## 4.7 Homework Help regarding signs of work

DHavo_1E
Posts: 118
Joined: Sat Aug 17, 2019 12:17 am

### 4.7 Homework Help regarding signs of work

Hello,

Could anyone explain how to determine the signs in the enthalpy equation: deltaH= deltaU + PdeltaV? In this question, work is found to be positive and when rearranging for deltaU, I would expect a negative internal energy. However, in the solution book, they wrote deltaU= deltaH + PdeltaV instead of deltaU=deltaH-PdeltaV. Do we account for the positive work after rearranging the equation then? Thank you!

KaleenaJezycki_1I
Posts: 127
Joined: Sat Aug 17, 2019 12:18 am
Been upvoted: 2 times

### Re: 4.7 Homework Help regarding signs of work

Hi! This comes from the original equation deltaH= q+w

Step1: isolate q
So q=DeltaU-w

Step2: To get Delta H we are assuming that the reaction is occurring under constant pressure (qp)
Therefore DeltaH= DeltaU-w

Step3: BUT w=-PexDeltaV so the we have - -Pex which becomes positive to get:

DeltaH= DeltaU+ PexDeltaV

Return to “Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)”