Midterm Question 6

[Ahmed Mahmood 4D]

For question six on the midterm, you add work and change in enthalpy to get change in internal energy. I calculated 19.1 kj for work and 11.5 kj for change in enthalpy—how was my change in internal energy -11.5 kj?

[Chem_Mod]

Please, provide the full question when possible since we don't have full access to the midterms.

Thanks in advance! :)

[Ahmed Mahmood 4D]

Please, provide the full question when possible since we don't have full access to the midterms.

Thanks in advance! :)

“4 Moles of butane (C4H10) are used for cooking. What mass butane id this? Will this combustion do expansion work? Calculate change in internal energy for the rxn. Assume rxn will occur at 1 atm and 300 deg celcius, and enthalpy of combustion of butane is -2878 kj/mol.”

If you don’t mind, I’m also curious to question 7D on the midterm; ‘Calculate rxn enthalpy per mole of methane combustion.’

Given is the following:
Hb(O-H)= 463 kj/mol
HB (C=O) = 743 kj/mol
Hb (C-H) = 412 kj/mol
Hb (O2) = 496 kj/mol

I got the answer -228 kj/mol and put exothermic, but still was docked four points.

[Helen Zhao 1F]

For Question 6, I first found delta V using (delta)nRT/P. Using delta V, I plugged the values given into delta U = delta H -PdeltaV. delta H = 4 * -2878 kJ/mol and PdeltaV = (-1atm)(282L)(8.314 J/Kmol )(1/8.206*10^-2Latm/Kmol)(1kJ/1000J) = 28.6kJ. Using those values I got -11541 kJ.

[Helen Zhao 1F]

For Question 7, I wrote out an equation and balanced it out. I got CH4 + 2O2 -> 2H2O + CO2. I then drew out diagrams for each molecule and calculated the delta H for each. CH4 was 1648kJ, O2 was 992kJ, H2O was 1852 kJ, and CO2 was 1486kJ. I finally did (delta H of CH4 + O2) - (delta H of H2O + CO2) and I got a resulting answer of -698kJ.

[lukezhang2C]

I think for this question on the midterm it was far easier to approach it by finding the net moles of gas produced than it was to calculate each bond enthalpy as you have delta U = delta H (which is found by multiplying the delta H given by 4 moles) - nRT where your n is just the net change in moles of gas.

[Roberto Gonzalez 1L]

I used delta n*r*t to solve for work with a change of moles of 6. Which resulted in 28.58 Kj of work. Subtracted from n*enthalpy= -11512 gave me -11540 Kj