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

explanation:

venn diagram is pictorial representation with of circles showing the relation between two finite sets or things. the intersection part of two circles conveys the relation present in between those sets or things.

here in the venn diagram of sand and salt two circles represent sand and salt. but the area falling under the intersection of these circle represents the mixture of sand and salt.

so, in a venn diagram the mixture will be placed in a middle that is in between the intersection of two circles.