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### Topic: Standard Internal Energy Question  (Read 6232 times)

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#### cvc121

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##### Standard Internal Energy Question
« on: February 10, 2017, 01:19:42 AM »
Hi,

For the reaction 2H2 + O2 ---> 2H2O, ΔH° is given at -571.6 kJ/mol. What would the energy change (ΔU°) of the reaction be at 25°C and 1.00 bar?

Here is my attempt:
ΔU° = q + w
At constant pressure: q = ΔH° = -571.6 kJ/mol
w = -PΔV = (-1.00 bar)(0) = 0

ΔU° = -571.6 kJ/mol + 0 = -571.6 kJ/mol
In this case, since there is no change in volume, ΔU° = ΔH°.

I am not sure about my approach and reasoning. Can anyone confirm? Thanks. All help is very much appreciated.

#### sjb

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##### Re: Standard Internal Energy Question
« Reply #1 on: February 10, 2017, 02:16:43 AM »
What are the states of reactants and products under the given conditions?

#### cvc121

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##### Re: Standard Internal Energy Question
« Reply #2 on: February 10, 2017, 02:18:45 AM »
2H2 (gas) + O2 (gas) ---> 2H2O (liquid)

#### cvc121

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##### Re: Standard Internal Energy Question
« Reply #3 on: February 10, 2017, 02:46:02 AM »
Is the phase change important in this case?

#### Borek

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##### Re: Standard Internal Energy Question
« Reply #4 on: February 10, 2017, 03:01:51 AM »
since there is no change in volume

Think again.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### cvc121

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##### Re: Standard Internal Energy Question
« Reply #5 on: February 10, 2017, 11:32:40 AM »
Ok. So would the ideal gas law be needed here?
PΔV = ΔngasRT
(1.00 bar)(ΔV) = (0 mol gas product - 3 mol gas reactant)(0.08206 L atm K-1 mol-1)(273 + 25)
ΔV = (-3 mol)(0.08206 L atm K-1 mol-1)(298 K) / 1.00 bar
ΔV = -73.36 L

Therefore, w = -PΔV = (-1.00 bar)(-73.36 L) = 73.36 bar·L
73.36 bar·L = 7336 J = 7.336 kJ x 6.022 x 1024 mol-1 = 4.42 x 1024 kJ/mol
ΔU° = -571.6 kJ/mol + 4.42 x 1024 kJ/mol = 4.42 x 1024kJ/mol

Am I on the right track here or completely approaching it wrong? I believe that the change in volume is much too large and did not calculate it correctly.
« Last Edit: February 10, 2017, 12:30:53 PM by cvc121 »

#### mjc123

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##### Re: Standard Internal Energy Question
« Reply #6 on: February 10, 2017, 12:54:17 PM »
ΔV looks OK, but why have you multiplied by NA (which you have got wrong)? What you have calculated is 7336 J per mole of reaction, i.e. for 2 moles of H2 and 1 mole of O2 giving 2 moles of water. Not per molecule!!!

#### cvc121

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##### Re: Standard Internal Energy Question
« Reply #7 on: February 10, 2017, 02:47:46 PM »
Ok. Thanks for the clarification. So ΔU° = -571.6 kJ/mol + 7.336 kJ/mol = -564.26 kJ/mol?

#### mjc123

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##### Re: Standard Internal Energy Question
« Reply #8 on: February 14, 2017, 08:04:55 AM »
I would say -564.3 kJ/mol, as your first number is only accurate to 1 d.p., so you're not really justified in going to higher precision, however precisely you have calculated Δ(PV). Otherwise it looks OK.