Given the following thermochemical data:
1) C(s) + O2(g) -> CO2(g) ∆H = -393.5 kJ
2) H2(g) + 1/2 O2(g) -> H2O(l) ∆H = -285.8 kJ
3) 3 C(s) + 3 H2(g) -> C3H6(g) ∆H = +20.4 kJ
4) C3H6 (cyclopropane)(g) -> C3H6(propene)(g) ∆H = -33.0 kJ
What is the ∆H for the following reaction?
C3H6 (cyclopropane)(g) + 9/2 O2(g) -> 3CO2(g)∆H = ______.
To find the ∆H for the given reaction, we need to use the thermochemical data provided.
The given reaction is:
C3H6 (cyclopropane)(g) + 9/2 O2(g) -> 3 CO2(g)
We can calculate the ∆H for this reaction using the Hess's Law, which states that if a reaction can be expressed as the sum of two or more other reactions, then the ∆H for the overall reaction is the sum of the ∆H values of the individual reactions.
In this case, we can break down the given reaction into multiple steps:
1) C3H6 (cyclopropane)(g) -> C3H6 (propene)(g) ∆H = -33.0 kJ (Given in thermochemical data)
2) C3H6 (propene)(g) + 3 O2(g) -> 3 CO2(g) + 3 H2O(l) ∆H = -393.5 kJ (Using the first thermochemical data for C(s) + O2(g) -> CO2(g))
3) 3 H2O(l) -> 3 H2(g) + 3/2 O2(g) ∆H = +285.8 kJ (Using the second thermochemical data for H2(g) + 1/2 O2(g) -> H2O(l))
Now, we can add up these three reactions to get the overall reaction:
C3H6 (cyclopropane)(g) + 9/2 O2(g) -> 3 CO2(g)
∆H = ∆H1 + ∆H2 + ∆H3
= (-33.0 kJ) + (-393.5 kJ) + (+285.8 kJ)
Calculating the sum, we get:
∆H = -421.2 kJ
Therefore, the ∆H for the given reaction is -421.2 kJ.