Consider the reaction below.
3 O2(g) --> 2 O3(g)
At 175°C and a pressure of 128 torr, an equilibrium mixture of O2 and O3 has a density of 0.168 g/L. Calculate the Kp for the above reaction of 175°C
To calculate the Kp for the given reaction at 175°C, we need to know the molar mass of the gases involved. The molar mass of oxygen (O2) is 32 g/mol, and the molar mass of ozone (O3) is 48 g/mol.
Next, we need to determine the number of moles of oxygen gas (O2) and ozone gas (O3) in the given density of the equilibrium mixture. The density is given as 0.168 g/L.
Let's start by calculating the number of moles of the equilibrium mixture. We can use the equation:
moles = mass / molar mass
moles of mixture = 0.168 g / (32 g/mol) = 0.00525 mol/L
Now, since we know the stoichiometry of the reaction, we can determine the number of moles of O2 and O3. According to the balanced equation, for every 3 moles of oxygen gas (O2), we have 2 moles of ozone gas (O3).
moles of O2 = (2/3) * moles of mixture = (2/3) * 0.00525 mol/L = 0.0035 mol/L
moles of O3 = (2/3) * 0.00525 mol/L = 0.0035 mol/L
Now, we can use the ideal gas law to relate the moles of gases to the pressure. The ideal gas law equation is:
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.
To calculate Kp, we need to convert the given pressure from torr to atm (since R is given in atm). 1 atm = 760 torr.
So, the pressure in atm is 128 torr / 760 = 0.1684 atm.
Now, we can calculate Kp using the expression:
Kp = (P O3)^2 / (P O2)^3
Kp = (0.0035 mol/L)^2 / (0.0035 mol/L)^3 = 0.2222
Therefore, the Kp for the given reaction at 175°C is approximately 0.2222.
To calculate the Kp for the reaction at 175°C, we need to use the ideal gas law and the given information. The ideal gas law equation is:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in L)
n = moles
R = gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, we need to convert the pressure from torr to atm:
1 atm = 760 torr
128 torr / 760 torr/atm = 0.1684 atm
Next, we need to find the number of moles of the equilibrium mixture. We can use the density and molar mass to calculate the moles. The molar mass of O2 is 32 g/mol, and O3 is 48 g/mol.
Density = mass/volume
0.168 g/L = mass (in g) / 1 L
mass = 0.168 g
moles = mass / molar mass
moles of equilibrium mixture = 0.168 g / 48 g/mol = 0.0035 mol
Now, we can plug the values into the ideal gas law equation:
PV = nRT
(0.1684 atm) * V = (0.0035 mol) * (0.0821 L·atm/mol·K) * (175 + 273 K)
Now, let's solve for V:
V = ((0.0035 mol) * (0.0821 L·atm/mol·K) * (175 + 273 K)) / (0.1684 atm)
V ≈ 0.531 L
Now that we have the volume, we can use the balanced equation to calculate the partial pressures of O2 and O3 in the equilibrium mixture:
For 3 moles of O2 and 2 moles of O3:
P(O2) = (3/5) * total pressure = (3/5) * 0.1684 atm ≈ 0.101 atm
P(O3) = (2/5) * total pressure = (2/5) * 0.1684 atm ≈ 0.068 atm
Now, we can calculate Kp using the partial pressures:
Kp = (P(O3))^2 / (P(O2))^3
Kp = (0.068 atm)^2 / (0.101 atm)^3
Kp ≈ 0.032
Therefore, the Kp for the reaction at 175°C is approximately 0.032.