Consider the reaction
2H2O(g) → 2H2(g) + O2(g) ΔH = +483.60 kJ/mol
at a certain temperature. If the increase in volume is 42.7 L against an external pressure of 1.00 atm, calculate ΔU for this reaction. (The conversion factor is 1 L · atm = 101.3 J.)
To calculate ΔU for the reaction, we can use the equation:
ΔU = ΔH - ΔnRT
Where:
ΔU: change in internal energy
ΔH: change in enthalpy
Δn: change in moles
R: ideal gas constant (0.0821 L·atm/(mol·K))
T: temperature in Kelvin
First, we need to determine the change in moles (Δn) of the reaction. By examining the balanced equation, we can see that 2 moles of water produce 2 moles of hydrogen gas (H2) and 1 mole of oxygen gas (O2). Therefore, Δn = (2 + 1) - 2 = 1 mole.
Next, we need to convert the given volume change (42.7 L) to moles using the ideal gas law equation:
PV = nRT
Where:
P: pressure (1.00 atm)
V: volume change (42.7 L)
n: number of moles
R: ideal gas constant (0.0821 L·atm/(mol·K))
T: temperature in Kelvin
Solving for n:
n = PV / RT
n = (1.00 atm) * (42.7 L) / (0.0821 L·atm/(mol·K)) * T
Now, we need to convert the volume from liters to moles at 0 degrees Celsius (273 K):
n = (1.00 atm) * (42.7 L) / (0.0821 L·atm/(mol·K)) * 273 K
After calculating n, we can substitute the values in the equation:
ΔU = ΔH - ΔnRT
ΔU = +483.60 kJ/mol - (1 mole) * (0.0821 L·atm/(mol·K)) * T
Note: We don't have the specific temperature given in the problem, so we cannot determine the exact value for ΔU without that information.
To calculate the change in internal energy (ΔU) for this reaction, we need to use the relationship:
ΔU = ΔH - ΔnRT
where ΔH is the change in enthalpy, Δn is the change in the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
First, let's determine the change in moles of gas (Δn). From the balanced equation, we can see that for every 2 moles of H2O, we get 2 moles of H2 and 1 mole of O2. So, Δn = (2 moles of H2 + 1 mole of O2) - (2 moles of H2O).
Now, let's calculate the value of Δn:
Δn = (2 + 1) - 2 = 1
We also need to convert the given increase in volume to moles using the ideal gas law:
PV = nRT
Rearranging the equation, we get:
n = PV / RT
Using the given values, n = (1.00 atm × 42.7 L) / (1.00 L atm/mol K × temperature in Kelvin) = (42.7 L) / (temperature in Kelvin)
Now, we can substitute the values into the equation ΔU = ΔH - ΔnRT:
ΔU = +483.60 kJ/mol - (1 mol × 1.00 L atm/mol K × temperature in Kelvin)
However, we need to convert the units to Joulse (J) because the conversion factor is given as 1 L · atm = 101.3 J.
ΔU = +483.60 kJ/mol - (1 mol × 101.3 J/mol K × temperature in Kelvin)
Finally, we have the expression for ΔU in terms of the temperature:
ΔU = +483.60 kJ/mol - 101.3 J/mol K × temperature in Kelvin
Simplifying further, we convert kJ to J:
ΔU = 483,600 J/mol - 101.3 J/mol K × temperature in Kelvin
Please note that the final answer will depend on the temperature provided.
du = dq + dw
du = +483.6 KJ/Mol + 42.7 L * 1*10^5
The above by Cameron is correct EXCEPT the dw portion.
In L*atm that is 42.7 L x 1 atm. Convert that to J by 42.7 x 1 x 101.3 = dW.