[BaO]

 today my teacher gave me this question:

 
    Two experiments were performed involving the following equilibrium.
The temperature was the same in both experiments.

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

In experiment A, 1.0 M  H2 and 1.0 M  I2 were initially added to a flask and
equilibrium was established. In experiment B, 2.0 M  HI was initially added to a
second flask and equilibrium was established. Which of the following statements
is always true about the equilibrium concentrations?

A.  [H2 ] =  [HI ] in experiment A.
B.  [HI ] =  2 [H2 ] in experiment A.
C.  [HI ] in experiment A = [ HI ] in experiment B.
D. [ HI ] in experiment A = 1/2 [I2] in experiment B

and she said this was a hard one so she gave us the answer too : C
but  i dont get it ,why C?
 
i need your *delete me*   please....


thank you very much

[arnyk]

Doesn't seem to bad, intuitively I can see that A, B and D don't make sense.  But I guess if you're going to quanititatively back it up, then you could probably just ICE it.

[BaO]

because my teacher asked a good explanation for that answer!

[BaO]

but intuitively i also see B is correct!

[Borek]

Is there any difference between TOTAL amounts of hydrogen and iodine in both systems?

[BaO]

i dont know!

what i dont get is how can [HI] of both systems are equal when one is produced later and one is initially added . and the question asks us to compare them at EQ!

[Borek]

i dont know!

Calculate!

[BaO]

dont get me wrong. i'm not a lazy person , but honestly, i dont know what you mean and have no ideal how to calculate it

[wereworm73]

For both experiments, lets just say both flasks are 1L. So...

In Experiment A, this means you start out with 1 mole of hydrogen & 1 mole of iodine gas.  Some of it combines to form HI.  If this was a completely one-way reaction, it can form 2 moles of HI
(H₂ + I₂ <--> 2 HI).

In Experiment B, you start out with 2 moles of HI, and some of it breaks down into hydrogen & iodine.  If this was a completely one-way reaction, this can form 1 mole of H₂ & 1 mole of I₂.

However, the reaction goes both ways and there is an equilibrium constant that determines the ratio of hydrogen & iodine to HI. 

This ratio will be somewhere between all HI/no H₂ & I₂ and all H₂ & I₂/no HI.




->-------------->----------->----X---<---------<-------<-
^                                              ^                               ^
all H₂                         equilibrium point                 all HI
& all I₂

start of                                   end of                   start of
Exp. A                                 Exp. A & B                 Exp. B


So, in both of those experiments, this same ratio is approached but from opposite directions.  In the end, the concentration of HI will be the same for both experiments.  So, the answer is C.

[arnyk]

Looking it over I think you don't even need the K value for that equilibrium. Since you have two trials, you effectively have a system of equations, which means you can sub out the K value.

Basically, you set up your ICE table, then you get for:

'A'

K = [blah blah]/[]

'B'

K = [blah]

so....

[blah blah]/[] = [blah]

solve.

*Hopefully you get what I'm saying, or else you'll need to learn the concept first and such...or I'm just horrible at explaining things lol.

[BaO]

hey i get it,  totally get it 
thanks a lot every one
espeacially the Keq explanation ! that helps me alot