## 8.49

Jaime_Chamberlain_3G
Posts: 23
Joined: Sat Jul 09, 2016 3:00 am

### 8.49

How would you go about doing number 49 for the chapter 8 homework? "Oxygen difluoride is a colorless, very poisonous gas that reacts rapidly with water vapor to produce O2, HF, and heat: OF2(g)+H2O(g)--> O2(g)+2HF(g) deltaH=-318J What is the change in internal energy for the reaction of 1.00 mol OF2?"

Lily_Dermendjian_2B
Posts: 10
Joined: Sat Jul 09, 2016 3:00 am

### Re: 8.49

You use the equation deltaU=q+w. You can use the given deltaH for q, and use the equation w=-PdeltaV=-deltanRT to find w. This gives you deltaU=q+w=deltaH-PdeltaV=-318kJ-2.48kJ=-320kJ.

Sunny Chera 1N
Posts: 20
Joined: Wed Sep 21, 2016 2:56 pm

### Re: 8.49

For this problem, you must look at the balanced equation and determine the change in moles, delta(n)=nfinal-ninitial. In this case, nfinal would be 3 moles since there are 3 moles on the products side of the equation, and ninitial would be 2 moles, as there are two moles on the reactants side of the equation. Therefore, delta(n)=3-2=+1 mol. Then you use the relationship between delta(H) and delta(U): delta(H)=delta(U)+delta(n)gas*R*T. You know that delta(H)=-318 kJ as listed in the problem for the reaction. Now you isolate that relationship for delta(U) to get -320 kJ as your change in internal energy!

Let me know if you need any further elucidation regarding any aspect of this problem!

Sunny

ntyshchenko
Posts: 21
Joined: Wed Sep 21, 2016 2:59 pm

### Re: 8.49

What would you plug in for T in this equation?