What amount of moles will be produced?

Question

the equilibrium constant for H2 I2 <-----> 2 HI

is Kc= 50.0 at 745 K. when 1 mole of I2 and 1 mole of H2 are allowed to equilibrate at 745 K in a flask of volume 8 L, what amount (in moles of HI will be produced?

Answer 1

FIRST:REACTION DOES NOT CHANGE THE TOTAL AMOUNT IN MOLES,BECAUSE 1 MOL H2+1 MOL I2 &gt; 2MOL HI

THIS IMPLIES Kc=Km=50

SECOND: Km=(2*X)^2/(1-X)(1-X),WHERE X IS THE QUANTITY IN MOL

FOR THE FORMED HI.

INITIAL:H2=1MOL,I2=1MOL;HI=0MOL

FINAL:H2=1-X,I2=1-X,HI=2X

THIRD:Km=50=(2X)^2/(1-X)(1-X),

OR:50=(2X)^2/(1-X)^2,SQRT50=2X/(1-X),

7.071=2X/(1-X), 2X=7.071-7.071X, 9.071X=7.071,X=0.7795MOL HI

IF U WANT TO VERIFY,PUT THE FINAL RESULT IN THE EXPRESSION OF Kc,BY DIVIDING MOLES TO VOLUME.

mob3 DID WROG HERE:[H2] = [I2] = 0.125 - x

YOU CANNOT SUBSTRACT FROM A CONCENTRATION ANOTHER CONCENTRATION

AND HERE: [HI] = 2 (0.0974) = 0.1948 mol;he just multiplied

a concentration with 2 by no reason

do not confuse 3MOL with 3M(MOL/L)

Answer 2

H2 + I2 &lt;-----&gt; 2 HI

initially there is 1 mol I2 and 1 mol H2 in the flask plus no HI

[H2] = [I2]= 1/8 = 0.125 mol assuming equal vol.

[HI] = 0

to attain an equilibrium x mol I2 and x mol H2 will react to produce 2x mol HI according to the mol ratio in the above eqn.

Therefore at equilibrium, assuming equal vol.

[H2] = [I2] = 0.125 - x

[HI] = 2x

Kc = [HI]^2/[I2] . [H2]

50 = (2x)^2 / (0.125-x)^2

50 = [2x/0.125-x]^2

rt50 = 2x/0.125-x

+/- 7.07 (0.125 - x) = 2x

0.88375 - 7.07x = 2x

0.88375 = 9.07 x

x= 0.0974

-0.88375 + 7.07x = 2x

-0.88375 = -5.07x

x = 0.1743

But x cannot be &gt; 0.125 , hence x = 0.0974 mol

Then [HI] = 2 (0.0974) = 0.1948 mol in 8 L at equilibrium