[Brian Lin]

Hello!

I have a thermochemistry- related question.

The standard enthalpy of formation of H2O(l) at 298 K is -285.8 kJ/mol. Calculate the change in internal energy for the following process at 298 K and 1 atm (basically at STP).

The equation for formation is
H2+ 0.5O2  H2O +285.8 kJ

Since we have to find internal energy.. the equation is..

ΔE= q+ w 

Q is definitely -285.8 kJ

W is derived from PΔV and PV can come from the ideal gas law equation.

PV= nRT

PV= (moles?)(0.08206 atmL/Kmol) (298k)(101.3 J/Kmol)

The question here is where do you get the number of moles for n? Is it 1.5 moles? I do not have the answer, so please *delete me*


Thanks! 

[Borek]

Sounds like you are asked to find the answer per mole of water produced.

[forrestgreen]

Borek is correct. You are asked to find internal energy change per mole of water produced.

I think your problem should be solved as follows:

ΔH = ΔE + Δ(PV)

ΔH = ΔE + (PV)_H2O – (PV)_gases

Because water volume is very small, we can only consider gases for the term (PV)

ΔH = ΔE  – (PV)_gases


ΔE = ΔH  +  (PV)_gases


ΔH = - 285.8 KJ

(PV)_gases = nRT = 1.5*R*298

ΔE = - 285.8 - 1.5*R*298