At 2200 degrees celsius,kp=0.050 for the reaction: n2(g)+O2(g) ->2NO(g)
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What is the partial pressure of NO in equilibrium with N2 and O2 that were placed in a flask at initial pressures of .80 and .20 atm, respectively?
To find the partial pressure of NO in equilibrium, we can use the equation for the equilibrium constant (Kp) and the initial pressures of N2 and O2.
The equilibrium constant expression for the given reaction is:
Kp = (partial pressure of NO)^2 / (partial pressure of N2 * partial pressure of O2)
Given that Kp = 0.050, and the initial pressures of N2 and O2 are 0.80 atm and 0.20 atm, respectively, we can set up the equation as follows:
0.050 = (partial pressure of NO)^2 / (0.80 * 0.20)
To find the partial pressure of NO, we can rearrange the equation and solve for it:
(partial pressure of NO)^2 = 0.050 * (0.80 * 0.20)
(partial pressure of NO)^2 = 0.008
Taking the square root of both sides, we get:
partial pressure of NO = √0.008
Therefore, the partial pressure of NO in equilibrium with N2 and O2 is approximately 0.089 atm.
0.082
N2 + O2 ==> 2NO
Kp = pNO^2/pN2*pO2 = 0.05
Set up an ICE chart.
initial:
NO = 0
N2 = 0.8
O2 = 0.2
change:
NO = 2x
N2 = -x
O2 = -x
equilibrium:
NO = 2x
N2 = 0.8-x
O2 = 0.2-x
Substitute into Kp and solve.
Post your work if you get stuck.