# Thermodynamics: internal energy change

1. Oct 14, 2011

### Dr. Science

1. The problem statement, all variables and given/known data

Enthalpy of formation of a mole of atomic hydrogen = 218kJ. Enthalpy changes when a mole of atomic hydrogen is formed by dissociating half a mole of molecular hydrogen. Calculate ΔU of the process of molecular hydrogen dissociation.

2. Relevant equations

ΔH = ΔU + Δ(PV)
PV=nRT

3. The attempt at a solution

ΔU = ΔH - Δ(PV)
ΔU = ΔH - Δ(nRT)
ΔU = 218kJ - (1/2 mol)(8.31 J/Kmol)(298K)
ΔU = 216kJ

I really don't think my answer is correct.

2. Oct 15, 2011

### Dr. Science

anyone?

3. Oct 16, 2011

### Dr. Science

guys...

4. Oct 16, 2011

### Andrew Mason

You will need to know the temperature at which the disassociation occurs. So I assume you are referring to the standard enthalpy of formation, which occurs T = 25C.

dU = dH - d(PV) = dH - PdV -VdP

If the process occurs at constant pressure (eg. at atmospheric pressure) the VdP term is zero. Alternatively, if it occurs at constant volume, the PdV term is zero. Let's assume that it occurs at constant pressure of one atmosphere (and temperature of 25C = 298K).

So:

dU = dH - PdV

Since there are double the number of molecules after disassociation, the volume must double, as you note. I think you should be able to take it from there.

AM

5. Oct 16, 2011

### Dr. Science

so then

ΔU = ΔH -P(dv)
ΔU = ΔH - ∫ nRT/V dv
ΔU = ΔH - nRT ln(vf/vi)
ΔU = 218kJ - 1 mol * 8.21 J/k Mol *298 K * ln2
ΔU = 216kJ ?

Last edited: Oct 16, 2011
6. Oct 16, 2011

### Dr. Science

right?

7. Oct 16, 2011

### Dr. Science

ΔH formation is +'ve

shouldn't my answer be negative since the system has less potential energy.
1/2 H2 -> H

8. Oct 16, 2011

### Andrew Mason

Since pressure is constant, ∫PdV = PΔV

AM

9. Oct 16, 2011

### Andrew Mason

The system has greater potential energy since it takes energy to separate the hydrogen atoms. (In forming H2 from H atoms, heat would be given off).

AM