At 25ºC and 1 atmosphere the following equilibrium is eatablished :
2NO (g) <=> N2O4 (g)
The partial equilibrium pressure of N2O4 is 0.7 atm.
A) calculate Kp of the above system at 25ºC.
B) calculate the partial pressure of the above two gases at equilibrium when the total pressure at equilibrium becomes 9atm and the temperature is still 25ºC
A) To calculate the equilibrium constant (Kp) of the system, you need to use the expression that relates the partial pressures of the reactants and products at equilibrium.
For the reaction:
2NO (g) <=> N2O4 (g)
The general expression for Kp can be written as:
Kp = (P(N2O4))^2 / (P(NO))^2
Given that the partial pressure of N2O4 is 0.7 atm, we can substitute the values into the expression:
Kp = (0.7)^2 / (P(NO))^2
B) To calculate the partial pressures of the two gases when the total pressure at equilibrium becomes 9 atm, we need to use the concept of the mole fraction.
The mole fraction of a gas is defined as the ratio of the number of moles of that gas to the total number of moles in the system. It can be calculated using the ideal gas law:
PV = nRT
Where:
P - Partial pressure of the gas
V - Volume of the gas
n - Number of moles of the gas
R - Ideal gas constant
T - Temperature in Kelvin
To find the partial pressure of each gas, we first need to calculate the number of moles of each gas.
For N2O4:
n(N2O4) = (P(N2O4) * V) / (RT)
For NO:
n(NO) = (P(NO) * V) / (RT)
Now, let's calculate the partial pressures of the two gases at equilibrium.
P(total) = P(N2O4) + P(NO)
Given that P(total) = 9 atm, we substitute the values:
9 = 0.7 + P(NO)
P(NO) = 8.3 atm
Therefore, at equilibrium, the partial pressures of N2O4 and NO are 0.7 atm and 8.3 atm, respectively, when the total pressure is 9 atm and the temperature is 25ºC.