Consider the following reaction:
4NH3(g) +5O2(g) -->4NO(g) +6H2O(g)
Suppose we have two flasks connected by a stopcock. In one flask, we have 2.0L of NH3(g) at a pressure of .500 atm. In the other flask, we have 1.00L of O2(g) at a pressure of 1.5atm. When the stopcock is opened, the gases mix according to the equation above.
Calculate the partial pressure of all gases after the reaction is complete.
Calculate the total pressure after the reaction is complete. You may assume the total volume is 3.00L.
To find the partial pressure of each gas after the reaction is complete, we can use the stoichiometry of the balanced equation.
First, let's calculate the moles of NH3 and O2 in the initial solutions:
Given:
Volume of NH3 = 2.0 L
Pressure of NH3 = 0.500 atm
Using the ideal gas law, we can calculate the number of moles of NH3:
PV = nRT
n(NH3) = (PV) / RT
Substituting the values into the equation:
n(NH3) = (0.500 atm * 2.0 L) / (0.0821 L.atm/mol.K * 298 K)
n(NH3) = 0.0407 mol
Given:
Volume of O2 = 1.00 L
Pressure of O2 = 1.5 atm
Using the ideal gas law, we can calculate the number of moles of O2:
n(O2) = (PV) / RT
Substituting the values into the equation:
n(O2) = (1.5 atm * 1.00 L) / (0.0821 L.atm/mol.K * 298 K)
n(O2) = 0.0612 mol
Now, let's analyze the balanced equation to determine the stoichiometry. The balanced equation shows that for every 4 moles of NH3 consumed, 4 moles of NO are produced, and for every 5 moles of O2 consumed, 6 moles of H2O are produced. Therefore, NH3 and O2 are in a 1:1 ratio.
Since we have 0.0407 mol of NH3 and 0.0612 mol of O2, we can determine that the limiting reactant is NH3.
Using the balanced equation, 4 moles of NH3 produce 4 moles of NO. Therefore, the number of moles of NO produced will also be 0.0407 mol.
Since the NH3 and O2 are completely consumed, there will be no excess of either gas remaining.
Now, let's calculate the partial pressure of each gas after the reaction is complete.
Partial pressure NH3 = (moles of NH3 * RT) / V
Partial pressure NH3 = (0.0407 mol * 0.0821 L.atm/mol.K * 298 K) / 3.00 L
Partial pressure NH3 = 0.672 atm
Partial pressure O2 = (moles of O2 * RT) / V
Partial pressure O2 = (0.0612 mol * 0.0821 L.atm/mol.K * 298 K) / 3.00 L
Partial pressure O2 = 0.734 atm
Partial pressure NO = (moles of NO * RT) / V
Partial pressure NO = (0.0407 mol * 0.0821 L.atm/mol.K * 298 K) / 3.00 L
Partial pressure NO = 0.672 atm
Partial pressure H2O = (moles of H2O * RT) / V
Partial pressure H2O = (0.0612 mol * 0.0821 L.atm/mol.K * 298 K) / 3.00 L
Partial pressure H2O = 0.734 atm
Now, to calculate the total pressure after the reaction is complete, sum up the partial pressures of all the gases:
Total pressure = Partial pressure NH3 + Partial pressure O2 + Partial pressure NO + Partial pressure H2O
Total pressure = 0.672 atm + 0.734 atm + 0.672 atm + 0.734 atm
Total pressure = 2.812 atm
Therefore, the partial pressures of the gases after the reaction is complete are as follows:
NH3: 0.672 atm
O2: 0.734 atm
NO: 0.672 atm
H2O: 0.734 atm
And the total pressure after the reaction is complete is 2.812 atm.
To calculate the partial pressure of all gases after the reaction is complete, we need to first determine the moles of each gas present in the initial conditions and then use the stoichiometry of the reaction to determine the moles of each gas consumed or produced.
1. Moles of NH3:
Using the ideal gas law, PV = nRT, we can rearrange the equation to solve for the number of moles (n). Given the volume (V) of NH3 = 2.0 L and pressure (P) = 0.500 atm, assuming temperature (T) is constant, we can calculate the number of moles of NH3:
n(NH3) = PV / RT
2. Moles of O2:
Similarly, using the given volume of O2 = 1.00 L and pressure of 1.5 atm, we can calculate the number of moles of O2 using the ideal gas law:
n(O2) = PV / RT
3. Calculate the number of moles of each gas produced or consumed:
From the balanced equation, we see that the stoichiometry of the reaction tells us that 4 moles of NH3 produces 4 moles of NO, and 5 moles of O2 produce 4 moles of NO. Thus, the molar ratio between NH3 and O2 is 4:5.
Since we have the number of moles for NH3 and the molar ratio, we can determine the number of moles of O2 consumed:
n(O2 consumed) = (4/5) * n(NH3)
4. Calculate the number of moles of each gas after the reaction:
For NH3 and O2, we subtract the consumed moles from the initial moles:
n(NH3 final) = n(NH3) - (4/5) * n(NH3)
n(O2 final) = n(O2) - n(O2 consumed)
5. Calculate the moles of NO and H2O produced:
Using the stoichiometric coefficients in the balanced reaction equation, we find that 4 moles of NH3 produce 4 moles of NO and 6 moles of H2O.
n(NO) = 4 * (n(NH3 final))
n(H2O) = 6 * (n(NH3 final))
6. Calculate the partial pressure of each gas:
To find the partial pressure of each gas, we use the ideal gas law again, utilizing the moles we calculated in step 5. Assume the same temperature (T) as before:
P(NH3 final) = (n(NH3 final) * RT) / V(total)
P(O2 final) = (n(O2 final) * RT) / V(total)
P(NO) = (n(NO) * RT) / V(total)
P(H2O) = (n(H2O) * RT) / V(total)
7. Calculate the total pressure:
To find the total pressure after the reaction, we sum up the partial pressures of all the gases:
P(total) = P(NH3 final) + P(O2 final) + P(NO) + P(H2O)
By following these steps and plugging in the appropriate values, you can calculate the partial pressure of all gases after the reaction is complete and the total pressure after the reaction is complete.
2.0 L flask:
Use PV = nRT and solve for n.
1.00L flask:
Use PV = nRT and solve for n.
Add to find total mols.
Then PV = nRT and use total mols and total volume to find total pressure.
Use PV = nRT and EACH n with total volume to find partial pressure of each.
I know you don't have a temperature listed but the easy way is to make up a temperature and use that throughout. I would use the number 1 to make calculation easier; the n you get will not be the right n but that won't make any difference.