# all reversible reactions

Kp= 3.5 X 10^4
temp: 1495K

N2(g)+3H2(g) reversible 2NH3(g)

1/2N2(g) + 3/2H2(g) reversible NH3(g)

what is the value of K at the temp.

what is the process for this?

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1. It's easier than it looks on the surface.
Let's call Kp = 3.5 x 10^4 the original Kp. The second equation is just 1/2 of the first one; therefore, new Kp = (original Kp)1/2.
If you had 2N2 + 6H2 <==> 4NH3, then
new Kp = (original Kp)2 . In other words, just raise the original Kp to whatever power the new equation is relative to the original coefficients. In these two examples, that is 1/2 and 2.

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2. where does the temperature factor in? I raised it to 1/2 and got 1.75, but that was not the right answer.
(I forgot to mention this, but the first reaction is the original one)
Could you put a example up? I am very much a visual learner and need to see it before I can do it.

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3. As long as the temperature doesn't change, K doesn't change. So for 1/2 the reactants then the new Kp = 3.5 x 10^4)1/2.
How did you get 1.75? I'm not surprised the key told you that was wrong.
I have square root of 3.5 x 10^4 = 187 and not 1.75.
Let me know if still have problems. (I just see how you got that number. You divided 3.5/2 = 1.75. You forgot the exponent of 10^4 I think.)

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4. Is there a good website that has good examples for ap chemistry ranging from easy to difficult?

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5. Here is one that I use often to direct students. Its' written very well and it's an AP site.

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6. calculate the standard enthalpy change for each of the following reactions. N2O4(g)+4H2(g)=N2(g)+4H2O(g)

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