1. Oct 26, 2004

### peter444

$$N_2_{(g)} + \frac{5}{2}O_2 \longrightarrow N_2O_5 + \mathfrm 533 kJ energy$$
533kJ energy is released as heat

We are given that 101.7g N2 , 102.97g O2 reacted. And this happens at T= 25C and P = 1atm

what is the delta H and delta U for this process per mol N2(g). And how much work is done by the surrounding?

Thanks
(H<- enthalpy , U<- internal energy of the system )

Last edited: Oct 26, 2004
2. Oct 26, 2004

### chem_tr

First of all, you need to write the coefficients correctly for the reaction.

$$N_2_{(g)} + \frac{5}{2}O_2 \longrightarrow N_2O_5 + \mathfrm 533 kJ energy$$

Then find the molar amounts of nitrogen and oxygen gas, by using N=14 and O=16 gram/mol.

If a thermal energy is given to the environment, then it is an exothermic reaction, thus $\Delta H$ must be negative, and must have the dimension $$\frac {mol}{L}$$.

I have no idea what $\displaystyle \Delta U$ is, so another friend will help you in the following days, I think.

Last edited: Oct 26, 2004
3. Oct 26, 2004

### peter444

i found the number of moles for each gas, but we are not given the volume;As you say $\Delta H$ = $$\frac {mol}{L}$$. And how is -533kJ usefull in this problem?
Thanks

4. Oct 26, 2004

### so-crates

H = U + pv, so therefore $$\Delta H = \Delta U + \Delta (pV)$$. First off, what is $$\Delta U$$?

You can't find $$\Delta (pV)$$ directly. But for an ideal gas, what is pV equal to ?

Last edited: Oct 26, 2004
5. Oct 26, 2004

### peter444

For an ideal gas: PV=nRT
but how do I get U, (H=U+PV ) if i also need H

6. Oct 26, 2004

### so-crates

I just realized something, you say the reaction occurs at 1 atm and 25 C, so it is both a constant temperature and constant pressure reaction ?

7. Oct 27, 2004

### wolfson_1123

1 mole of a gas = 24 L (24dm^3)

So you have your mass, so just change that into number of moles, then multiply by 24 L. That will find your Volume.

8. Oct 27, 2004

### so-crates

OK I don't know how you did on your test but

1) For an isothermal reaction for an ideal gas, $$\Delta U = C_v\Delta T$$ and since presumably $$\Delta T = 0$$, it follows that $$\Delta U = 0$$

2) Under constant pressure, $$\Delta H = q$$. If you don't know what $$\Delta H$$ is, go back and look at the thermodynamics section in your first-year chemistry book.

3) Well now since you know q and $$\Delta U$$ , you can solve for w, right ?