Sapling HW Week 5&6 Question #7

[idella 1D]

Hello all. I'm having a bit of difficulty in trying to solve question #7 from the sapling homework. I would really appreciate a step-by-step approach since I think that all the tiny things we have to do are tripping me up.

Three liquid samples of known masses are heated to their boiling points with the use of a heater rated at 500.0 W. Once the boiling points of each sample are reached, the samples are heated for an additional 4.02 min, which results in the vaporization of some of each sample. After 4.02 min, the samples are cooled and the masses of the remaining liquids are determined. The process is performed at constant pressure. The results are recorded in the table.

Liquid Boiling point (°C) Initial mass (g) Final mass (g)
CH3OH 64.5 416.0 314.84
C4H10 −1.00 438.6 114.04
C3H6O 56.2 532.1 298.62

Calculate the molar enthalpy of vaporization, ΔHvap, and the molar entropy of vaporization, ΔSvap, for each sample. Assume that all of the heat from the heater goes into the sample.

ΔHvap of CH3OH:

ΔSvap of CH3OH:

ΔHvap of C4H10:

ΔSvap of C4H10:

ΔHvap of C3H6O:

ΔSvap of C3H6O: ??

[Edward Tang 1k]

For calculating the molar enthalpy of vaporization you use q=nC, where n is the mole of element vaporized and C is the enthalpy of vaporization, the value we are looking for. We know what the value of q is, from the fact that a 500W heater is used for 4.02 minutes during the phase change. So q would be 500*54.02*60 since 1W=1J/S so you need to convert minutes to seconds. You also know the mass of each element vaporized from calculating initial minus final mass, and divide that by the molar mass you get the number of moles vaporized . Last step is just to do q/n to solve for C. Note C is in Kj/mol so you need to multiply you answer by 10*-3.

Once you have the enthalpy, you just do delta S = q/T= delta H/T, which means dividing each enthalpy of vaporization in joules by their corresponding boiling points in kelvin

[Shruti Kulkarni 2I]

The enthalpy for the reaction is given by the heater, which is running at 500 W (J/s). So, you would convert the W into J by multiplying by the 4.02 minutes over which each compound boils, as that would reference the time in which the compound is vaporized. This is the enthalpy in the reaction.
Then, for each compound, find how much of it vaporized by doing final - initial g. This is the amount that has vaporized in that time. Convert this value to moles.
Now that you have the amount vaporized and the enthalpy that caused it to vaporize, you can divide the enthalpy by the number of moles to determine the molar enthalpy of vaporization. Divide that by 1000 to put it in kJ/mole.
You can find the entropy by dividing the molar enthalpy by the boiling point, in Kelvins.

Hope this helped!

[Francesca_Borchardt_2D]

Hello,
To calculate the molar enthalpy of vaporization, ΔHvap, the equation q=nC is used. In the equation q is equal to the amount of absorbed heat, n is the amount of moles vaporized, and C is the enthalpy of vaporization. For the q value, we can find this because the problem states that the heater is rated at 500.0 W. We need to convert this to joules per minute and since 1W=1J/S we can get the value of q by multiplying 500W by the amount of time which in this case is 4.02 min and then multiply by 60 seconds. So, 500W * 4.02 * 60 gives us the value of q.
Next, in order to find the amount of moles vaporized, we take the initial mass - final mass and we divide that by the molar mass of the element. This gives us the n value. So to find C, which is the molar enthalpy of vaporization, we plug in the value of q and n we found into the q=nC equation. q/c gives the molar enthalpy of vaporization.

To find the molar entropy of vaporization, we can use the equation deltaS= deltaH/T. T is the boiling point of the element. Temperature must be in kelvins so we have to convert the boiling point in celsius to kelvin and we do this simply by adding 273. Next, we found the deltaH in previous question, so in order to find deltaS, we divide deltaH by the boiling point in kelvins.

I hope this helps!

[idella 1D]

thank you!!

[WS405590915]

One tip I found when calculating the molar entropy of vaporization if you are going to calculate it using this equation

Once you have the enthalpy, you just do delta S = q/T= delta H/T, which means dividing each enthalpy of vaporization in joules by their corresponding boiling points in kelvin

is to remember to multiply by 1000 so that your answer is in Joules and not Kilojoules. Hope this helps