Question

1. Calculate Kp for the reaction below at 298K

2𝐶𝑙2(𝑔)+10𝑁𝑂(𝑔)⟷5𝑁2(𝑔)+2𝐶𝑙2𝑂5(𝑔)𝐾𝑝=?

Given at 298K:
2𝐶𝑙2𝑂5(𝑔)⟷5𝑂2(𝑔)+2𝐶𝑙2(𝑔) 𝐾𝑝1=0.095
2𝑁𝑂(𝑔)⟷𝑁2(𝑔)+𝑂2(𝑔) 𝐾𝑝2=476

2. Below is a list of reactions. Match the rearranged equations with how you would calculate the new reaction coefficient.

reaction 1: A <---> C K_1

reaction 2: A <---> E K_2

reaction 3: B <---> D K_3

-C <---> E
-2A <---> 2C
-A + B <---> C + D
-C <---> A

Choose:
K_1/K_3
K_1
Kc = K_2/K_1
K_3/K_1
K_2*K_1
Kc = K_1*K_3
Kc = 1/K_1
Kc = K_1^2
1/K_1^2
K_1/K_2

Answer 1

I have no clue what you want. For the first part I understand you want the Kp for this reaction. 2𝐶𝑙2(𝑔)+10𝑁𝑂(𝑔)⟷5𝑁2(𝑔)+2𝐶𝑙2𝑂5(𝑔) 𝐾𝑝=? which I will call eqn 1. You have these two eqns (eqn 2 and eqn 3) to help.
2𝐶𝑙2𝑂5(𝑔)⟷5𝑂2(𝑔)+2𝐶𝑙2(𝑔) 𝐾𝑝1=0.095
2𝑁𝑂(𝑔)⟷𝑁2(𝑔)+𝑂2(𝑔) 𝐾𝑝2=476
Do this:
Reverse eqn 2 and add to 5 times eqn 3. I suggest you do this on paper and make SURE you get eqn1 since I did this in my head and may have made an error. If you do that, then kp1 reverse = 1/Kp1 = (1/0.095) and new Kp2 = (Kp2)^5 so Kp for the reaction you want (eqn 1) will be
(476)^5/(0.095)
Post your work if you get stuck.
The remainder of what you wrote is gibberish. What are A, B, C, D and to which reactions so K1, K2, K3 etc refer?

Answer 2

well, the second question is that there's these reactions 1-3 given. i need to solve for reaction coefficaiton of the rearranged equations: C <---> E. the answer options are K_1/K_3 you see.
Thank you, I got the Kp2^5, but i was confused with the Kp1 since i thought it meant that i had to make it -Kp1 not 1/kp1

Answer 3

For part 1. If Kforward = Kf then Kreverse = Kr = 1/Kf and if you make
A + B = C and Kc = ? and you multiply by 2 then
2A + 2B = 2C, Then K'c = (Kc)^2 and
2C = 2A + 2B then K"c = (1/K'c) = (1/Kc)^2
When adding equations for delta H or delta G or delta S calculations you are correct. The reversed equation is the negative of the forward but for K it is 1/K.

I'm not sure I understand completely for part 2. As I understand it you have these three following equations.
reaction 1: A <---> C K_1
reaction 2: A <---> E K_2
reaction 3: B <---> D K_3
I don't know what "reaction coefficaiton" means but I think you want the K for the reaction E<===> C. I assume k1 is Kc but I don't know that.
To do that you use eqn 1 and 2 (You don't need eqn 3 since that has B and D and neither show up in what you want).
Reverse eqn 1 like this.
A ==> C Kc = k1 so reversed it is

C ---> A K'c = 1/k1. Now add in eqn 2 like this
A ---> E K = k2
----------------------------------
C + A ==> A + E. The A cancels and you're left with
C ==> E Then Kc = k2/k1 so I would pick the third answer from the top.
Hope all of this helps.