## HW Q 6th ed. 11.53

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### HW Q 6th ed. 11.53

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

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

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.

chari_maya 3B
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### Re: HW Q 6th ed. 11.53

Why would you not convert to P and calculate Kp?