C---An important step in the synthesis of nitric acid is the reaction of ammonia
ID: 924032 • Letter: C
Question
C---An important step in the synthesis of nitric acid is the reaction of ammonia (NH3) and molecular oxygen to form nitrogen monoxide gas and water vapor.
--Write a balanced equation for this reaction. (2 points)
--Calculate DHo for this reaction, using standard heat of formation values from Appendix G in the Open Stax Chemistry text. (3 points)
D----When an electron in a 2p orbital of a particular atom makes a transition to the 2s orbital, a photon of approximate wavelength 629.1 nm is emitted. What is the energy difference between these 2p and 2s orbitals?
E---Provide a numerical value (whole number) for each of the following questions. Briefly explain your reasoning (1 point each).
How many electrons can be contained in all of the orbitals withn = 4?
Among the orbitals with n = 4 (see part a), which one has the highest energy and how do you know this?
What is the l quantum number for a 4s orbital?
How many electrons can be described by the quantum numbersn = 3, l = 3, ml = 1?
l = 3, how many electrons can be contained in all the possible orbitals of a given principal quantum number?
Explanation / Answer
C. (a) Balanced equation : 4NH3 + 5O2 ---> 4NO + 6H2O
(b) dHorxn = dHoproducts - dHoreactants
= (4 x 90.29 + 6 x -285.8) - (4 x -45.9)
= -1170/04 kJ/mol
D. E = hc/lambda
= 6.626 x 10^-34 x 3 x 10^8/6.29 x 10^-7 = 3.16 x 10^-19 J is the required energy
E. (a) 4 would have : 4s, 4p, 4d, 4f = 2 + 6 + 10 + 14 = 32 electrons
(b) The highest energy would be for 4f orbitals. The energy levels can be calculated by counting n+l for each.
4s = 4 + 0 = 4
4p = 4 + 1 = 5
4d = 4 + 2 = 6
4f = 4 + 3 = 7
So as we can see the highest energy is of 4f value with highest value of n+l
(c) For 4s, l = 0
(d) When, n= 3, l = 3, ml = 1 would be for 3f with 2 electrons
(e) Number of electrons possible, 2l+1 with l = n-1
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