5G 9 HW C VS. D AND E

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Phuong Tran 1G
Posts: 33
Joined: Sat Sep 14, 2019 12:16 am

5G 9 HW C VS. D AND E

Postby Phuong Tran 1G » Tue Jan 14, 2020 4:52 pm

Hi! Can someone explain why c wouldn’t be the same? I know that d and e are the same in both containers because they’re the ratio of the equilibrium constant and the reverse of it, but what about C?

The only explanation I thought of was that c isn’t in a balanced equation so it can’t reach equilibrium so the amount you start with (.1 vs .5) will change the ratio.

AArmellini_1I
Posts: 107
Joined: Fri Aug 09, 2019 12:15 am

Re: 5G 9 HW C VS. D AND E

Postby AArmellini_1I » Tue Jan 14, 2020 5:09 pm

basically the only thing the remains unchanged in the equilibrium constant, and since d is the equilibrium constant and e is the reciprocal, they won't change. And while c does have products/reactants, it lacks the correct exponents so a change in concentration of a reactant or product would result in a different value.

Sartaj Bal 1J
Posts: 101
Joined: Thu Jul 25, 2019 12:17 am

Re: 5G 9 HW C VS. D AND E

Postby Sartaj Bal 1J » Wed Jan 15, 2020 5:16 pm

The above explanation is correct. A way to try this mathematically is to come up with different concentrations/partial pressures of O2 and O3 that allow for the same ratio. Then, substitute these values into two separate equations for the equilibrium constant that includes the respective exponents. Due to these exponents, the values for the equilibrium constant should NOT be the same in each instance. Since we know the equilibrium constant has to be the same, we have proved that the ratio cannot be the same. Hope that makes sense!


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