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).
delta Hrxn = (sum products)-(sum reactants)
This is not a calculation. You can see the products are zero since both are in their standard state and heat formation water is a negative number so delta H rxn is + and that means the left side is favored. Delta S should favor the right side.
Good job
1. To determine which side enthalpy favors in the given reaction, we need to examine the enthalpy change (∆H) of the reaction. If the ∆H is negative, it means the reaction is exothermic, and if the ∆H is positive, it means the reaction is endothermic.
In this reaction:
2 H2O (l) → 2 H2 (g) + O2 (g)
Water (H2O) is being converted into hydrogen gas (H2) and oxygen gas (O2). As water is being split into its component parts, it requires energy input, making it an endothermic reaction. Thus, the enthalpy change (∆H) is positive, and enthalpy favors the right side (the product side) of the reaction.
Regarding entropy, without performing calculations, we cannot determine the direction in which entropy favors. We would need to calculate the entropy change (∆S) of the reaction to determine whether it increases or decreases.
2. To calculate the volume of water in gallons required to store the same energy as found in 10.0 gallons of gasoline, we need to convert the energy units and consider the energy density.
Given:
Energy content of gasoline = 32.0 MJ/L
Density of gasoline = 0.72 g/mL
1 L = 0.2641775 gal
Step 1: Calculate the energy content of gasoline in 10.0 gallons.
Energy of 1 gallon of gasoline = 32.0 MJ/L * 1 L = 32.0 MJ/gal
Energy of 10.0 gallons of gasoline = 32.0 MJ/gal * 10.0 gallons = 320 MJ
Step 2: Calculate the mass of gasoline in 10.0 gallons.
Density of gasoline = 0.72 g/mL
Volume of gasoline = 10.0 gallons * 0.2641775 L/gallon = 2.641775 L
Mass of gasoline = Volume * Density = 2.641775 L * 0.72 g/mL = 1.90352 kg
Step 3: Convert the mass of gasoline to the equivalent energy content in water.
Energy content of water = Energy content of one gram of water = 4.18 J/g * 1000 g = 4180 J/g
Energy stored in water = Energy content * Mass of gasoline = 4180 J/g * 1.90352 kg = 7941 kJ
Step 4: Convert the energy stored in water to volume in gallons.
Energy content of water = 4180 J/g
Energy of 1 gallon = 32.0 MJ = 32,000,000 J
Volume of water (in gallons) = Energy stored in water / Energy content of 1 gallon = 7941 kJ / (32,000,000 J/gal) = 0.248 gal
Therefore, approximately 0.248 gallons of water would need to be split to store the same amount of energy as found in 10.0 gallons of gasoline.