## Sapling #5 on finding k

Blake Ballew 1H
Posts: 100
Joined: Wed Sep 30, 2020 9:35 pm

### Sapling #5 on finding k

The problem states:

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

I found all the other problems in this homework to be pretty easy but I am totally lost on where to start with this one. How do I combine the information from two of these equations in order to determine k without any given moles or concentrations or anything? Any help is greatly appreciated :)

Posts: 173
Joined: Wed Sep 30, 2020 9:32 pm
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### Re: Sapling #5 on finding k

You basically just combine equations to get the equation you want. Equations can be reversed, in which case the K value would now be equal to 1/K, and the equation can be multiplied by a number, in which case the K value would be raised to the power of that number. For this question, multiplying the first equation by 3 and reversing the third equation will result in original equation, 2NH3(g)+3I2(g)↽−−⇀N2(g)+6HI(g). From here you can just adjust the K values and multiply the new K values of both equations to find the K value of the original reaction.

Hope this helps!

Charlene D 3H
Posts: 110
Joined: Wed Sep 30, 2020 9:54 pm

### Re: Sapling #5 on finding k

Hi! I approached this problem by rearranging the given reactions and using the following knowledge.
1) know that when you are adding/combining reactions together the K values are multiplied. ie) rxn1 + rxn2 has a K value of K1*K2.
2) the K of a reverse reaction is just 1/K
3) If a molecule has the same amount of moles on both sides of a reaction, it cancels
4) when you add a coefficient (n) to a reaction, the K's value is raised to the nth power
After knowing the above, you should be able to find your K!
Let me know if you have any more questions.