The equilibrium constant for the reaction 2NO(g) + O2(g) <--> 2NO2(g) is Kp=1.48x10^4 at 184C. Calculate Kp for 2NO2(g) <--> 2NO(g) + O2(g).

I think you are missing the boat.

Kp for 2NO + O2 ==> 2NO2 Kp = 1.48E3
You want Kp for
2NO2 ==> 2NO which is just the reverse of the original equation. For that the new Kp = 1/old Kp (i.e., just the reciprocal of old Kp).

Oh wow. I really need to stop over thinking this stuff, I tend to do that a lot. Thank you very much.

Well, it seems like you have set up the equation correctly! However, there's a small issue with the value of delta n.

Delta n represents the change in the number of moles of gas molecules between the products and the reactants. In this case, we have:

2NO(g) + O2(g) <--> 2NO2(g)

So, on the reactant side, we have 2 moles of NO and 1 mole of O2, giving a total of 3 moles of gas molecules. On the product side, we have 2 moles of NO2, giving a total of 2 moles of gas molecules.

Therefore, the change in the number of moles of gas molecules (delta n) is 2 - 3 = -1.

Now, using the equation Kp = Kc(RT)^delta n, with R = 0.0821 and T = 184 + 273 = 457 K, we can plug in the values:

Kp = Kc(0.0821 * 457)^(-1)

Unfortunately, we still need the value of Kc to proceed further. However, we don't have that information available.

If you can provide the value of Kc, I would be happy to help you calculate Kp. Otherwise, I'm afraid we cannot proceed without that missing piece of information.

You are on the right track! To solve for Kp in the equation 2NO2(g) <--> 2NO(g) + O2(g), you can use the equation:

Kp = Kc(RT)^(Δn)

Let's break it down step by step:

1. Given that Kp = 1.48x10^4, R = 0.0821 (atm⋅L/mol⋅K), T = 184°C (which is equivalent to 457 K), and Δn = 1.

2. First, we need to find the value of Kc using the equation:

Kp = Kc(RT)^(Δn)

We will rearrange the equation to solve for Kc:

Kc = Kp / (RT)^(Δn)

Substituting the known values:

Kc = (1.48x10^4) / ((0.0821)(457))^1

Kc ≈ 0.1849

3. Now that we have the value of Kc, we can calculate Kp using the equation:

Kp = Kc(RT)^(Δn)

Substituting the known values:

Kp = (0.1849) * ((0.0821)(457))^1

Kp ≈ 0.1849 * 1.188 = 0.2196 (rounded to four significant figures)

Therefore, the calculated value of Kp for the reaction 2NO2(g) <--> 2NO(g) + O2(g) is approximately 0.2196.