Please help! Benzene is a organic solvent with the formula C6H6. It boils at 80.1 degrees C and melts at...?

Question

5.5 degrees C. Benzene's density is 0.88 gm/ml, its heat of vaporization is 20.0 KJoules per mole, its heat of fusion is 10.0 KJoules per mole, its heat capacity is 0.50 KJoules per mole-degree C for liquid benzene, 0.25 KJoules per mole-degree C for solid benzene, and 0.125 KJoules per mole-degree C for gaseous benzene. Calculate the following:
A. The energy required to heat 390. grams of benzene from 300 degrees C to 600 degrees C.
B. The energy required to cool 3.90 micrograms of benzene from 60 degrees C to 20 degrees C.
C. The energy removed to heat 39.0 kilograms of benzene from -100 degrees C to -50 degrees C.
D. The volume occupied by 1 kilomole of liquid benzene at 25 degrees C and the volume occupied by 1 nanomole of benzene as a vapor at STP (assume an ideal gas)

Answer 1

A] Since t is 300 to 600 C everything is in vapor state.

Number of moles benzene x heat capacity of VAPOR x temperature CHANGE

390g/78g/mol x 0.125kj/mol C x [600-300 C] = 187.5kJ

B] Since t is 60 to 20 C everything is liquid:

3.9 x 10^-6mol/78g/mol x 0.50 kJ/mol C x [60-20] =

1 x 10^-6 kJ or 1 x 10^-3J

C] Note: you need to add energy to raise temperature.
Since t is from -100 to -50 C everything is solid:

39,000g/78g/mol x 0.25kJ/mol C x [-100 - -50] =

6250 kJ or 6.25 mJ or 6.25 x 10^6J

D] part 1] determine volume of 1 mole then multiply:
[78g/mol]/[0.88g/ml] x 1000mol = 88636ml or 88.6liters

part2] Remember; 1 mole of any gas at STP is 22.4 liters!

22.4l/mol x 1 x 10^-9 mol = 2.24 x 10^-8 liters or
2.24 x 10^-5ml

Note: I'm surprized they didn't ask any questions that involved a phase change! Are you sure you listed the proper temperatures!

Just in case, here is an example with a phase change; lets take part A and change starting temperature to 0 C (everything else the same).

On its way to 600 C it will melt, then boil; we have to account for the energy we use for melting and boiling. here is how we do it (I am leaving the units off this time except for heat of fusion/heat of vaporization to make the math less scary).

Since we already know that 390/78 = 5, I will just write 5 for the moles part.

Note that during phase changes the temperature is constant hence there is no temperature term for those parts of the equation.

5 x 0.25 x [5.5-0] + 5 x 10kj/mol + 5 x 0.50 x [80.1-5.5] +

5 x 20kj/mol + 5 x 0.125 x [600-80.1] = 668.4kJ

Here's what the 5 terms are accounting for: first, warming of 5 moles of solid benzine to its melting point (but not melting it!); second, melting 5 moles of benzine to liquid at its melting point (solid to liquid phase change only); third, warming of the liquid benzene to its boiling point (but not boiling it), fourth, the energy to boil the benzene to vapor (benzine liquid at 88.1 C to benzine vapor at 88.1 C); heating of the vapor to the final temperature.

One final thing, the calculator I used doesn't do scientific notation so just double-check the exponents to see that they're OK, I had to keep track of them in my head. Your book (or teacher) may have some rules about what to do when you are subtracting negative temperatures ex [-100 - -50] (do you keep the negative sign all through the calcualtions or just take the absolute value. Be certian you know how to do this on your own, I won't be on line with you during your test!