A sample of air was passed through an electrical discharge causing the following reaction
N2(g) + O2(g)->2NO(g) .
Assuming that 1.00 mole of air (78% N2, 21% O2) was originally present at 1.00 atm, determine the partial pressure (in atm) of NO at equilibrium. Kp = 0.0123 at 4200 K (the temperature of the discharge).
I know Kp=[NO]^2/[N2][O2]. Also, I need to set up an ICE table, but do I use .78 for initial of N2 and .21 for Initial of O2? I'm confused.
Yes, that's what you do.
So would the ICE table look this this:
N2 O2 NO
.78 .21 0
-x -x 2x
.78-x .21-x 2x
(2x)^2/(.78-x)(.21-x)=Kp
Is that the correct set up? Thanks
You have it. Congrats!
To solve this problem, you will need to set up an ICE (Initial, Change, Equilibrium) table to determine the partial pressure of NO at equilibrium.
First, let's determine the initial moles of each component present in the reaction.
Given that 1.00 mole of air (composed of 78% N2 and 21% O2) is originally present, the initial moles of N2 and O2 can be calculated as follows:
Initial moles of N2 = 1.00 mol x 0.78 = 0.78 mol
Initial moles of O2 = 1.00 mol x 0.21 = 0.21 mol
Now, let's set up the ICE table:
N2(g) + O2(g) -> 2NO(g)
-----------------------------------------------------
Initial: 0.78 0.21 0
Change: -2x -x +2x
Equilibrium: 0.78-2x 0.21-x 2x
Here, 'x' represents the change in moles of NO at equilibrium.
According to the balanced chemical equation, 1 mole of N2 reacts with 1 mole of O2 to produce 2 moles of NO. Therefore, the change in moles of N2 is -2x, the change in moles of O2 is -x, and the change in moles of NO is +2x.
Now, we can express the equilibrium partial pressures of N2, O2, and NO in terms of 'x' and calculate the equilibrium expression.
Partial pressure of N2 at equilibrium (P(N2)): P(N2) = (moles of N2) / (total moles) = (0.78-2x) / (0.78-2x+0.21-x)
Partial pressure of O2 at equilibrium (P(O2)): P(O2) = (moles of O2) / (total moles) = (0.21-x) / (0.78-2x+0.21-x)
Partial pressure of NO at equilibrium (P(NO)): P(NO) = (moles of NO) / (total moles) = 2x / (0.78-2x+0.21-x)
Using the given equilibrium constant (Kp = 0.0123), we can now set up the equation:
Kp = (P(NO))^2 / (P(N2) * P(O2))
0.0123 = (2x / (0.78-2x+0.21-x))^2 / ((0.78-2x) / (0.78-2x+0.21-x)) * ((0.21-x) / (0.78-2x+0.21-x))
Simplifying the equation and solving for 'x' will give you the value for the change in moles of NO at equilibrium. Once you have 'x', you can substitute it back into the equilibrium expressions to calculate the partial pressure of NO at equilibrium.