given 2C6H6+15O2->12CO2+6H2O
1. Starting with a confined volume of 4.80L and O2 with a partial pressure of 223.1 torr at 25 degrees C. C6H6 is in excess. The reaction is allowed to proceed to completion. The container is returned to 25C. Both liquid water and benzene are now present. The vapor pressure for benzene is 96.0 torr at 25C and vapor pressure of water at 25C is 24.0 torr. Calculate the total pressure. (you first need to calculate the CO2 pressure, which is its partial pressure)
See your other post above.
To calculate the total pressure in the container, we first need to calculate the partial pressure of CO2. We are given the equation 2C6H6 + 15O2 -> 12CO2 + 6H2O, which tells us that the molar ratio of O2 to CO2 is 15:12.
Since O2 is in excess, we can assume that all the O2 is consumed in the reaction and is converted to CO2. Therefore, the moles of CO2 formed will also be in a 15:12 ratio with the moles of O2 initially present.
To find the number of moles of CO2, we need to convert the given volume of 4.80L and the partial pressure of O2, 223.1 torr, to moles.
Using the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin, we can rearrange the equation to solve for n:
n = PV / RT
Given:
Partial pressure of O2 (P) = 223.1 torr
Volume (V) = 4.80 L
Temperature (T) = 25°C = 25 + 273.15 = 298.15 K
The ideal gas constant (R) is 0.0821 L·atm/(K·mol).
n = (223.1 torr)(4.80 L) / (0.0821 L·atm/(K·mol))(298.15 K)
n = 42.87 mol O2 (rounded to two decimal places)
Since the moles of CO2 will be in a 15:12 ratio with the moles of O2, we can calculate the moles of CO2 formed:
moles of CO2 = (12/15) * 42.87 mol
moles of CO2 = 34.30 mol CO2 (rounded to two decimal places)
Now, let's calculate the partial pressure of CO2. The total volume in the container is still 4.80 L, but remember that both liquid water and benzene are present, causing the total pressure to be higher than just the pressure of O2.
To calculate the partial pressure of CO2, we can use Dalton's law of partial pressures, which states that the total pressure in a mixture of non-reactive gases is equal to the sum of the partial pressures of the individual gases.
Taking into account the vapor pressures of benzene and water, we can calculate the partial pressure of CO2 as follows:
Total pressure = partial pressure of CO2 + vapor pressure of benzene + vapor pressure of water
Total pressure = partial pressure of CO2 + 96.0 torr + 24.0 torr
Given:
Vapor pressure of benzene = 96.0 torr
Vapor pressure of water = 24.0 torr
Total pressure = partial pressure of CO2 + 96.0 torr + 24.0 torr
Since the total pressure is what we are trying to calculate, let's rearrange the equation to solve for the partial pressure of CO2:
partial pressure of CO2 = Total pressure - vapor pressure of benzene - vapor pressure of water
partial pressure of CO2 = Total pressure - (96.0 torr + 24.0 torr)
Now, substitute the known values:
partial pressure of CO2 = Total pressure - 120.0 torr
Therefore, the total pressure in the container is equal to the partial pressure of CO2 plus the vapor pressures of benzene and water, which is (partial pressure of CO2) + 120.0 torr.