Thermodynamics: internal energy change

[Dr. Science]

Homework Statement

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.

Homework Equations

ΔH = ΔU + Δ(PV)
PV=nRT

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.

 

[Dr. Science]

anyone?

 

[Dr. Science]

guys...

 

[Andrew Mason]

Homework Statement

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.

Homework Equations

ΔH = ΔU + Δ(PV)
PV=nRT

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.

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

 

[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 ?

 

[Dr. Science]

right?

 

[Dr. Science]

ΔH formation is +'ve

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

 

[Andrew Mason]

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 ?

Since pressure is constant, ∫PdV = PΔV

AM

 

[Andrew Mason]

ΔH formation is +'ve

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

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