3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g)
standard enthalpy:
-133.5
Find the change in internal energy for this reaction. Enter in kJ.
chem is for fuvking losers i hate my life
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.
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
lasted
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.