Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: cvc121 on February 10, 2017, 01:19:42 AM
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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.
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What are the states of reactants and products under the given conditions?
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2H2 (gas) + O2 (gas) ---> 2H2O (liquid)
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Is the phase change important in this case?
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since there is no change in volume
Think again.
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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.
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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!!!
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Ok. Thanks for the clarification. So ΔU° = -571.6 kJ/mol + 7.336 kJ/mol = -564.26 kJ/mol?
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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.