A 10.0-L stainless steel container was charged with 2.00 atm of hydrogen gas and 3.00 atm of oxygen gas. A spark ignited the mixture, producing water. What is the pressure in the tank at a constant temperature of 25°C?
Well, I hope the tank didn't get too steamed up! Anyway, to find the total pressure in the tank, we can add the partial pressures of hydrogen and oxygen. So, let's get to it!
The pressure of hydrogen is 2.00 atm and the pressure of oxygen is 3.00 atm. Adding these together, we get a total pressure of 5.00 atm. Ta-da!
So, the pressure in the tank at a constant temperature of 25°C is 5.00 atm. Now, that's a lot of pressure! But hey, at least we got some water out of it. Stay hydrated!
To determine the pressure in the tank after the reaction, we need to calculate the moles of reactants present in the container and use the ideal gas law.
Given:
Volume (V) = 10.0 L
Hydrogen gas pressure (P_H2) = 2.00 atm
Oxygen gas pressure (P_O2) = 3.00 atm
Temperature (T) = 25°C or 298 K
1. Calculate the moles of hydrogen gas (H2):
We can use the ideal gas law equation:
PV = nRT
R = 0.0821 L.atm/(mol.K) (gas constant)
T = 298 K (temperature)
Rearranging the equation, we get:
n = PV / RT
n_H2 = (2.00 atm) * (10.0 L) / [(0.0821 L.atm/(mol.K)) * (298 K)]
n_H2 = 0.805 mol
2. Calculate the moles of oxygen gas (O2):
Using the same ideal gas law equation:
n_O2 = (3.00 atm) * (10.0 L) / [(0.0821 L.atm/(mol.K)) * (298 K)]
n_O2 = 1.207 mol
3. Determine the limiting reactant:
To find the limiting reactant, we compare the number of moles of hydrogen and oxygen gas. The reactant that has fewer moles will limit the reaction and be fully consumed.
Since we have 0.805 mol of H2 and 1.207 mol of O2, hydrogen gas is the limiting reactant.
4. Determine the number of moles of water produced:
The balanced chemical equation for the reaction is:
2H2 + O2 -> 2H2O
From the equation, each mole of O2 reacts with 2 moles of H2 to produce 2 moles of H2O.
Since the number of moles of hydrogen gas is the limiting reactant, the moles of water produced will be equal to twice the number of moles of hydrogen gas.
n_H2O = 2 * n_H2
n_H2O = 2 * 0.805 mol
n_H2O = 1.610 mol
5. Calculate the total moles of gas in the container after the reaction:
The moles of water produced are not considered in the pressure calculation since it is in the liquid phase.
The total moles of gas in the container after the reaction would be the sum of the remaining moles of hydrogen gas and oxygen gas:
Total moles of gas = (n_H2 + n_O2) - n_H2O
Total moles of gas = (0.805 mol + 1.207 mol) - 1.610 mol
Total moles of gas = 0.402 mol
6. Calculate the pressure in the tank:
Now we can apply the ideal gas law again to get the pressure:
P = nRT / V
P = (0.402 mol) * (0.0821 L.atm/(mol.K)) * (298 K) / (10.0 L)
P ≈ 0.989 atm
Therefore, the pressure in the tank at a constant temperature of 25°C is approximately 0.989 atm.
To determine the pressure in the tank after the reaction, we need to use the ideal gas law. The ideal gas law formula is given by:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
In this case, we are given the volume (V) of the container, which is 10.0 L, and the temperature (T), which is 25°C. However, to use the ideal gas law, the temperature needs to be in Kelvin, so we'll need to convert it.
To convert Celsius to Kelvin, we use the formula:
T(K) = T(°C) + 273.15
So let's convert the temperature:
T(K) = 25°C + 273.15
T(K) = 298.15 K
Now that we have the volume and temperature in the correct units, we need to calculate the number of moles of each gas. To do this, we'll use the ideal gas law rearranged to solve for the number of moles (n):
n = PV / RT
For hydrogen gas:
P(H2) = 2.00 atm
n(H2) = (2.00 atm) * (10.0 L) / [(0.0821 L·atm/(K·mol)) * (298.15 K)]
n(H2) ≈ 0.0826 mol
For oxygen gas:
P(O2) = 3.00 atm
n(O2) = (3.00 atm) * (10.0 L) / [(0.0821 L·atm/(K·mol)) * (298.15 K)]
n(O2) ≈ 0.124 mol
Now we can calculate the total pressure of the gases after the reaction by adding the pressures of hydrogen and oxygen:
P(total) = P(H2) + P(O2)
P(total) = 2.00 atm + 3.00 atm
P(total) = 5.00 atm
Therefore, the pressure in the tank at a constant temperature of 25°C is 5.00 atm.
Since these are gases, we can dispense with converting to moles; we may use pressure directly.
2 atm H2 will produce 2 atm H2O (as a gas which it isn't at 25C)
3 atm O2 will produce 6 atm H2O (as a gas but it isn't); therefore, H2 is the limiting reagent and O2 will be used but some will remain. How much will remain.
2 atm H2 reacts with 1 atm O2 to produce water (which as a liquid at 25 C we assume will not occupy much volume or change the pressure very much.)
We started with 3 atm O2, we used 1 atm, so 2 atm must remain. No pressure from H2 since all of it was used and no pressure from H2O since it is a liquid. I believe the answer is 2 atm pressure will exist in the container at 25 C. Check my thinking. You can double check the numbers by converting pressures to moles H2 and moles O2, redetermining the limiting reagent from moles, then convert back to pressure.