Consider the equilibrium N₂(𝑔) + O₂(𝑔) + Br₂(𝑔) ⇌ 2 NOBr(𝑔) Calculate the equilibrium constant 𝐾𝑝 for this reaction, given the following information at 298 K:
2 NO(𝑔) + Br₂(𝑔) ⇌ 2 NOBr(𝑔) 𝐾_𝑐= 2.02
NO(𝑔) ⇌ N₂(𝑔) + O₂(𝑔) 𝐾_𝑐= 2.1×10³⁰

Question

Consider the equilibrium N₂(𝑔) + O₂(𝑔) + Br₂(𝑔) ⇌ 2 NOBr(𝑔) Calculate the equilibrium constant 𝐾𝑝 for this reaction, given the following information at 298 K:

2 NO(𝑔) + Br₂(𝑔) ⇌ 2 NOBr(𝑔) 𝐾_𝑐= 2.02

NO(𝑔) ⇌ N₂(𝑔) + O₂(𝑔) 𝐾_𝑐= 2.1×10³⁰

Accepted Answer

everyone said you were asked at the equilibrium constant K. P. For the following reaction at 373.15 Kelvin. Ask you the following the reactions below in there, K. C. At the same temperature. Since we're asked for the K. P. Of the overall reaction, we need to first find the K. C. Of the reaction and then we're going to use the inclusion K. P equals K. C. R. T. The power of dr an to find the K. P. So in this equation, r. Is the gas constant. T. Is the temperature. That's the end is the sum of the malls the gaseous product. This is sam of the malls of the gasses reacted. So we're giving the two reactions with the K. C. Values. We need to find A K. C. For the overall reaction above. We need to look for each of the reactions and the products and the two actions below Here, we have nitrogen oxide as a reactant here. That is a product right here in this reaction, we're gonna have to reverse this reaction. We're gonna get n. 0. 2. Yes, in the shields and gas plus one half 02. Gas. Since we reverse the reaction, the K. C. It's gonna be one Over the K. c. value, which is eight times 10 To the 4th Coward. Now, we have nitrogen gas as a product here and it's also a product in this reaction as well and has the same number of moles in both. So we don't need to do anything to the reaction here, let me have oxygen gas as a product in the supply. And the new reaction and a product in this reaction as well. But we have a one half in front of the oxygen here. We need the most part. This whole reaction by two. Let me get to In 0. 2. Yes, getting to and no gas Plus 0. 2 gas. To not be out of the amount of auction gas we have in both reactions. It should equal how much we have in the overall reaction. We have one here and one here for a total of two. We're gonna raise our K. C. To the power of two And this to give us 1.56. I'm sensing a negative 10 power. We're gonna leave this reaction like it is. Now we can add up both reactions and cancel out any common species to get the overall reaction coffee right to N. O. Gas here. Getting into gas Plus 0. 2 gas. The K. C bias over here into N. O cancels out on both sides. We're left with two and 02. Yes. Building into gas Plus 2. 0. 2 gas. And for the K. C. We need to apply these two K. C values. Once we do that, we get 7.8 times 10 to the 14 power. Now the key is this K. C. Body to find the K. P. Using the equation we talked about this equation is K. P equals K. C. R. T. The power of delta, N K P is what we're looking for. KC equals 7.8. I was 10 to the or 0.8206 Leaders Times atmosphere. But about moles Times kelvin in the temperature 373.15 Kelvin. And for Delta end we have one ball up into Plus two miles of here -2 moles of n. 0 2. Let me get one more. And after taking the values we get K P Equal 7.8 times 10 to the 14. From 0.082 are set leader, some atmosphere about about morals times kelvin On stage of 73 115 Kelvin The power of one and for the K P Get 2.39, 10 to 16. Since the K P does not have any units. We can disregard the units. Thanks for watching my video and I hope it was helpful

Verified Solution

Key Concepts

Equilibrium Constant (K)

Relationship Between Kp and Kc

Le Chatelier's Principle

Consider the equilibrium N2(𝑔) + O2(𝑔) + Br2(𝑔) ⇌ 2 NOBr(𝑔) Calculate the equilibrium constant 𝐾𝑝 for this reaction, given the following information at 298 K: 2 NO(𝑔) + Br2(𝑔) ⇌ 2 NOBr(𝑔) 𝐾𝑐 = 2.02 NO(𝑔) ⇌ N2(𝑔) + O2(𝑔) 𝐾𝑐 = 2.1×1030

Verified Solution

Key Concepts

Equilibrium Constant (K)

Relationship Between Kp and Kc

Le Chatelier's Principle