The pollutant NO is formed in diesel engines. The reaction fixes atmospheric nitrogen with oxygen to form NO. The reaction is: N2(g) + O2(g) = 2NO(g). If the equilibrium constant for this reaction at elevated temperatures is 5.60E-11, then what is the partial pressure of NO gas if nitrogen is 1.50 atm and oxygen is 0.500 atm? The units are atmospheres.

I assume you mean you START with pN2 = 1.50 atm and START with pO2 = 0.5 atm?

..........N2 + O2 ==> 2NO
I........1.50..0.5.....0
C.........-x....-x.....2x
E.......1.50-x..0.5-x...2x

Substitute the E line into K expression and solve for x, and the othr values.

BTW, just a reminder that THERE ARE NO UNITS in Kp, Kc, Ksp, Keq, etc. I know some profs use units (some because they don't know better) and if they call them provisional units that makes it ok, but technically, activities go in those constants and activities don't have units so K can't have units.

To find the partial pressure of NO gas, we need to use the equilibrium constant (K) and the given partial pressures of nitrogen (N2) and oxygen (O2). The equilibrium constant expression for the given reaction is:

K = [NO]^2 / ([N2] x [O2])

Here, [NO], [N2], and [O2] represent the partial pressures of each gas.

Given that K = 5.60E-11, [N2] = 1.50 atm, and [O2] = 0.500 atm, we can rearrange the equation to solve for [NO]:

K = [NO]^2 / (1.50 atm x 0.500 atm)

Now, let's substitute the known values into the equation:

5.60E-11 = [NO]^2 / (1.50 atm x 0.500 atm)

To solve for [NO], we need to isolate it. First, multiply both sides of the equation by (1.50 atm x 0.500 atm):

5.60E-11 x (1.50 atm x 0.500 atm) = [NO]^2

Now, evaluate the left side of the equation:

5.60E-11 x (1.50 atm x 0.500 atm) = 4.2E-11 atm^2

Taking the square root of both sides of the equation will give us the value of [NO]:

[NO] = sqrt(4.2E-11 atm^2)

Now, calculate the value of [NO]:

[NO] = 6.48E-6 atm

Therefore, the partial pressure of NO gas is approximately 6.48E-6 atm.