In a vehicle, H2 could be used in a fuel cell to generate energy.
Hydrogen is not an energy source, but a means of transporting energy because the extraction of H2 from the logical source, H2O, takes as much energy as is returned when the gas is used.
2 H2O (l) → 2 H2 (g) + O2 (g)
1. Without doing any calculations, which side will enthalpy favor in this reaction? Entropy?
2. What volume of water in gallons would need to be split to store the same amount of energy as found in 10.0 gallons of gasoline? (Gasoline contains 44.4 MJ/kg, or 32.0 MJ/L, and has a density of 0.72 g/mL where 1 L = 0.2641775 gal).
Not good Andrew, not good....
OMG! Professor Moore!!
1. To determine the side that enthalpy favors in this reaction, we need to consider the enthalpy changes associated with the formation of H2 and O2 from H2O. Without any calculations, we can make the following predictions based on our understanding of enthalpy and the reactions involved:
- Entropy: The reaction involves the breakdown of water (H2O) into hydrogen (H2) and oxygen (O2) gases. Since there is an increase in the number of gas molecules, it is likely that the entropy will increase. Therefore, the entropy will favor the product side, i.e., 2 H2 (g) + O2 (g).
- Enthalpy: The values of enthalpy changes (ΔH) for the reactions are not provided, so we cannot determine this without doing calculations. However, the extraction of H2 from water typically requires energy input, so it is expected that ΔH for the reaction will be positive, meaning that enthalpy favors the reactant side, i.e., 2 H2O (l).
It should be noted that these predictions are based on general trends and assumptions. Actual calculations would be needed to accurately determine the enthalpy changes.
2. To find the volume of water required to store the same amount of energy as 10.0 gallons of gasoline, we need to convert the energy content of gasoline into the equivalent energy content of water:
- Energy content of gasoline: 32.0 MJ/L
- Density of gasoline: 0.72 g/mL
- Conversion factor from liters to gallons: 1 L = 0.2641775 gal
First, we need to convert the energy content of gasoline from MJ/L to MJ/gal:
32.0 MJ/L * 0.2641775 gal/L = 8.45736 MJ/gal
Next, we need to convert the energy content of gasoline from MJ/gal to MJ to match the units of water's energy content:
8.45736 MJ/gal * 10.0 gal = 84.5736 MJ
Now, we need to convert the energy content of water into volume using its density:
1 g/mL * X mL = 84.5736 MJ
Rearranging the equation to solve for X (the volume of water in milliliters):
X = 84.5736 MJ / 1 g/mL
Finally, we need to convert the volume of water from milliliters to gallons:
X mL * (1 L / 1000 mL) * (1 gal / 0.2641775 L)
Performing the calculations will give us the volume of water required to store the same energy as 10.0 gallons of gasoline.