I don't understand how to even start this problem:
A mixture of H2(g), O2(g), and 2 mL of H2O is present in a 0.5 L rigid container at 25 degrees Celsius. The number of moles of H2 and the number of moles of O2 are equal. The total pressure is 1146 mmHg. Vapor pressure of water at 25 degrees Celsius is 24 mmHg. The mixture is sparked and H2 and O2 react until one reactant is completely missing.
1) Identify the reactant remaining and calculate the number of moles of the reactant.
2) Calculate the total pressure in the container at the conclusion of the reaction if the final temperature is 90 degrees Celsius. The equilibrium pressure of water at 90 degrees Celsius is 526 mmHg.
3) Calculate the number of moles of water present as vapor in the container at 90 degrees Celsius.
Partially answered at the original post.
To solve this problem step-by-step, let's break it down and address each part individually.
1) Identify the reactant remaining and calculate the number of moles of the reactant.
First, we know that the number of moles of H2 and O2 are equal. Let's assume that x represents the number of moles of each gas.
Therefore, the initial moles of H2 gas are x, and the moles of O2 gas are also x.
Next, we need to determine the reaction taking place. Since H2 and O2 are being sparked, they will react to form water (H2O) according to the equation:
2H2(g) + O2(g) -> 2H2O(g)
From the equation, we can see that 2 moles of H2 are needed to react with 1 mole of O2.
Since the number of moles of H2 and O2 are equal at the start, we can conclude that O2 is the limiting reactant, and H2 is the reactant remaining.
To calculate the number of moles of H2 remaining, we need to find the number of moles of O2 reacted. Since 2 moles of H2 reacts with 1 mole of O2, the number of moles of O2 reacted is x/2.
Therefore, the number of moles of H2 remaining is x - x/2 = x/2.
2) Calculate the total pressure in the container at the conclusion of the reaction if the final temperature is 90 degrees Celsius. The equilibrium pressure of water at 90 degrees Celsius is 526 mmHg.
To find the total pressure at the conclusion of the reaction, we need to consider the partial pressures of the remaining H2 and the water vapor.
The partial pressure of H2 is given by the ideal gas law equation:
PV = nRT
Where P is the partial pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Since the number of moles of H2 remaining is x/2 (as calculated in step 1), we can find the partial pressure of H2 by substituting the values into the ideal gas law equation:
P_H2 = (x/2)RT
The equilibrium pressure of water vapor at 90 degrees Celsius is given as 526 mmHg. Therefore, the partial pressure of water vapor is 526 mmHg.
The total pressure is the sum of the partial pressures:
Total Pressure = Partial Pressure of H2 + Partial Pressure of Water Vapor
Total Pressure = (x/2)RT + 526 mmHg
3) Calculate the number of moles of water present as vapor in the container at 90 degrees Celsius.
To find the number of moles of water present as vapor, we can use the ideal gas law equation again.
We know that the total pressure of the system is given as the sum of the partial pressures of H2 and water vapor:
Total Pressure = (x/2)RT + 526 mmHg
By rearranging the ideal gas law equation, we can solve for the number of moles:
n = PV/RT
Substituting the values into the equation:
n_water_vapor = (526 mmHg) * (0.5 L) / (0.0821 L·atm/mol·K * (273+90) K)
Simplifying the equation and calculating the result will give you the number of moles of water vapor present in the container at 90 degrees Celsius.
Please note that it's essential to convert any units to match the units used in the gas constant (R), which is shown in the equation as L·atm/mol·K.
To solve this problem, we need to use the ideal gas law equation, which is given by:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
1) To identify the reactant remaining and calculate the number of moles of the reactant, we can start by calculating the number of moles of water vapor present in the container.
First, we need to convert the given volumes to liters. 2 mL is equal to 0.002 L.
Since the total volume of the container is 0.5 L, the volume of the gas mixture (H2 and O2) is 0.5 L - 0.002 L = 0.498 L.
Next, we can calculate the moles of water vapor present in the container using the ideal gas law equation.
First, we rearrange the equation to solve for moles:
n = PV / RT
The vapor pressure of water at 25 degrees Celsius is given as 24 mmHg. We need to convert it to atmospheres (atm) using the conversion factor: 1 atm = 760 mmHg.
So, the vapor pressure of water is 24 mmHg / 760 mmHg/atm = 0.03158 atm.
Substituting the values into the equation:
n (water vapor) = (0.03158 atm) * (0.002 L) / (0.0821 L*atm/mol*K) * (298 K)
= 0.000244 mol
Since the number of moles of water in the mixture is extremely small compared to the number of moles of H2 and O2, we can assume that the number of moles of water remains constant throughout the reaction.
Since the number of moles of H2 and O2 are equal initially, and the reaction continues until one is completely missing, it means that both reactants are being consumed in equal amounts.
Therefore, the reactant remaining is water, and the number of moles of the reactant remaining is 0.000244 mol.
2) To calculate the total pressure in the container at the conclusion of the reaction when the final temperature is 90 degrees Celsius, we can assume that the reaction has gone to completion and all the H2 and O2 have formed water vapor.
First, we need to convert the given temperatures to Kelvin. 25 degrees Celsius is equal to (25 + 273) K = 298 K, and 90 degrees Celsius is equal to (90 + 273) K = 363 K.
We also need to convert the equilibrium pressure of water at 90 degrees Celsius to atmospheres. Given equilibrium pressure of water is 526 mmHg, we divide it by 760 mmHg/atm to get 0.6908 atm.
Since the number of moles of water vapor remains constant throughout the reaction, the total number of moles of gas in the container at the conclusion of the reaction is 0.000244 mol.
Using the ideal gas law equation, we can calculate the total pressure in the container:
P * V = n * R * T
Substituting the values:
P * 0.5 L = 0.000244 mol * (0.0821 L*atm/mol*K) * (363 K)
Solving for P:
P = (0.000244 mol * 0.0821 L*atm/mol*K * 363 K) / 0.5 L
= 0.00468844 atm
Adding the equilibrium pressure of water vapor:
Total pressure = 0.00468844 atm + 0.6908 atm
= 0.6955 atm
Therefore, the total pressure in the container at the conclusion of the reaction, when the final temperature is 90 degrees Celsius, is 0.6955 atm.
3) To calculate the number of moles of water present as vapor in the container at 90 degrees Celsius, we can use the ideal gas law equation again.
Using the same equation as before:
n = PV / RT
Substituting the values:
n (water vapor) = (0.6908 atm) * (0.002 L) / (0.0821 L*atm/mol*K) * (363 K)
= 0.01857 mol
Therefore, the number of moles of water present as vapor in the container at 90 degrees Celsius is 0.01857 mol.