Some simple calculations in organic chemistry

[3] The creation of significant amounts of nitrogen monoxide from human
activities is of concern because it is thought to lead to loss of ozone
from the stratosphere. Reactions D&E show how this loss can occur

Reaction D = NO. + O3 → NO2. + O2 ∆H = -100kj.mol-1

Reaction E = NO2. + O → NO. + O2 ∆H = +192kj.mol-1

Name one human activity which leads to the production of significant
quantities of NO. Driving internal combustion engine vehicles

Are reactions D&E endothermic or exothermic. D is endothermic, E is
exothermic

Calculate the value of ∆H for the overall process. ∆H = 292kj.mol-1

Explain why we should be worried about the effects which might arise
from the emission of small amounts of NO into the atmosphere.
Ground-level ozone, unlike other pollutants, is not emitted directly
into the atmosphere, but is a secondary pollutant produced by reaction
between oxides of nitrogen such as NO, hydrocarbons and sunlight.
Sunlight provides the energy to initiate ozone formation; consequently,
high levels of ozone are generally observed during hot, still sunny,
summertime weather. This leads to Health Effects such as irritation of
the human airways and lungs, particularly affecting those suffering from
asthma or other lung diseases

[4] The following reactions occur when ethane is pyrolysed at 620ºC:

C2H6→ 2CH3.

CH3. + C2H6→ CH4 + C2H5.

C2H5. → C2H4 + H.

H. + C2H6→ H2+ C2H5.

2C2H5. → C2H4+ C2H6

2C2H5. → C4H10

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ᔌﱨᘀ䭨扉㈀nor) iii) propagation (major) iv) propagation
(major) v) termination vi) termination

Two of the reactions can be classified by inspection

Which reaction is endothermic? Equation i) is endothermic

Which reaction is exothermic? Equation ii) is exothermic

Which of the chemical species in the reaction sequence are radicals?
Give their names and formulae

The radicals are

H. Hydrogen

CH3. named methyl

C2H5. named ethyl

[4] Assignment 4

a) What is the bond energy in joules per molecule of the oxygen-oxygen
bond in dioxygen? 498kJ.mol-1

b) Use the expression E=hν to calculate the frequency of radiation that
would break this bond. [Plank’s Constant h = 6.63x10-34j.Hz-1] E=hν
so 498kJ = 6.63x10-34j.Hz-1 x ν so ν = 75.1 x 1031 (seems wrong!)

c) What you have calculated is the minimum frequency to cause the bond
to break. Explain why it is a minimum value (not sure of the answer
here, so have not given one)