Calculate the equilibrium constant Kc for the reaction at this temperature

[ahhppull]

Homework Statement

H2 + I2 -> 2HI

A reaction mixture in a 3.67 L flask at a certain temperature initially contains 0.763 g H2 and 96.9 g I2. At equilibrium, the flask contains 90.4 g HI.

Calculate the equilibrium constant Kc for the reaction at this temperature.

Homework Equations

The Attempt at a Solution

I used the ICE chart to try and solve this.

First, I divided the grams of each substance by the molar mass of the compound, and divide it by 3.67 L to get molarity.

.763g H2 / 2 g / 3.67 = .104 M
96.9g I2 / (126.9x2) /3.67 = .104 M
90.4 HI / (126.9 + 1) / 3.67 = .193 M

In my ICE chart, I'm solving for x because .193 M is the concentration at equilibrium. Since [HI] is 0 initially, I found x to be .0965. I subtracted this number from the concentration of H2 and I2 to get its concentration at equilibrium.

Then I used the rate law k = [product]^2/[reactants]

(.193^2)/(.0075^2)

and my answer comes out to be 662

(CORRECT ANSWER IN BOOK IS 764)

 

[Borek]

No idea how you used ICE table here. This is a simple stoichiometry.

 

[ahhppull]

No idea how you used ICE table here. This is a simple stoichiometry.

How would you use stoich? I'm given 2 initial mass and 1 equalibrium mass..

 

[Borek]

Use equilibrium mass to calculate how much of the initial reacted - that will let you calculate how much was left unreacted.

This is not much different from using ICE table, after all, ICE table is just a simple of way of tracing stoichiometry. But, ICE table is used to construct reaction quotient using unknown concentrations expressed in terms of x, and I don't see how you can solve for x if you don't know Kc value. Unless you used ICE table but solved just for stoichiometry in some twisted way.

Shwo details of your work.

 

[ahhppull]

I still don't know how to do this problem, so I would be really happy is someone showed the work for this problem.

I solved for x by assuming that initial concentration of HI is 0 because it has not been formed yet. The change would be .193 M and since there's 2 moles, x is actually .0965.

 

[Borek]

I solved for x

I asked you to show your work. I have no idea what x is nor how you solved for it, so I can't help you.

And no, nobody will show you how to solve the question. We can help you solve it on your own.

 

[ahhppull]

Thanks...I got it but I'm just a bit off. Maybe due to my sig figs.