Oxidation Number vs Formal Charge Acetone, CHsCOCHa, in atmosphere of oxygen is
ID: 589226 • Letter: O
Question
Oxidation Number vs Formal Charge Acetone, CHsCOCHa, in atmosphere of oxygen is combusted according to the following unbalanced chemical equation: i H H(g) +02(g) H CO2(g) H2O(g) + Fill in the gap and answer the questions. a) when properly balanced, the equation indicates that mole(s) 002 are required for l each mole of CH3COCH3. (10 p.) b) Once initiated, is this reaction exothermic or endothermic? (5 p.) c) Justify your answer. (10 p.) d) Is this a redox reaction? (5 p.) e) Justify your answer. (10 p.) f) Determine the oxidation number of each atom in the reactants and the products. (1 p each) Products Reactants: C (in CO in acetone) C (in CH3 in acetone)-- O (in CO in acetone) H (in CH3 in acetone)-- O (in O2) C (in CO2) O (in CO.) H (in HO) O (in H2O)-Explanation / Answer
a) Balanced euation for combustion of acetone
CH3COCH3 + 4O2 ---> 3CO2 + 3H2O
So 4 moles of O2 are needed for complete combustion
b) The reaction is exothermic in nature
c) This can be confirmed by calculating enthalpy change for the reaction,
dHrxn = (3 x -393.5 + 3 x -285.8) - (-249.4) = -1788.5 kJ/mol
So dHrxn is -ve, that means the reaction is exothermic in nature.
d) Yes, this is a redox reaction
e) C in acetone goes from -3 oxidation state to +4 oxidation state in CO2, so oxidized
Oxygen in O2 goes from 0 oxidation state to -2 oxidation state, so reduced
f) Oxidation numbers
Reactants
C (in CO in acetone) = +2
C (in CH3 in acetone) = -3
O (in CO in acetone) = -2
H (in CH3 in acetone) = +1
O (in O2) = 0
Products
C (in CO2) = +4
O (in CO2) = -2
H (in H2O) = +1
O (in H2O) = -2
g) knowing oxidation number we can identify which atoms or species are undergoing oxidation and which are undergoing reduction reactions.
h) formal charge
Reactants
C (in CO in acetone) = 0
C (in CH3 in acetone) = 0
O (in CO in acetone) = 0
H (in CH3 in acetone) = 0
O (in O2) = 0
Products
C (in CO2) = 0
O (in CO2) = 0
H (in H2O) = 0
O (in H2O) = 0
No charge on any atom
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