The combustion of ethane in the presence of excess oxygen yields carbon dioxide and water:

2c2h6 (g) + 7o2 (g) → 4co2 (g) + 6h2o (l)

the value of δg° for this reaction is /mol. compare results for both methods of finding δg as done in class. is this difference significant?

ΔG° = - 2936.604 KJ/mol

Explanation:

⇒ ΔG°t = 4ΔG°fCO2(g) + 6ΔG°fH2O(l) - 7ΔG°fO2 - 2ΔG°fC2H6(g)

∴ ΔG°fCO2(g) = - 394.359 KJ/molfrom literature

∴ ΔG°fH2O(l) = - 237.178 KJ/mol

∴ ΔG°fO2(g) = 0.00 KJ/mol

∴ ΔG°fC2H6(g) = - 31.95 KJ/mol

⇒ ΔG°t = 4(- 394.359 KJ/mol) + 6(- 237.178 KJ/mol) - 7(0.00 KJ/mol) - 2( -31.95 KJ/mol)

⇒ ΔG°t = - 1577.436 KJ/mol - 1423.068 KJ/mol + 63.9 KJ/mol

⇒ ΔG°t  - 2936.604 KJ/mol

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

i think slash and