2H2(g) + O2(g) = 2H2O (l)
Calculate the volume of Hydrogen gas at STP required to produce an amount of energy equivalent to that produced by the combustion of a gallon of octane. The density of octane is 2.66 kg/gal, and its standard enthalpy of formation is -249.9 kJ/mol.
HELPPP!!
This problem requires many steps, and I will just outline the process of solution.
(1) Each mole of H2O formed will release 57.8 kcal if carried out at STP with liquid H2O as the product. One mole of H2 will be consumed in the process. The actual process will have H2O (g) as a product, with the result that less usable heat is usually generated.
(2) Write the balanced reaction for the complete oxidation of octane.
(3) Use your stated heat of formation of octane (C8H18), and those of H2O and CO2, to determine the heat of reaction per mole of octane burned. Assume liquid H2O and gaseous CO2 as products
(4) Compare the heats of reaction per mole of H2 and mole of CO2 burned
(5) Convert 1 gallon of octane to moles and grams.
(6) Compute the heat release from burning one gallon of octane
(7) Compute the number of moles of H2 needed to release the same amount of heat
(8) Convert (7) to STP volume of H2
no! im taking a timed quiz!!!
is it 2.25 L??
To calculate the volume of Hydrogen gas required to produce an amount of energy equivalent to that produced by the combustion of a gallon of octane, we need to follow these steps:
Step 1: Calculate the amount of energy produced by the combustion of octane.
We are given the standard enthalpy of formation of octane (-249.9 kJ/mol) and the density of octane (2.66 kg/gal). We need to convert the density of octane from kg/gal to kg/mol.
1 gallon of octane = 2.66 kg
Molar mass of octane (C8H18) = 12.01 * 8 + 1.01 * 18 = 114.22 g/mol
1 kg = 1000 g
Therefore, 2.66 kg of octane is equal to (2.66 kg / (114.22 g/mol)) = 0.0233 mol.
Now, we can calculate the amount of energy produced:
Energy = -249.9 kJ/mol * 0.0233 mol = -5.81 kJ
Step 2: Calculate the volume of Hydrogen gas required to produce the same amount of energy.
We can use the balanced equation for the combustion of Hydrogen gas:
2H2(g) + O2(g) -> 2H2O(l)
From the balanced equation, we can see that for every 2 moles of Hydrogen gas, we get 2 moles of water. Thus, the molar ratio between Hydrogen gas and water is 1:1.
Since we need to produce the same amount of energy as that produced by the combustion of octane, we can assume that the enthalpy change is the same. In other words, the enthalpy change for the combustion of Hydrogen gas is also -5.81 kJ/mol.
Now, we need to calculate the volume of Hydrogen gas required at STP (Standard Temperature and Pressure). STP is defined as 0 degrees Celsius (273.15 K) and 1 atmosphere (atm) pressure.
Using the ideal gas law: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant (0.0821 L.atm/mol.K), and T is the temperature in Kelvin.
At STP, the pressure (P) is 1 atm, and the temperature (T) is 273.15 K.
We can rearrange the ideal gas law to solve for volume (V):
V = nRT / P
Substituting the known values, we get:
V = (0.0233 mol)(0.0821 L.atm/mol.K)(273.15 K) / (1 atm)
V ≈ 0.52 L
Therefore, the volume of Hydrogen gas at STP required to produce an amount of energy equivalent to that produced by the combustion of a gallon of octane is approximately 0.52 liters.