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)
The "following reaction" is not balanced for H atoms.
-1530.4 kJ
To find the ∆H for the given reaction, you can use Hess's Law, which states that the overall enthalpy change for a reaction is the sum of the enthalpy changes of the individual reactions that lead to it.
Here's how you can calculate ∆H:
1) Start with the given reaction: C6H10(g) + 2H2(g) -> C6H12(g)
2) Identify which reactions involve the same substances on both sides of the equation. In this case, Reaction 2 involves the same products, CO2(g) and H2O(g), as the given reaction. However, Reaction 2 has the wrong stoichiometric coefficients, so it needs to be reversed.
Reversed Reaction 2: 6CO2(g) + 5H2O(g) -> C6H10(g) + 17/2 O2(g)
3) Obtain the balanced chemical equation for the given reaction by adding Reaction 1, Reaction 2 (reversed), and Reaction 3:
C6H10(g) + 2H2(g) -> C6H12(g)
(multiply Reaction 2 by 2)
2(C6H10(g) + 2H2(g) -> C6H12(g))
+ 16CO2(g) + 15H2O(g)
- 16CO2(g) - 16H2O(g)
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C6H10(g) + 2H2(g) -> C6H12(g) + 16CO2(g) + 15H2O(g)
4) Determine the change in enthalpy (∆H) by summing up the enthalpy changes (∆H) for each reaction:
∆H = -241 kJ + 2*(-3800 kJ) + (-3920 kJ)
∆H = -241 kJ - 7600 kJ - 3920 kJ
∆H = -11861 kJ
Therefore, the ∆H for the given reaction, C6H10(g) + 2H2(g) -> C6H12(g), is -11861 kJ.