11.37 6th Edition

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Alexa Tabakian 1A
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Joined: Fri Sep 28, 2018 12:20 am

11.37 6th Edition

Postby Alexa Tabakian 1A » Tue Jan 08, 2019 11:23 am

I was wondering if anyone could explain how to find the partial pressures of N2, H2, and NH3 when only given that K=41. Do we have to involve constants from a table?

Cody Do 2F
Posts: 62
Joined: Fri Sep 28, 2018 12:23 am

Re: 11.37 6th Edition

Postby Cody Do 2F » Tue Jan 08, 2019 1:26 pm

We know that for the equation N2 + 3H2 <-> 2NH3 K= 41. When we write it out:
K= PNH32/(PN2)(PH2)3 = 41

For part a, the equation given is 2NH3 <-> N2 + 3 H2. Thus they are looking for K where:
K= (PN2)(PH2)3/PNH32 = 41
When you look at it, this is essentially the inverse of the K value given (41). Thus, to solve part a, you need to do 1/41, which will give a value of 0.024.

For part b, the equation provided is the original equation square rooted, so solve for the square root of 41 = 6.4.

For part c, the equation provided is the original equation squared, so solve for the square of 41, which gives 1.7 x 103

Kayla Vo 1B
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Joined: Fri Sep 28, 2018 12:26 am

Re: 11.37 6th Edition

Postby Kayla Vo 1B » Tue Jan 08, 2019 4:42 pm

In other words for part A, you are looking for the equilibrium constant of the reverse reaction. Since we know that K=41 for the forward reaction, K for the reverse reaction would be the inverse of that of the forward reaction. Thus, the equilibrium constant of the reverse reaction would be aka
1/41.

Since the coefficients of the reactants and products are halved, the equilibrium constant would equal to . Therefore, the equilibrium constant would equal to .

For part c, the coefficients of the reactants and products are doubled. Therefore, the equilibrium constant equals to where n equals to 2 since that is the factor. As a result, the equilibrium constant would equal to .

Matthew Tran 1H
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Joined: Fri Sep 28, 2018 12:16 am

Re: 11.37 6th Edition

Postby Matthew Tran 1H » Tue Jan 08, 2019 4:52 pm

Remember that the equilibrium constant is a ratio. That means that you can technically have an infinite number of partial pressures, but their ratio will be the same. Basically, you can have high partial pressures or low partial pressures, but at equilibrium their ratio is equal to the equilibrium constant. So, if you are given only the value of K then you cannot know the partial pressures. If you are given 2 partial pressures then you can find the other partial pressure if there are 3 total products and reactants.


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