HW 5I.35

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Dina 2k
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Joined: Wed Sep 18, 2019 12:17 am

HW 5I.35

Postby Dina 2k » Wed Jan 22, 2020 5:23 pm

can someone explain how to approach that hw problem. the answer is k=p^2/(1-2p)^2. I don't understand the reasoning behind the answer for the denominator.

Alicia Lin 2F
Posts: 83
Joined: Wed Sep 18, 2019 12:17 am

Re: HW 5I.35

Postby Alicia Lin 2F » Wed Jan 22, 2020 7:16 pm

If you make an ice table, the initial value for NO is 1.0 bar. The change would be -2x because the mole ratio. And the equilibrium value of NO would therefore be 1.0-2x. In this case, we are told p is the equilibrium value of N2. Since N2 starts with 0 concentration, p would be equal to x. So we can say the equilibrium value for NO would be 1.0-2p. Because the stoichiometric coefficient is 2 for NO, you would square the NO concentration in the denominator.

Posts: 105
Joined: Sat Jul 20, 2019 12:16 am

Re: HW 5I.35

Postby Julie_Reyes1B » Wed Jan 22, 2020 7:26 pm

So looking at NO specifically, the initial would be 1.0, and the change would be -2p. The two comes from the 2 coefficient in the reaction. So the equilibrium concentration is 1.0-2p. The whole quantity is squared in the K equation because of the 2 coefficient as well.

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