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HW Problem 5H.1

Posted: Thu Jan 10, 2019 9:29 pm
by Gisela F Ramirez 2H
N2 (g) + 3H2 (g) <---> 2NH3 (g) at 400. K, K=41

Find the value of K:
a) 2NH3 (g) <---> N2 (g) + 3H2 (g)
answer: 1/K : 0.024
But how do I find the K value for the other reactions?

b) 1/2 N2 (g) + 3/2 H2 (g) <---> NH3 (g)
c)2N2 (g) + 6H2 (g) <---> 4NH3 (g)

Re: HW Problem 5H.1

Posted: Thu Jan 10, 2019 9:37 pm
by Chem_Mod
If a reaction is multiplied by a number X, the equilibrium constant K' of the new reaction will be K^(X)

Re: HW Problem 5H.1  [ENDORSED]

Posted: Thu Jan 10, 2019 9:55 pm
by Swetha Ampabathina1I
For part B, since you are technically working with the exponents, you'd have to square root the K to get the answer. Whereas, for part c the coefficients (which become exponents) are doubled so you would have to square the K.

Re: HW Problem 5H.1

Posted: Fri Jan 11, 2019 12:59 am
by Andie Jian 1D
Here's a helpful table from the book (pg. 410 7th edition) summarizing these rules. It also explains these concepts with more examples on that page if you need more clarification.
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Re: HW Problem 5H.1

Posted: Fri Jan 11, 2019 9:27 am
by abbydouglas1K
Swetha Ampabathina1I wrote:For part B, since you are technically working with the exponents, you'd have to square root the K to get the answer. Whereas, for part c the coefficients (which become exponents) are doubled so you would have to square the K.

So does this mean that every time you have molar coefficients that are less than one. i.e; (1/2)N2 + (3/2)H2 <--> NH3 K is always just the square root?