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
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To calculate the Kp for the reaction, we first need to determine the concentrations of O2 and O3 in the equilibrium mixture.
Given:
Temperature (T) = 175°C = 448 K
Pressure (P) = 128 torr
Density (d) = 0.168 g/L
Step 1: Convert pressure to partial pressures
Since we have the total pressure, we need to determine the partial pressures of O2 and O3.
Partial pressure of O2 (P(O2)) = (density of O2) × (molar mass of O2) × (R constant) × (temperature)
= (0.168 g/L) × (32 g/mol) × (0.0821 L atm/mol K) × (448 K) / 1 atm
= 19.86 atm
Partial pressure of O3 (P(O3)) = (density of O3) × (molar mass of O3) × (R constant) × (temperature)
= (0.168 g/L) × (48 g/mol) × (0.0821 L atm/mol K) × (448 K) / 1 atm
= 29.79 atm
Step 2: Calculate the Kp
The Kp expression for the given reaction is: Kp = (P(O3))^2 / (P(O2))^3
Kp = (29.79)^2 / (19.86)^3
= 2.264
Therefore, the Kp for the reaction at 175°C is 2.264.
To calculate the equilibrium constant Kp for the given reaction at 175°C, we need to know the partial pressures of O2 and O3 in the equilibrium mixture and the stoichiometric ratio of the reactants and products.
Since we are given the density of the mixture, we can use the ideal gas law to find the partial pressures. The ideal gas law is given by:
PV = nRT
Where:
P is the pressure (in atmospheres),
V is the volume (in liters),
n is the number of moles,
R is the gas constant (0.0821 L·atm/(mol·K)), and
T is the temperature (in Kelvin).
We can rearrange this equation to solve for the number of moles:
n = PV / RT
Since we are given the density (mass per unit volume) of the mixture, we can find the molar mass (M) of the mixture by dividing the mass (m) by the volume (V):
M = m / V
Now, we can substitute the values given in the problem:
Density (d) = 0.168 g/L
Pressure (P) = 128 torr = 128/760 atm (converting torr to atm)
Temperature (T) = 175°C + 273.15 = 448.15 K (converting from Celsius to Kelvin)
Gas constant (R) = 0.0821 L·atm/(mol·K)
First, let's calculate the number of moles (n) of the mixture:
n = (P * V) / (R * T)
n = (128/760) * 1 / (0.0821) * 448.15
n = 0.0227149 mol
Next, we calculate the molar mass (M):
M = m / V
M = 0.168 g / 1 L
M = 0.168 g/mol
Since the reaction is given in moles, we need to convert the molar mass of the mixture to the molar mass of just O3. The molar mass of O3 is 48 g/mol. So, we can calculate the number of moles of O3 (n(O3)):
n(O3) = (0.168 g/mol * 0.0227149 mol) / 48 g/mol
n(O3) = 0.00007955 mol
Comparing the stoichiometric coefficients in the balanced reaction, we find that the ratio of the number of moles of O3 to O2 is 2:3. Therefore, the number of moles of O2 (n(O2)) can be calculated as follows:
n(O2) = (0.00007955 mol * 3) / 2
n(O2) = 0.00011933 mol
Now that we have the moles of O2 and O3, we can calculate the partial pressures. The partial pressure of O2 (P(O2)) is given by:
P(O2) = (n(O2) * R * T) / V
P(O2) = (0.00011933 mol * 0.0821 L·atm/(mol·K) * 448.15 K) / 1 L
P(O2) = 3.6098 atm
Similarly, the partial pressure of O3 (P(O3)) is given by:
P(O3) = (n(O3) * R * T) / V
P(O3) = (0.00007955 mol * 0.0821 L·atm/(mol·K) * 448.15 K) / 1 L
P(O3) = 2.3736 atm
Finally, we can calculate the equilibrium constant Kp using the partial pressures:
Kp = (P(O3))^2 / (P(O2))^3
Kp = (2.3736 atm)^2 / (3.6098 atm)^3
Kp = 0.175
Therefore, the equilibrium constant Kp for the reaction at 175°C is 0.175.