HW Q 6th ed. 11.53

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Jerome Mercado 2J
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HW Q 6th ed. 11.53

Postby Jerome Mercado 2J » Mon Jan 14, 2019 2:30 pm

A reaction mixture that consisted of 0.400 mol H2 and 1.60 mol I2 was introduced into a 3.00-L flask 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?

I have already calculated the concentration of H2 at equilibrium but I am unsure how to calculate the concentration of I2 and HI at equilibrium. Any help is appreciated.

Hai-Lin Yeh 1J
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Re: HW Q 6th ed. 11.53

Postby Hai-Lin Yeh 1J » Mon Jan 14, 2019 2:37 pm

If you have already calculated the concentration of H2 at equilibrium, which should be 0.053, then you know that the change in concentration of H2 should be 0.08. Note that how I got 0.08 was 60% (.60) of the gas reacted so .60 x 0.400 which equals .24 and to find concentration, divide by 3, getting 0.08. And looking at the equation, for every 1 mole of H2, 1 mole of I2 reacts with it, so you know that the change in concentration of I2 should also be 0.08. Similarly, for HI, 2 moles of HI is required, so 2(0.08) is 0.16.
So your ICE table should look somewhat like this:
H2 I2 HI
I .133 .533 0
C -0.08 -0.08 +0.16
E 0.053 0.453 0.16

Then, just calculate Kc, using the equilibrium molar concentrations: [HI]^2/[H2][I2] = (0.16)^2/(0.053)(0.453)

Kc = 1.1
Last edited by Hai-Lin Yeh 1J on Mon Jan 14, 2019 2:46 pm, edited 1 time in total.

Laura Gong 3H
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Re: HW Q 6th ed. 11.53

Postby Laura Gong 3H » Mon Jan 14, 2019 2:44 pm

In this question, you will have to set up an ICE table.
[H2]=0.400mol/3.00L=.133mol/L and [I2]=1.60mol/3.00L=.533 mol/L
If 60.0% of hydrogen gas reacted than X (change in molarity of reactants and products) will be equal to .133mol/L * (.60)=0.08

From there you can calculate the equilibrium concentrations of H2, I2, and HI:
[H2] at EQ = .133-x = .133-.08=.053
[I2] at EQ = .533-.08 = .533-.08=.453
[HI] at EQ = 2*X= 2(0.08)= .16

Then plug these numbers into the Kc expression for this reaction which is:
Kc= [HI]^2/[H2][I2]= 0.16^2/(.053*.453)=1.06
Since there are two significant figures, Kc=1.1.


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