ATOC/CHEM 5151 – Fall 2016 (updated web link Aug 31, 2016)
Problem 7
Oxidation States of Carbon and Reaction Enthalpy
Answers: to be posted Thursday, September 1, 2016
Environmental chemists often use oxidation states of important nutrients (e.g., carbon,
nitrogen, and sulfur) to track energy flow in the earth system. In Earth’s atmosphere, which is
rich in oxidants (O3, O2, OH, HO2, and NO3, for example), oxidation reactions tend to
dominate over other reactions when it comes to extracting energy from compounds emitted
to the atmosphere. As a result, many compounds are oxidized to forms that produce acids
(which are highly oxidized species). Examples include CO2 (which forms carbonic acid, of
H2CO3, in water), HNO3, HONO, HSO3, and H2SO4.
(1) Using the simple rule that each H counts as +1 and each O counts as -2, determine the
oxidation state of the carbon atom and the nitrogen atom in each of the following
compounds.
____-4__ Methane (CH4)
____-2__ Methanol (CH3OH)
____0___ Formaldehyde (CH2O)
___+2___ Formic acid (HCOOH)
___+1___ Glyoxal (C2H2O2 or CHO-CHO, where each C is singly bonded to H and
doubly bonded to O)
___+4___Carbon dioxide (CO2)
___+4___ Carbonic acid (H2CO3)
___+4___ Bicarbonate ion (HCO3-)
___+4___ Carbonate ion (CO32-)
(2) Determine the energy released by the composite reaction to form formaldehyde from OH
and CH4. Use the following source for “enthalpy of formation” for each of the reactants
and products (http://jpldataeval.jpl.nasa.gov/pdf/JPL_Publication_15-10.pdf). Note – see
Table 6-2 starting on page 6-3, and use the value for H(298).
OH + CH4  CH3 + H2O
CH3 + O2  CH3O2
CH3O2 + NO  CH3O + NO2
CH3O + O2  HO2 + CH2O
HO2 + NO  OH + NO2
There are two ways to do this problem – let’s start with the easy way, and just add up all the
steps.
Composite: CH4 + 2O2 + 2NO  2NO2 + CH2O + H2O
From the table, we have for H of the reactants and products
H(CH4) = –17.88 kcal mol-1 (all same units throughout)
H(O2) = 0.0 (defined in the standard state)
H(NO) = 21.57
H(NO2) = 7.9
H(CH2O) = –26
H(H2O) = –57.8
So for the overall reaction,
H(reaction) = 2 x H(NO2) + H(CH2O) H(H2O)
– H(CH4) – 2 x H(O2) – 2 x H(NO)
(2 x 7.9) – 26 – 57.8 – (–17.88) – (2 x 0) – (2 x 21.57)
= –93.26 kcal/mole
We can also do this for individual steps. We need additional information.
H(OH) = 9.3 kcal mol-1 (all same units throughout)
H(CH3) = 35
H(CH3O2) = 4
H(CH3O) = 4
H(HO2) = –2.8
OH + CH4  CH3 + H2O,
(–57.8 + 35 – 9.3 + 17.88) = –14.22
CH3 + O2  CH3O2
(4 – 35 – 0) = –31
CH3O2 + NO  CH3O + NO2
(4 + 7.9 – 4 – 21.57) = –13.67
CH3O + O2  HO2 + CH2O
(– 26 – 2.8 – 4 – 0) = –32.8
HO2 + NO  OH + NO2
(9.3 + 7.9 + 2.8 – 21.57) = –1.57
Summing up the steps,
–93.26 kcal mol-1