point and freezing points of ethanol. I wouldn't want you to become homeless. ;-)
Your plight reminds me of this old proverb:
"For Want of a Nail
For want of a nail the shoe was lost.
For want of a shoe the horse was lost.
For want of a horse the rider was lost.
For want of a rider the message was lost.
For want of a message the battle was lost.
For want of a battle the kingdom was lost.
And all for the want of a horseshoe nail."
And for want of an answer the home was lost.