Question

Enter your answer in the provided box. From the data below, calculate the total heat (in J) needed to convert 0.803 mol of gaseous ethanol at 300.0°C and 1 atm to liquid ethanol at 25.0°C and 1 atm: b.p. at 1 atm: 78.5°C AH®: 40.5 kJ/mol vap Cgas: 1.43 J/g.°C Cliquid: 2.45 J/g °C -4849|o1 J

My answer is wrong for the 6th time what am I doing wrong and how do i do this

Answer 1

First convert 0.803 mol to g as the specific heats are in J/g oC.

Amount in g = number of moles * molar mass

0.803mols of C2H5OH = 0.803 x 46 = 36.938 g

Now calculate the heat as follows:

1. The heat required to cool down the gas from 300 C --> 78.5 C (This will be negative -heat out)

heat change = mass x C x change in temperature

= 36.938 g x 1.43 J/g C x -221.5 C

= - 11699.93 J

2. The heat change required to condense the gas to liquid

at 78.5 C = mols x heat of vaporisation/mol

= 0.803 mols x - 40.5 kJ/mol = -32.52 kJ (32521.5 -J)

3. The heat required to cool down the liquid from 78.5 C to 25 C

This = mass x C x Change in temp = 36.938 g x 2.45J/g C x - 53.5 C

= - 4841.65 J

Then add up : Total heat required = - 11699.93 + - 32521.5 + - 4841.65 = - 49063.08 J

= - 49.06 kJ