Question

The equilibrium constant, Kc, for the reaction H2(g) + I2(g) 2HI(g), is 60 at 450 °C.
What is the number of moles of hydrogen iodide in equilibrium with 2mol of hydrogen and 0.3mol
of iodine at 450°C?
A) 1/100 B) 1/10 C)6 D)36

Answer 1

@Abhisar can you please help?

Answer 2

@aaronq can you please help with this question

Answer 3

So i think you have to find the Kp first, then use an ICE table. Did you get the Kp yet?

Answer 4

Oh its the same because \(\Delta n=0\)

Answer 5

i think its not related much to Kp that's what provided only.

Answer 6

if we have such a big value of Kc the reactants must be in a very small quantity

Answer 7

[Products]^n/[Reactants]=Kc

Answer 8

\(H_2+ I_2 \rightleftharpoons 2HI\)

I 2 0.3 0
C -x -x +2x
E 2-x 0.3-x 2x

\(K_p=\dfrac{ (2x)^2}{ (2-x ) (0.3-x)}\)

solve for x

Answer 9

and i eliminated options C and D because we cant have negative number of moles thats what i did

Answer 10

when i solved for x, my solution was not even related to any of the options

Answer 11

yeah, my answers arent matching the options either

Answer 12

60=4x^2/(2-x)(0.3-x)
60=4x^2/0.6-2x-0.3x+x^2
36-138+60x^2=4x^2
4x^2-60x^2+138-36
i got x=0.3

Answer 13

should we plug 0.3 to the (2x)^2

Answer 14

yeah, thats what i got too. Yep but thats gonna give you a huge number

The limiting reagent is iodine, so the most you could possibly make is 0.6 mol

Answer 15

oh wait no, i got 0.351786

Answer 16

still none of the options

Answer 17

but @aaronq but how is that possible we would have a negative number of moles
0.3-0.351786=
i got 0.29655807875501894

Answer 18

i got 0.29 for x too but 0.35 for HI

It's possible because your limiting reagent is I2, but look at the stoichiometric coefficients,
their ratio is 1:2, the theoretical yield is

\(\dfrac{0.3}{1}=\dfrac{n_{HI}}{2}\rightarrow n_{HI}=0.3*2=0.6 mol\)

Answer 19

Are you sure these options are 100% correct?

Answer 20

yes that was in my chemistry question paper and i chose B

Answer 21

yeah B is what i would go with as well, it's the closest to anything that would be believable

Answer 22

but still i am not sure based on calculations

Answer 23

thanks for your help :D

Answer 24

yeah i dont know, everything we did was right :S

no problem!

Answer 25

@aaronq If we assume a 1.00L container, then the equilibrium concentrations are the same as the number of moles.

H2(g) + I2(g) <==> 2HI(g) .................. Kc = 60.
2M .......0.3M ............?M

Kc = [HI]² / ([H2] [I2])
60 = x² / 2 / 0.3
x = 6
[HI] = 6M
moles of HI = 6
i got this as an answer on yahoo

Answer 26

thats incorrect, they're assuming the concentrations of reactants wont change (what you can when K is very small) but you can't do it here.

Answer 27

Think about it, how are you gonna get 6 moles from 0.3 moles of the limiting reagent

Answer 28

i get what you are saying. we cant assume that volume is1 L in both cases and yes the Kc value is big. (Y)

Answer 29

we could assume that the volume was 1L, but recall these are gases, not liquids. So this would be in terms of pressure. Regardless, it works out to the same thing.

Answer 30

@abb0t

Answer 31

?

Answer 32

i am stuck on getting the number of moles of HI based on the 2 mol of H2 and the 0.3 mol of I2

Answer 33

1.3, if you are using the correct quadratic that I see above.

Answer 34

but 1.3 is not included in any of the choices @abb0t

Answer 35

Kc value is given i.e 60.Kc=concentration i.e moles of product/volume raised to stoichiometric coefficient of product divided by concentration of reactant raised to its stoichiometric coefficient .you will have to note one thing that volume for both reactants and product will be same as reaction is always performed in a testtube or vessel containing both reactants as well as products.so lets assume x to be moles of HI and V volume.know put values 60=(x/v)^2/0.3/v multiplied by 2/v
so 60=(x/v)^2/o.6 multiplied by v^2
v^2 cancel out hence we get x^2/o.6=60
x^2=36
x=6

Answer 36

Zareen as reaction progresses the amounts of reactants decrease so it should be 2-x and 0.3-x

Answer 37

i think that x cant equal 6 because we cant have number of moles which are negative.

Answer 38

why 2-x or 0.3 -x.2 and 0.3 are moles at equilibrium not initial moles which we have taken before reaction .because these are moles of hydrogen and iodine at 450 degree celsius and kc i.e equilibrium constant is also t 450 degrees ccelsius.in your question what is the number of moles of hydrogen iodide in equilibrium with 2 moles of hydrogen and 0.3 moles of iodine ,mark the line 'in equilibrium with 'it simply suggest that 2and 0.3 moles are moles at equilibrium not initial.

Answer 39

^He's right, the wording in the problem didn't make it clear that those were the concentrations of reactants AT eq.

"What is the number of moles of hydrogen iodide in equilibrium with 2mol of hydrogen and 0.3mol I2" \(\sf could~have ~been~worded~clearer~as\): "At eq. there are 2mol of hydrogen and 0.3mol I2, How many moles of HI are there?".

Answer 40

Equilibrium constants aren't changed if you change the concentrations of things present in the equilibrium. The only thing that changes an equilibrium constant is a change of temperature. now i understand because the concentration does not change with time.

Answer 41

because the temperature has not changed.

Answer 42

Yeah, thats true. My confusion with the question arose from the wording. I assumed the concentrations of reactants given were initial concentrations, not equilibrium concentrations.

Answer 43

me too, thanks @aaronq,@Abhisar ,@abb0t and @zareen