Page 1 of 1

HW Q 6th ed. 11.53

Posted: Mon Jan 14, 2019 2:30 pm
by Jerome Mercado 2J
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.

Re: HW Q 6th ed. 11.53

Posted: Mon Jan 14, 2019 2:37 pm
by Hai-Lin Yeh 1J
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

Re: HW Q 6th ed. 11.53

Posted: Mon Jan 14, 2019 2:44 pm
by Laura Gong 3H
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.

Re: HW Q 6th ed. 11.53

Posted: Sat Jan 18, 2020 6:55 pm
by chari_maya 3B
Why would you not convert to P and calculate Kp?