two most abundant atmospheric gases react to a small extent at 298K in the presence of a catalyst to achieve dynamic equilibrium:
N2(g) + O2(g) ->/<- 2NO(g) Kp= 0.30
Partial pressure of N2 and O2 = 0.78atm
1dm3 flask and equilibrium established after 30min.
temperature of system raised to 350K at 40min and equilibrium re-established at 60min. calculate the presssure of N2 and NO at 70min
1. Is the 0.78 atm the partial pressure to start or after equilibrium is established?
2. What's the new Kp (or Kc) at 350K?
I suggest you repost the question at the top of the page and include the information in answer to the questions 1 and 2.
To calculate the pressure of N2 and NO at 70 minutes, we need to determine the changes in the amounts of N2 and NO as the temperature is increased.
First, let's determine the moles of N2 and NO at equilibrium at 30 minutes.
Given:
Partial pressure of N2 and O2 = 0.78 atm
Kp = 0.30
In the balanced equation, the stoichiometry between N2 and NO is 1:2. Therefore, if x moles of N2 react, then 2x moles of NO will be formed.
At equilibrium, the partial pressure of NO is 0.78 * 2 = 1.56 atm. (using the stoichiometry)
From the given Kp expression: Kp = (P(NO))^2 / (P(N2) * P(O2))
Since the partial pressure of N2 and O2 is the same, let's denote it as x. Therefore, we can write the Kp expression as: 0.30 = (1.56)^2 / (x * x)
Simplifying the equation:
0.30 * (x^2) = 1.56^2
x^2 = (1.56^2) / 0.30
x β 2.51
So there are approximately 2.51 moles of N2 and 2 * 2.51 = 5.02 moles of NO at equilibrium at 30 minutes.
Now, let's consider the change in temperature. The temperature is raised to 350K at 40 minutes, and the new equilibrium is re-established at 60 minutes.
Since the temperature increases, according to Le Chatelier's Principle, the reaction will shift in the direction that consumes heat. In this case, it will be the reverse reaction, which is the formation of N2 and O2 from NO.
Now, let's calculate the pressure of N2 and NO at 60 minutes.
From 30 to 60 minutes, the equilibrium concentration of NO decreases by 5.02 moles. Therefore, at 60 minutes, the concentration of NO is (5.02 - 5.02) moles.
At 60 minutes, the concentration of N2 will be (2.51 + 5.02) moles, as 2 moles of N2 are formed for every mole of NO consumed.
Now, at 70 minutes, we need to consider the final change in pressure due to the change in temperature.
From 60 to 70 minutes, the equilibrium concentration of NO will remain the same because the reaction has reached a new equilibrium. Therefore, the concentration of NO at 70 minutes will be (5.02 - 5.02) moles.
Similarly, the concentration of N2 at 70 minutes will be (2.51 + 5.02) moles.
Finally, we calculate the pressure of N2 and NO at 70 minutes by using the ideal gas law:
Pressure(N2) = (moles(N2) * R * T) / V
Pressure(NO) = (moles(NO) * R * T) / V
Where:
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature in Kelvin (350K)
V = volume of the flask (1 dm^3)
Substituting in the values, we can calculate the pressures of N2 and NO at 70 minutes.