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Prelab was worth 10 pts (answers online, point values were 1, 1, 2, 1, 2, 1, 2 for the 7 questions) Postlab was worth 80 pts Total out of 90, divide score by 3 to get the contribution towards your overall grade Post lab answers: 1. (10 pts) C N + H H H H H C N + H H H H H + Alpha cleavage reaction 2. (15 pts) C N + H H H H H C N + H H H H H + C N + H H H H H C N H H H H H -22.5 kJ/mol e - + 2e - + 8.9 eV use value from your graph overall energy 10.55 eV, energy for process is difference, so 10.55-8.9 or 1.65 eV again use values from your graphs +218 kJ/mol X X Y C N H H H H H -22.5 kJ/mol e - + C N + H H H H H + +218 kJ/mol Y 2e - + 10.55 eV use value from your graphs This can be treated as 2 step process, so first calculate X, then calculate Y. For the 2 step process: dHrxn = dHprod – dHreact, thus for ionization reaction: 8.9 eV = 858.7 kJ/mol

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858.7 = X - -22.5 (dH of electron is assumed to be 0) X = +836.3 kJ/mol Alpha cleavage reaction: 1.65 eV = 159.2 kJ/mol 159.2 = [Y + 218] – 836.3 159.2 = Y - -618.3 Y = 777.5 kJ/mol Alternatively, a 1 step process (3 rd equation above) 10.55 eV = 1017.9 kJ/mol 1017.9 = [Y + 218] - -22.5 1017.9 = Y + 240.5 Y = 777.4 kJ/mol (0.1 kJ difference is likely a rounding-induced error) 3. (3 pts) Use data from your graph. Points given for agreement with graph and justification 4. (10 pts) It is critically important that you make sure to use SI units all the way through or the problem gets VERY difficult. This is a multi-step problem. The first few steps convert all the units into SI.
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