HW Problem 5H.1  [ENDORSED]

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Gisela F Ramirez 2H
Posts: 61
Joined: Fri Sep 28, 2018 12:27 am

HW Problem 5H.1

Postby Gisela F Ramirez 2H » Thu Jan 10, 2019 9:29 pm

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)

Chem_Mod
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Re: HW Problem 5H.1

Postby Chem_Mod » Thu Jan 10, 2019 9:37 pm

If a reaction is multiplied by a number X, the equilibrium constant K' of the new reaction will be K^(X)

Swetha Ampabathina1I
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Re: HW Problem 5H.1  [ENDORSED]

Postby Swetha Ampabathina1I » Thu Jan 10, 2019 9:55 pm

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.

Andie Jian 1D
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Re: HW Problem 5H.1

Postby Andie Jian 1D » Fri Jan 11, 2019 12:59 am

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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abbydouglas1K
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Re: HW Problem 5H.1

Postby abbydouglas1K » Fri Jan 11, 2019 9:27 am

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?


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