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Chloe Alviz 1E
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Joined: Sat Aug 17, 2019 12:17 am


Postby Chloe Alviz 1E » Tue Jan 14, 2020 9:39 pm

A reaction mixture that consisted of 0.400 mol H2and 1.60 mol I2 was introduced into a flask of volume 3.00 L and heated. At equilibrium, 60.0% of the hydrogen gas had reacted. What is the equilibrium constant K for the reaction H2(g) + I2(g) <=> 2 HI(g) at this temperature?

Can someone help me through this question? I'm not sure where to start.

Daria Azizad 1K
Posts: 116
Joined: Thu Jul 25, 2019 12:15 am

Re: 5I.19

Postby Daria Azizad 1K » Tue Jan 14, 2020 9:55 pm

To solve this problem, you need to make an ICE table. The initial concentrations of H2 and I2 can be found by dividing the number of moles by the volume of the flask. Then, we can find H2's equilibrium concentration by multiplying its initial value by 0.6. If we say that the concentration of H2 changed by x, we can then solve for x:
initial - x = final
Once we have our x value, we can also solve for the eq concentration of I2, which decreased by x, and HI which increased from 0 to 2x.

Harry Zhang 1B
Posts: 101
Joined: Sat Sep 14, 2019 12:16 am

Re: 5I.19

Postby Harry Zhang 1B » Tue Jan 14, 2020 10:06 pm

First, calculate the molar concentration of H2 and I2 in the flask, which is n/v. From there, since the question says that 60% of the hydrogen gas reacted at equilibrium, it means that to reach equilibrium, H2 reduced by 0.6 of its original concentration. Therefore you will multiply 0.6 by the molar concentration of H2 to find out the C(change) in your ICE box; this value would then also represent the change for I2 since their stoichiometric coefficients are the same. The change for HI will be two times the change in reactants. From there, you can use the [HI]^2/[H2][I2] to calculate Kc.

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