Homework #3

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Homework #3

Postby 805377003 » Thu Jan 07, 2021 10:12 am

At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3.


At this temperature, 0.300 mol H2 and 0.300 mol I2 were placed in a 1.00 L container to react. What concentration of HI is present at equilibrium?

Can someone explain this?

Wil Chai 3D
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Re: Homework #3

Postby Wil Chai 3D » Thu Jan 07, 2021 10:44 am

For this question, it would help to set up an ICE table. I think Prof Lavelle is going to go over this type of problem on Friday. It looks like this:

Initial. .3 .3 0
Change. -x -x +2x
Equilib. .3-x .3-x 2x

(Formatting looks off but each number should line up left to right with each compound)

And you use K to set up the equation 53.3=(2x)^2 / (.3-x)(.3-x) and solve for x, then go back and find 2x which is the equilib concentration of HI.

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Re: Homework #3

Postby Chem_Mod » Thu Jan 07, 2021 10:45 am

For these types of problems, it is most helpful to use the ICE table technique (Initial concentration, Change in concentration and Equilibrium concentration). You were given initial moles of both reactants, divide that by the volume of the container to get initial concentrations. There is 0 product initially. The given reaction tells you that for every mole of H2 that reacts, 1 mole of I2 also reacts and 2 moles of HI are formed. Thus change in concentration would be " -x", where x is a variable for an unknown value, for both reactants and "+2x" for the product. Adding those "change" values to the initial concentration gives you values for the equilibrium concentrations in terms of x. Set up your equilibrium constant expression. All terms in this reaction are in gaseous phase, so include all of them in the K equation. K = [HI]^2/([H2][I2]) = 53.3 using the reaction and stoichiometric coefficients. Plug in the equilibrium values you got from the table into the respective places in the K equation and solve for x. At equilibrium, [HI] = 2x, so multiply x by 2 to get your answer

Benjamin Chen 1H
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Re: Homework #3

Postby Benjamin Chen 1H » Thu Jan 07, 2021 11:08 am

You'll want to use the ICE or Equilibrium table described in Toolbox 5I.1.

When the reaction proceeds, [H2] and [I2] will each decrease by a certain change called "x" while [HI] will increase by "2x" since there is a 1:2 molar ratio between reactants and products.

The resulting concentrations:
[H2] = 0.300 - x
[I2] = 0.300 - x
[HI] = 2x

You then want to put it into the equilibrium constant ratio:
Kc = [HI]^2 /[H2] [I2]

Solve for x.
The concentration of HI will be 2x.
I got [HI] = 0.471 M

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