Question

3 moles of H_2 and 4 moles of I_2 were heated in a sealed tube at a given temperature at which Kc is 50. If the volume of the tube is 1 dm^3 , determine the composition of the equilibrium.

H_2 + I_2 ⇌ 2HI

Answer 1

help plzz!! its been 3 hours !!

Answer 2

what's the form for the equilibrium expression for this reaction?

Answer 3

product over reactants

Answer 4

what about the equation? K = ?? for this reaction?

Answer 5

i just need help on the table !! rest i can do

Answer 6

K=50

Answer 7

what's the equation for K for this reaction?

Answer 8

\[Kc=\frac{ [HI]^2 }{ [H_2][I_2] }\]

Answer 9

that's the one

Answer 10

You know you start with 3 moles of H2 and 4 moles of I2, and you know the equilibrium value has to be 50, so plug into the equation for Kc and solve for the equilibrium concentrations

Answer 11

so could u help me in the table ?

Answer 12

H2 + I2 --> 2HI
Moles at initial state = 3 4 0

Answer 13

Moles at equilibrium state ??

Answer 14

R \(H_2 + I_2 \rightleftharpoons 2HI\)
I 3M 4M "0"
C "-x" "-x" "+2x"
E

what are the equilibrium values going to be?

Answer 15

thats what i need to know

Answer 16

ok i'll try

Answer 17

H2 + I2 --> 2HI
moles at e = 3-x 4-x 7-x

Answer 18

is this correct ? coz i m not sure about this

Answer 19

what happened ??

Answer 20

hello u there ?

Answer 21

the equilibrium concentration of HI is 2x, since you start with nothing, and gain twice the amount that the H2 loses.

Look at the coefficients of the balanced reaction

Answer 22

u mean 7-2x

Answer 23

no, i mean 2x

Answer 24

where does the 7 come from? from adding the H2 and I2 together? why would you do that?

Answer 25

oh i was mistaken

Answer 26

so this is the equation

Answer 27

\[Kc=\frac{ [2x]^2 }{ [3-x][4-x] }\]

Answer 28

yes, and the whole value of Kc is 50.

Answer 29

so solve for X, and you'll get the equilibrium concentrations

Answer 30

for all 3 ?

Answer 31

solve for x, then plug each x back into the equilibrium concentration of each species

Answer 32

oh yes ..i remember !! thanks

Answer 33

yw

Answer 34

x^2-35x+600=0

Answer 35

sorry it is
46x^2-350x+600=0

Answer 36

then solve for x

Answer 37

which value should i use for eq. conc. x=5 or x=60/23 ??

Answer 38

can the equilibrium concentration of H2 be LARGER thanthe amount of H2 you start with?

Answer 39

no

Answer 40

x=60/23 ? right ?

Answer 41

so which solution for x is the only one that makes sense?

Answer 42

x=60/23

Answer 43

right, now find the equilibrium concentration of H2

Answer 44

0.391

Answer 45

and the concentrations of the rest of the reaction

Answer 46

I2 = 1.391

Answer 47

and 2HI = 5.217

Answer 48

look right to me. to check, plug those values into K and make sure you get 50

Answer 49

YES IT IS CORRECT IT IS COMING 50.04