What is the enthalpy for reaction 1 reversed?
reaction 1 reversed: 2 CO2 + 3 H2O → C2H5OH + 3 O2
Express your answer numerically in kilojoules per mole.
The change in enthalpy, ΔH, is the heat absorbed or produced during any reaction at constant pressure. Hess's law states that ΔH for an overall reaction is the sum of the ΔH values for the individual reactions. For example, if we wanted to know the enthalpy change for the reaction
3 Mn + 3 O2 → 3 MnO2
we could calculate it using the enthalpy values for the following individual steps:
Step 1: 4 Al + 3 O2 → 2 Al2O3
Step 2: 3 Mn + 2 Al2O3 → 3 MnO2 + 4 Al
Overall: 3 Mn + 3 O2 → 3 MnO2
If the enthalpy change is −3352 kJ/mol for step 1 and 1792 kJ/mol for step 2, then the enthalpy change for the overall reaction is calculated as follows:
ΔH = −3352 + 1792 = − 1560 kJ/mol
It is also important to note that the change in enthalpy is a state function, meaning it is independent of path. In other words, the sum of the ΔH values for any set of reactions that produce the desired product from the starting materials gives the same overall ΔH.
Now consider the following set of reactions:
1. C2H5OH + 3 O2 → 2 CO2 + 3 H2O, ΔH = −1370 kJ/mol
2. C2H4 + 3 O2 → 2 CO2 + 2 H2O, ΔH = −1410 kJ/mol
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