I would do this.
dE = q+w
dE = -2252 kJ + w.
Modifying PV = nRT and remembering that pdV = work, then
pdV = dnRT and dn is 11 + 23 = 34 mols.
pdV = 34*0.08205*298 = ? L*atm.
Convert to J by multiplying by 101.325 and convert that to kJ.
What't the sign of work in this case.
Work = -pdV = -p(Vfinal-Vinitial)
work = -[(pVfinal - pVinitial)]
work = -p*0 -(-34) = +p*34... so work is +.
Looking at work another way, the volume changes from 34 mols to zero mols gas so the surroundings will do work on the system and that makes it + work.
Substitute into dE and calculate.
The enthalpy change for the reaction below at 25oC is -2,252 kJ (per mole of C11H24). What is the internal energy change for the reaction at 25oC?
11 CO(g) + 23 H2(g) --> C11H24(l) + 11 H2O(l)
2 answers
The first part of my response is ok. It's the work that isn't. As you've probably already noticed I used mols for volume and that's a no, no.
Work = is pdV with the proper sign.
pdV = dn RT. Substitute R, T and delta n (delta n is 34 mols) and solve for pdV which is work. Since you have 34 mols on the left and zero gas on the right, it means the surroundings are doing work on the system so the sign of work is +.
Looking at work another way,
Vfinal is zero and Vinitial is whatever it will be vor 34 mols gas (substantial), then work = -p(Vfinal-Vinitial) = -p*0 -(-p*some value) = + work.
I hope this correction doesn't come too late.
Substitute into dE and calculate.
Work = is pdV with the proper sign.
pdV = dn RT. Substitute R, T and delta n (delta n is 34 mols) and solve for pdV which is work. Since you have 34 mols on the left and zero gas on the right, it means the surroundings are doing work on the system so the sign of work is +.
Looking at work another way,
Vfinal is zero and Vinitial is whatever it will be vor 34 mols gas (substantial), then work = -p(Vfinal-Vinitial) = -p*0 -(-p*some value) = + work.
I hope this correction doesn't come too late.
Substitute into dE and calculate.
0 answers