Question
Given the following thermochemical data:
1) H2(g) + 1/2 O2(g) -> H2O(g) ∆H = -241kJ
2) C6H10(g) + 17/2 O2(g) -> 6CO2(g) + 5H2O(g) ∆H = -3800kJ
3)C6H12(g) + 9O2(g) -> 6CO2(g) + 6H2O(g) ∆H = -3920 kJ
What is ∆H for the following reaction? C6H10(g) + 2H2(g) -> C6H12(g)
1) H2(g) + 1/2 O2(g) -> H2O(g) ∆H = -241kJ
2) C6H10(g) + 17/2 O2(g) -> 6CO2(g) + 5H2O(g) ∆H = -3800kJ
3)C6H12(g) + 9O2(g) -> 6CO2(g) + 6H2O(g) ∆H = -3920 kJ
What is ∆H for the following reaction? C6H10(g) + 2H2(g) -> C6H12(g)
Answers
DrBob222
The "following reaction" is not balanced for H atoms.
Anonymous
-1530.4 kJ