## Basic Thermodynamics; Change in U at Constant Pressure

1. The problem statement, all variables and given/known data
What is the ΔU° at 25°C for the following reaction at constant pressure:

C$_{2}$H$_{2}$ (g) + 5/2O$_{2}$ (g) $\rightarrow$ 2CO$_{2}$ (g) + H$_{2}$O (g) ΔH° = -1299.5kJ

2. Relevant equations

ΔU = Q ± W
PV = nRT
W = PΔV

3. The attempt at a solution

Since we have constant pressure, ΔH° = Q, and since we are at standard conditions P = 1atm = 101.3kPa. Additionally T = 298K (which is given anyway). I try and find the work done on the system by finding the reduction in volume;

ΔV = ΔnRT/P

ΔV = (0.5 x 8.31 x 298)/101.3 = 12.2L

W$_{on system}$ = 101.3kPa * 12.2L = 1238.2kJ

ΔU = -1299.5kJ + 1238.2kJ = -61.3kJ

Interestingly, this answer is what I would get if I had accidently calculated the W$_{on system}$ was in joules and forgotten to convert - but I don't see how this can be the case - I'm dealing with kPa not Pa as my unit for pressure. Any help would be much appreciated!
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 Quote by Silvius 1. The problem statement, all variables and given/known data What is the ΔU° at 25°C for the following reaction at constant pressure: C$_{2}$H$_{2}$ (g) + 5/2O$_{2}$ (g) $\rightarrow$ 2CO$_{2}$ (g) + H$_{2}$O (g) ΔH° = -1299.5kJ 2. Relevant equations ΔU = Q ± W PV = nRT W = PΔV 3. The attempt at a solution Since we have constant pressure, ΔH° = Q, and since we are at standard conditions P = 1atm = 101.3kPa. Additionally T = 298K (which is given anyway). I try and find the work done on the system by finding the reduction in volume; ΔV = ΔnRT/P ΔV = (0.5 x 8.31 x 298)/101.3 = 12.2L W$_{on system}$ = 101.3kPa * 12.2L = 1238.2kJ ΔU = -1299.5kJ + 1238.2kJ = -61.3kJ However, the answer is -1298.3kJ. Interestingly, this answer is what I would get if I had accidently calculated the W$_{on system}$ was in joules and forgotten to convert - but I don't see how this can be the case - I'm dealing with kPa not Pa as my unit for pressure. Any help would be much appreciated!
That is because ##E_{\mbox{int}}=Q+W## applies when work is done on the system.

 Quote by dimension10 That is because ##E_{\mbox{int}}=Q+W## applies when work is done on the system.
Work is being done on the system in this case; the pressure and temperature is constant, and so a decrease in the amount of gas results in a decrease of volume of the system. That is, the surrounding constant pressure is doing work in compressing the system.

Either way, I have found my error - it was simply an error in dimensional analysis! 1kPa * 1L = 1J, not kJ! Whoops!