Question

Can someone help me in explaining why is this the answer?

2. Given the equilibria:
N2(g) + ½ O2(g) ⇔ N2O(g) Kc = 2.40 x 10-18
N2(g) + O2(g) ⇔ 2NO(g) Kc = 4.1 x 10-31
What is the equilibrium constant for N2O(g) + ½ O2(g) ⇔ 2NO(g)? answer= 1.708x10-13

I know that for Kc = (2.4 x10^-18)(4.1 x 10^-49)
then from there I'm lost on how 1.078 x10^-13 came to be.

Answer 1

Kc = (2.4 x10^-18)(4.1 x 10^-49) ??
how?

Answer 2

sorry my mistake , it would be: Kc= (2.40 x 10-18)(4.1 x 10-31) = 9.84E-49

Answer 3

Kc= product/ reactant

Answer 4

yes

Answer 5

why are you multiplying then?

Answer 6

@Kainui

Answer 7

Alright so I'm gonna call the first one K1 and the second one K2. Check it out, it's kind of a fun one! =)

Alright so to get the reaction we want, we rearrange them. \[N_2O \rightarrow N_2+ \frac{1}{2}O_2 \ :\ \frac{1}{K_1} \\ N_2+ O_2 \rightarrow 2NO \ :\ K_2\]
Let's combine the reactions by adding the left stuff to the left and right stuff to the right. \[N_2O + N_2+ O_2 \rightarrow N_2+ \frac{1}{2}O_2 + 2NO\] But remember, equilibrium constants are about the net reaction. If you transfer 1 mol N2 from the left, you get 1 mol N2 on the right! There's no change there, so we can safely remove them. Similarly, we can remove 1/2 O2 from both sides as well.

\[N_2O +\frac{1}{2}O_2 \rightarrow 2NO\] What does the rate constant for this look like? It is both of those rate constants multiplied together. But why? If we call this new reaction's rate constant K3, look at this:

\[\frac{1}{K_1}K_2=\frac{[N_2][O_2]^{1/2}}{[N_2O]}\frac{[NO]^2}{[N_2][O_2]}=\frac{[NO]^2}{[N_2O][O_2]^{1/2}}=K_3\]

Mathematically, it makes sense! If you take the logarithm of the equilibrium constant, then you can see that there's a similarity there.

\[\log(K_3)= - \log [N_2O] - \frac{1}{2}\log [O_2] + 2\log[NO] \\ N_2O + \frac{1}{2}O_2 \rightarrow 2 NO\]

Forget about this last part if it doesn't help you, but it sort of explains how our addition and multiplication connects.

Answer 8

N2(g) + ½ O2(g) ⇔ N2O(g)
you made it this way -> N2O(g) ⇔N2(g) + ½ O2(g)
because N2O is reactant in that equation?

Answer 9

Yup, exactly. I wanted to make a reaction out of the reactions I already know.

Answer 10

next you just added up 2 equations as they actually look
meaning as they were reactants
they stayed reactants same with products
right?

Answer 11

Yep, you got it. =)

Answer 12

next you just canceled out same stuff and made the equation that we r asked to find Kc for

Answer 13

Yes, I just wanted to make sure I had the right rate constants, which weren't K1 and K2, but instead we needed 1/K1 and K2 which is only slightly different.

Answer 14

next you just wrote Kc for those 2 reactions that you added up and equate it with the original equation

Answer 15

now let me check if i will get the answer

Answer 16

Sorry I was saying rate constant when I meant to say equilibrium constant whoops.

Answer 17

O!!! wait
so basically saying all we do is 1/K1 * K2 ?

Answer 18

goooot it :D thnx

Answer 19

Yeah! The whole chemical equation stuff was just extra to help us out. Really, just make the equilibrium constant by doing algebra with the equations of the other equilibrium constants. =P

Answer 20

algebra is quite fun <3
its similar to me playing with 10 formulas to make just one xD