Posted by maryanne on Saturday, November 8, 2008 at 8:49pm.
How much heat is absorbed.
mass H2O x specific heat water x delta T.
35.0g x 4.18 J/g*C x (22.7-19.4) = ??
That is delta H/1.50 g NH4NO3.
Change that to delta H/mol NH4NO3 and change to kJ/mol.
Post your work if you get stuck.
I got:
delta H= 482.79J
482.79/150gNH4NO3 x 80gNH4NO3/mol NH4NO3
=2574.88J/mol
=2.575 kJ/mol
Is this right?
i have one more question:
I know this one has to do with hess's law but i'm not sure of how to get the enthalpy of vaporization?
Clculate the enthalpy of vaporization of solid potassium bromide to a gas of ions, the process KBr--->K{+1}(g)+Br{-1}(g), from the following information and the enthalpy of formation of KBr(s), which is -394kJ/mol
Atomization of K: K(s)--->K(g) delta H given
Ionization of K: K(g)--> K{+1}(g)+ e(g) delta H given
VAporization of bromine: Br2(l)-->Br2(g) delta H given
Dissociation of bromine: Br2(g)-->2Br(g) delta H given
Electron attachment of bromine: Br(g)+e--->Br{-1}(g) delta H given
subl = energy to sublime K solid.
IP = energy to ionize K(g).
BDE = bond dissociation energy to dissociate Br2 liquid to Br2 gas. Take 1/2 of the value unless that has been done already. Also, note that this should include the vaporization, also. If you have been given that separately, add them together before taking the 1/2.
EA = electron affinity. Energy (exothermic) for adding an electron to Br(g).
U = lattice energy = energy to combine gaseous ions to form the solid. This is exothermic, also. This is what you want to calculate.
delta Hf = formation of
K(s) + 1/2Br2(l) ==> KBr(s)
Here is what you want to use.
delta Hf = subl +IP+ BDE(after taking the 1/2)+ EA (use the negative number) + U. Solve for U. What you are calculation is K^+ + Br^- ==> KBr(s) and the value will be negative. You want the reverse of this; therefore, just write the reverse reaction and change the sign. I hope this helps.
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I worked the problem and obtained a different answer. Isn't that 1.50 g (not 150)? I used 80.04 for the molar mass of NH4NO3 and obtained 25.762 kJ/mol which I would round to 25.8 kJ/mol.
ya you're right i accidently put the wrong value in!
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