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I am having great difficulty with the following questions. Any and all help will be greatly appreciated. I have read the chapter and even looked up online tutorials. I still do not understand it.

2 NO (G) + O2 (G) > 2 NO2 (G)

Write the equilibrium constant expression for the reaction.

What effect would an increase in temperature have on the reaction rate?

What would be the effect on the reaction rate if the [O2] were doubled?

What would be the effect on the reaction rate if the pressure on the reaction were doubled?

Assume the reaction is reversible and at equilibrium. What shifts in the equilibrium of this reaction are suggested by Le Chatelier's Principle?

IF AT EQUILIBRIUM THE
[NO2] = 0.400 M
[O2] = 0.875 M
AND THE KEQ OF THE EQUILIBRIUM IS .65



[NO] = ?

IF AT EQUILIBRIUM THE
[NO2] = 0.400 M
[O2] = 0.875 M
[NO] = 0.125 M



Keq = ?

3 answers

Can you explain in detail what you don't understand. I'm have difficult knowing what your trouble is if you've done all that research and reading. For example, the expression for Keq is straight forward.
Okay, I think I have all but the last two figured out but those last two are defeating me. I just don't know what to plug in where. For instance, for the next to last one, I think that I am supposed to add the [NO2] and the [O2] together and then divide by .65 to get the amount of [NO}. Is that right?
IF AT EQUILIBRIUM THE
[NO2] = 0.400 M
[O2] = 0.875 M
AND THE KEQ OF THE EQUILIBRIUM IS .65

I wouldn't do that. You don't add in anything. Just plug the numbers into the Keq expression and solve for (NO).
2 NO (G) + O2 (G) > 2 NO2 (G)

Keq = (NO2)^2/(NO)^2(O2)
0.65 = (O.4)^2/(NO)^2(0.875)
0.65*(NO)^2*(0.875)= (0.4)^2
(NO)^2 = (0.4)^2/0.65*(0.875)
(NO)^2 = 0.281
(NO) = sqrt (0.281) = 0.530 M

For the last one.
IF AT EQUILIBRIUM THE
[NO2] = 0.400 M
[O2] = 0.875 M
[NO] = 0.125 M
Keq = (NO2)^2/(NO)^2(O2)
K = (0.4)^2/(0.125)^2(0.875)
Kq = 11.7