At 2200°C, K = 0.050 for the following reaction.
N2(g) + O2(g) 2 NO(g)
What is the partial pressure of NO in equilibrium with N2 and O2 that were placed in a flask at initial pressures of 0.81 atm and 0.20 atm, respectively
...........N2 + O2 ==> 2NO
initial.0.81.....0.20....0
change...-p.....-p.....+2p
equi.....0.81-p..0.20-p..2p
Substitute into Kp expression and solve for x, then evaluate concns elements/compounds at equil conditions.
To find the partial pressure of NO in equilibrium, we need to use the equilibrium constant expression and set up an equation.
The equilibrium constant expression for the given reaction N2(g) + O2(g) → 2NO(g) is:
K = [NO]^2 / ([N2] * [O2])
Given that K = 0.050, we can rearrange the equation to solve for [NO]:
[NO]^2 = K * ([N2] * [O2])
Now we can substitute the given initial pressures for [N2] and [O2]:
[N2] = 0.81 atm
[O2] = 0.20 atm
Substituting these values into the equation, we get:
[NO]^2 = 0.050 * (0.81 atm * 0.20 atm)
Simplifying the right-hand side:
[NO]^2 = 0.050 * 0.162 atm^2
[NO]^2 = 0.0081 atm^2
Now we can solve for [NO] by taking the square root:
[NO] = √0.0081 atm
[NO] ≈ 0.09 atm
Therefore, the partial pressure of NO in equilibrium with N2 and O2 in the given conditions is approximately 0.09 atm.