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Topic: Enthalpy of formation of Unknown and Quantums  (Read 4095 times)

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Kdub

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Enthalpy of formation of Unknown and Quantums
« on: November 16, 2005, 01:14:26 AM »
Have a few questions:

1. The standard enthalpy change for the reaction below is -905.4 kJ (for the reaction as written). What is the standard enthalpy of formation of NO(g)?
4NH3(g) + 5O2(g) --> 4NO(g) + 6H2O(g)

Data
Substance                          Standard Enthalpy of formation (in kJ mol-1)
NH3(g)                        -46.1
H2O(g)                      -241.8

4mol(-46.1kJ/mol) + 5 mol(0kJ/mol)    
=-184.4 (kJ)    REACTANTS

4mol(x) + 6mol (-241.8kJ)
= 4mol(x) + -1450.8kJ     PRODUCTS

delta H = n x delta Hformation
           = 4mol x (-905.4 kJ/mol)
           = -3621.6 kJ

-3621.6kJ = 4mol(x) - 1450.8kJ + 184.4kJ
-1986.5kJ = 4mol(x)
            x = -588.8kJ/mol

Does this seem right?

2. What is the maximum kinetic energy of an ejected electron if silver metal is irradiated with 246nm of light? The threshold wavelength for a silver metal is 267nm.

My question is, do I need to use the threshold energy for something? This is what I did:
v = c/wavelength = 2.998x108 ms-1 / 246.0x10-9
m
  = 1.219x1015s-1

Then E = hv
          =  6.626x10-34 J photon-1x 1.219x1015s-1
          = 8.075x10-19 J/photon


Thanks for the help.

Kdub

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Re:Enthalpy of formation of Unknown and Quantums
« Reply #1 on: November 16, 2005, 12:15:12 PM »
*Ignore me, I am impatient*

Offline Donaldson Tan

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Re:Enthalpy of formation of Unknown and Quantums
« Reply #2 on: November 16, 2005, 07:16:07 PM »
1. With reference to diagram attached, it can be inferred that:
-(-905.4) + 4(46.1) = -[4Y + 6(-241.8)]
4Y = 361 kJ
Y = 361 / 4 = 90.25kJ/mol

2. regarding question(2), you failed to take in account of the threshold frequency. the threshold frequency corresponds to the work function of the metal. In fact the energy equation for this phenomena is:

energy of radiation = work function + kinetic energy of electron
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