Question
3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g)
standard enthalpy:
-133.5
Find the change in internal energy for this reaction. Enter in kJ.
standard enthalpy:
-133.5
Find the change in internal energy for this reaction. Enter in kJ.
Answers
1. Is the reaction 133.5 FOR THAT REACTION or per mole NO2?
2. And tht's 133.5 what? kJ/mol or kJ/reaction; i.e., for 3 mols NO2.
2. And tht's 133.5 what? kJ/mol or kJ/reaction; i.e., for 3 mols NO2.
for that reaction and its kj
-133.5 kJ for the reaction.
You have 3 mols gas on the left (volume wise if H2O occupies no volume and I suppose we suppose it doesn't) and 1 mol gas on the right.
3 mols gas on the left at STP is 3*22.4 and on the right is 1*22.4L. In the absence of other information I suppose p is 1 atm.
work = -p*dV = -p(Vfinal-Vinitial) and multiply by 101.325 to convert to J and divide by 1000 to convert to kJ. Then
dE in kJ = q(in kJ)+w(in kJ)
dH is q and you can calculate w. Substitute and calculate dE in kJ.
You have 3 mols gas on the left (volume wise if H2O occupies no volume and I suppose we suppose it doesn't) and 1 mol gas on the right.
3 mols gas on the left at STP is 3*22.4 and on the right is 1*22.4L. In the absence of other information I suppose p is 1 atm.
work = -p*dV = -p(Vfinal-Vinitial) and multiply by 101.325 to convert to J and divide by 1000 to convert to kJ. Then
dE in kJ = q(in kJ)+w(in kJ)
dH is q and you can calculate w. Substitute and calculate dE in kJ.
Why 22.4?
you don't multiply the 3 moles on the left and the 1 mole on the right by 22.4, you divide. this is because the key to the conversion is realizing that 1 mole of any gas occupies 22.4L. This gives us a conversion factor of 1 mol/22.4 L. To go from liters to moles you essentially are dividing by 22.4
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