Question

. Gaseous Hydrogen and Iodine: are placed in a closed container at 425 C, producing Hydrogen, which partially decomposes to Hydrogen and Iodine.

H2 (g) + I2 (g) <---->2HI (g)

The following are given:

[HI] = 2.53 * 10-4 M [H2] = 3.79 * 10-4 M [I2] = 3.79 * 10-4 M

Show the equilibrium Constant formula

What is the value of Kc at this temperature?

Answer 1

Pretty sure I understand how to do everything except factor the temperature into this. Correct me if I"m wrong, but the formula would be:

[HI]^2 / [H2] [ I2]

Then would I substitute those things for the moles like this?

[2.53 * 10^-4]^2 / [3.79 * 10^-4] [3.79 * 10^-4]

Answer 2

@TheSmartOne

Answer 3

@JFraser

Answer 4

@Kryten @levinelevine

Answer 5

You don't need to factor the temp into the calc. Just know that at another temp, the Kc would be different. :-)

Answer 6

So is the way I set the equation up correct then?

Answer 7

Yep, looks good.

Answer 8

^that's the way to do it

Answer 9

So just to confirm, it would be:

[2.53 * 10-4 M HI]^2 / [3.79 * 10^-4 M H2] [ 3.79 * 10^-4 M I2]

This would equal 4.6 * 10^-1

But shouldn't equillibrium constant of this system be 54.34?

Answer 10

Hm, why do you think it should be 54.34?

Answer 11

Well, I'm looking at my chemistry textbook, and it says

"At 425°C, the equilibrium constant for this equilibrium reaction system has the average value of 54.34. This value for K is constant for any system of H2, I2, and HI at equilibrium at this temperature. If the calculation for K yields a different result, there must be a reason. Either the H2, I2, and HI system has not reached equilibrium or the system is not at 425°C."

Isn't this reaction supposed to have reached equilibrium?

Answer 12

if the problem tells you the value of Kc, then the value you've calculated is for a reaction that has not yet reached equilibrium. If it's asking for the Kc value, then we assume the concentrations given are already at equilibrium

Answer 13

if you've calculated a value from concentrations and get a value other than what it tells you the Kc value is, you've calculated a \(Q\) value, a reaction quotient, which will tell you what direction the reaction will shift in order to \(get \space to\) equilibrium

Answer 14

So then my answer is correct? Because the question above does ask for the value of Kc.

Answer 15

The problem doesn't explicitly state that the values given are equilibrium concentrations. I don't really know what to tell you, sorry! It doesn't seem specific enough.