Week 1 Sapling 5

[Agustina Santa Cruz 2F]

Consider the reaction of NH3 and I2 to give N2 and HI.
2NH3(g)+3I2(g)↽−−⇀N2(g)+6HI(g)K

Using two or more of the given equations, determine the equilibrium constant, K , for the reaction of NH3 with I2.

H2 (g) +I2 (g) >>> 2HI (g) Ka = 160

I2 (g) >>>2I (g) Kb = 2.1 * 10-3

N2 (g) + 3H2 (g) >>>2NH3 (g) Kc = 3.6 * 10-2

H2 (g) + Cl2 (g) >>> 2HCl (g) Kd = 4.0 * 1018

Could someone explain this please?

[Sahiti Annadata 3D]

You need to use the 1st and 3rd equations because these equations can be manipulated to form the initial equation. You need to do 1 over the K of the first equation and raise it to the power of 3 since 3 of the inverse of first equation form one part of the initial equation. For the third equation, you want to also do the inverse, so your new K value is 1/(3.6 x 10^-2). Then multiply the 2 new K values to get the final one for the initial equation.

[Jessie Hsu 1C]

Does anyone know in which lecture did Dr. Lavelle go over this concept?

[luludaly2B]

You need to use the 1st and 3rd equations because these equations can be manipulated to form the initial equation. You need to do 1 over the K of the first equation and raise it to the power of 3 since 3 of the inverse of first equation form one part of the initial equation. For the third equation, you want to also do the inverse, so your new K value is 1/(3.6 x 10^-2). Then multiply the 2 new K values to get the final one for the initial equation.

I was having a lot of trouble with this question as well. Do you disregard the H₂ when doing 1/K of the first equation? If so, why are you allowed to do that. Correct me if I'm wrong, but the H₂'s do not cancel out from Ka and Kb right? Does that impact the solution at all? Thanks!

[Arti_Patel_3H]

You need to use the 1st and 3rd equations because these equations can be manipulated to form the initial equation. You need to do 1 over the K of the first equation and raise it to the power of 3 since 3 of the inverse of first equation form one part of the initial equation. For the third equation, you want to also do the inverse, so your new K value is 1/(3.6 x 10^-2). Then multiply the 2 new K values to get the final one for the initial equation.

I was having a lot of trouble with this question as well. Do you disregard the H₂ when doing 1/K of the first equation? If so, why are you allowed to do that. Correct me if I'm wrong, but the H₂'s do not cancel out from Ka and Kb right? Does that impact the solution at all? Thanks!

The H₂'s should cancel out from Ka and Kb. This would allow you to find the proper final K value of the initial equation.