Is the Equilibrium Calculation for HI Correct?

[myvow]

Kc1 = (5.8*2/5)^2 / (14/5)(1.4/5) = 6.865

6.865 = [HI]^2 / (45/253.8/100)(0.5/2/100)
[HI] = 5.52x10^-3 M
mass no of HI = [HI] x (126.9+1) x 100 = 70.6g
is it correct?
And how to do 3bii
I GOT 3bi Kc2 = [HCl(g)]^2/ / Kc(1) [H2(g)] [Cl2(g)]

 

[Bystander]

70.6g is it correct?

Think about it --- you've got 45 g I₂ to start --- what's the largest mass of HI you can make?

 

[myvow]

6.865 = x^2 / ((45/253.8/100)-0.5x)((0.5/2/100)-0.5x)
This equation is OK?because the mole ratio of H2 to Hi is 1:2
So [HI]=x=5.33*10-4M
HI =6.813g?

 

[Bystander]

The 6.865 is okay for an equilibrium constant. The notation you're using for the equation is unconventional. What're the "100s" doing in there?

 

[myvow]

The 6.865 is okay for an equilibrium constant. The notation you're using for the equation is unconventional. What're the "100s" doing in there?

100L
The volume

 

[Bystander]

And the HI is not also occupying that volume?

 

[myvow]

Yes, but molarity = mole/volume
Does not volume is necessary?

 

[Bystander]

You're using it in the denominator for hydrogen and iodine concentrations. Don't you think you should be using it in the numerator as well? After all, you did it that way when finding the equilibrium constant.

 

[myvow]

((45/253.8)/100)
？？this way？？

 

[Bystander]

((45/253.8)/100)
？？this way？？

No, the way you did it for the very first step when you solved for the equilibrium constant.

Kc1 = (5.8*2/5)^2 / (14/5)(1.4/5) = 6.865

 

[myvow]

You're using it in the denominator for hydrogen and iodine concentrations
What it mean...

 

[Bystander]

Do you see where you inserted the number "5" for five liters in the very first calculation you did? Do you see it in both numerator, HI, and denominator, H and I?

 

[myvow]

Do you see where you inserted the number "5" for five liters in the very first calculation you did? Do you see it in both numerator, HI, and denominator, H and I?

For example
14 is the mole of h2,so 14/5=2.8 which is concentration of h2, right?
OK then there is 0.5g of h2,
0.5/2=0.25mol
0.25/100=1.25*10^-3M
What wrong with me?

 

[Bystander]

6.865 =( x^2/100²) / ((45/253.8/100)-0.5x)((0.5/2/100)-0.5x)

So, you will have x moles of HI in 100 liters. Clear?

 

[myvow]

So, you will have x moles of HI in 100 liters. Clear?

Does numerator will not be wrong?

 

[Bystander]

Does numerator will not be wrong?

If you are using the 100 liter volume in the denominator, you HAVE to include it in the numerator.

 

[myvow]

If you are using the 100 liter volume in the denominator, you HAVE to include it in the numerator.

I know what you mean sir
if I use (x/100)^2, .x will be the mole of HI right?
But in the numerator,did you see((0.5/2/100)-0.5x)
if x =mole, minus 0.5x
this is the mole instead of concentration

 

[Bystander]

Okay. We are making progress. You also need to be dividing the 0.5x by 100 if you're going to work with concentration. It is your option or choice whether to express x the amount actually reacting subtracted from the amount you begin with as just moles, or to divide by the volume and handle amounts as concentrations. The one thing you do have to do is treat reactants and products and fraction reacted all the same way, either as number of moles, or as concentrations.