Given the equation
NO(g) + 1/2)O^2(g) <-> NO(g) at 457 Kelvin
with a K = 7.5 x 10^2
Find the K value for the equation
2NO^2(g) <-> 2NO(g) + O(g)
I think you have some typos in your post but in general this is what you want.
Easy to remember stuff.
If eqn is doubled, K' is K^2.
If eqn is halved, K' is square root K.
If eqn is tripled, K' is K^3
etc.
If eqn is reversed, K' = 1/K.
If you want to show subscripts on Jishka, it CAN be done but is a lot of trouble. Most of us write
NO2 as NO2 and all understand the 2 is a subscript. If we want to show a squared term we write NO^2 for NO2. The French write NO2 for the English of NO2
Hi. Thanks for your help. So this is how I solved the equation.
2NO2(g) <-> 2NO(g) + O2(g)
Since the 2NO2 is reversed from the 1st equation, I got (1/7.5x10^2) and then since the 2NO(g) and the O2(g) are doubled from the 1st equation I got (7.5x10^2)^2 for 2NO(g) and the same for O2(g)
So in the end I get
(1/7.5x10^2)(7.5x10^2)^2(7.5-10^2)2 and my final answer is 4.22 x 10^12
Is that correct?
No. I think you have misunderstood my response.
K = 7.2E2
So for the reversal, you are correct, in 1/7.5E2. The remainder is what you didn't understand properly. You are correct that the equation is doubled BUT you square K, (you squared it twice, once for NO2 and once for O2 and multiplied again by the original K which effectively is K^3. Remember K is the whole term and that's what you are squaring, taking reciprocal, etc and not the individual parts of K.
So what you want to do is for reversing you get 1/7.5E2 and for doubling you get
the squared so the final is 1/(7.5E2)^2 or 1.7777E-6 which rounds to 1.8E-6 rounded to 2 s.f.
Thanks so much for your help.
To find the K value for the equation 2NO^2(g) <-> 2NO(g) + O(g), we can start by writing the expression for the equilibrium constant (K) in terms of the concentrations of the species involved.
For the equation NO(g) + 1/2O^2(g) <-> NO(g), the equilibrium constant expression is:
K = [NO(g)] / ([NO(g)] * [O^2(g)]^(1/2))
Given that the equilibrium constant for this reaction is 7.5 x 10^2, we have:
7.5 x 10^2 = [NO(g)] / ([NO(g)] * [O^2(g)]^(1/2))
Simplifying the equation, we can cross-multiply and rearrange the terms:
[NO(g)] * [O^2(g)]^(1/2) = [NO(g)]
Now, let's find the K value for the equation 2NO^2(g) <-> 2NO(g) + O(g).
The equilibrium constant expression for this equation is:
K' = ([NO(g)]^2 * [O(g)]) / ([NO^2(g)]^2)
Since we want to find the K' value for this equation, we can substitute [NO(g)] and [O(g)] from the previous equation into the K' equation:
K' = ([NO(g)]^2 * [O(g)]) / ([NO^2(g)]^2)
= ([NO(g)] * [NO(g)] * [O(g)]) / ([NO(g)] * [NO(g)])^2
= [O(g)] / ([NO(g)])^2
Now, we can express the K' value in terms of the given K value:
K' = [O(g)] / ([NO(g)])^2 = (1/2 * [O^2(g)]^(1/2)) / ([NO(g)])^2
Since [O^2(g)]^(1/2) is equal to (2[NO(g)]), we can substitute it into the expression:
K' = (1/2 * (2[NO(g)])) / ([NO(g)])^2
= [NO(g)] / ([NO(g)])^2
= 1 / [NO(g)]
Therefore, the K' value for the equation 2NO^2(g) <-> 2NO(g) + O(g) is 1 / [NO(g)].
Please note that we don't have the concentration values [NO(g)] and [O(g)] given, so we cannot determine the exact value of K'.