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
My work:
.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.
I understand that you are frustrated with this problem, but don't worry, I'm here to help you figure it out. Let's go through the calculations step by step and see if we can identify any errors.
1) The first calculation looks correct. You calculated the heat required to raise the temperature of gaseous ethanol from the boiling point (78.5°C) to the given temperature (351°C). The mass you used, however, seems to be incorrect. The number of moles you provided (0.172 mol) does not correspond to 0.003734 g. To convert moles to grams, you need to multiply by the molar mass of ethanol (C2H6O), which is 46.07 g/mol. So, the correct mass should be:
0.172 mol * 46.07 g/mol = 7.92 g
Now let's recalculate the first step:
q1 = mcΔT = 7.92 g * 1.43 J/g·°C * (351 - 78.5)°C = 1.12516 × 10^3 J
2) In the second step, you correctly calculated the heat required for the phase change from gaseous ethanol to liquid ethanol using the molar heat of vaporization (ΔH°vap). However, you made a mistake in the calculation. The value you used for ΔH°vap is given in kJ/mol, so you need to convert it to J/mol:
ΔH°vap = 40.5 kJ/mol = 40.5 × 10^3 J/mol
Now let's recalculate the second step:
q2 = nΔH°vap = 0.172 mol * 40.5 × 10^3 J/mol = 6.951 × 10^3 J
3) In the third step, you calculated the heat required to lower the temperature of liquid ethanol from 78.5°C to the given temperature (25.0°C). However, you made an error in the mass calculation like in the first step. The mass is not 0.003734 g, it should be:
0.172 mol * 46.07 g/mol = 7.92 g
Now let's recalculate the third step:
q3 = mcΔT = 7.92 g * 2.45 J/g·°C * (78.5 - 25.0)°C = 8.46444 × 10^2 J
Now, let's add up all the heat values:
Total heat = q1 + q2 + q3 = 1.12516 × 10^3 J + 6.951 × 10^3 J + 8.46444 × 10^2 J = 8.233 (to 3 significant figures)
The correct answer to 3 significant figures is 8.23 × 10^3 J (not negative, since it represents the amount of heat needed).
I hope this explanation helps you understand the corrections needed in your calculations and obtain the correct answer. If you have any more questions, feel free to ask!