anonymous
  • anonymous
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
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
help plzz!! its been 3 hours !!
JFraser
  • JFraser
what's the form for the equilibrium expression for this reaction?
anonymous
  • anonymous
product over reactants

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More answers

JFraser
  • JFraser
what about the equation? K = ?? for this reaction?
anonymous
  • anonymous
i just need help on the table !! rest i can do
anonymous
  • anonymous
K=50
JFraser
  • JFraser
what's the equation for K for this reaction?
anonymous
  • anonymous
\[Kc=\frac{ [HI]^2 }{ [H_2][I_2] }\]
JFraser
  • JFraser
that's the one
JFraser
  • JFraser
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
anonymous
  • anonymous
so could u help me in the table ?
anonymous
  • anonymous
H2 + I2 --> 2HI Moles at initial state = 3 4 0
anonymous
  • anonymous
Moles at equilibrium state ??
JFraser
  • JFraser
R \(H_2 + I_2 \rightleftharpoons 2HI\) I 3M 4M "0" C "-x" "-x" "+2x" E what are the equilibrium values going to be?
anonymous
  • anonymous
thats what i need to know
anonymous
  • anonymous
ok i'll try
anonymous
  • anonymous
H2 + I2 --> 2HI moles at e = 3-x 4-x 7-x
anonymous
  • anonymous
is this correct ? coz i m not sure about this
anonymous
  • anonymous
what happened ??
anonymous
  • anonymous
hello u there ?
JFraser
  • JFraser
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
anonymous
  • anonymous
u mean 7-2x
JFraser
  • JFraser
no, i mean 2x
JFraser
  • JFraser
where does the 7 come from? from adding the H2 and I2 together? why would you do that?
anonymous
  • anonymous
oh i was mistaken
anonymous
  • anonymous
so this is the equation
anonymous
  • anonymous
\[Kc=\frac{ [2x]^2 }{ [3-x][4-x] }\]
JFraser
  • JFraser
yes, and the whole value of Kc is 50.
JFraser
  • JFraser
so solve for X, and you'll get the equilibrium concentrations
anonymous
  • anonymous
for all 3 ?
JFraser
  • JFraser
solve for x, then plug each x back into the equilibrium concentration of each species
anonymous
  • anonymous
oh yes ..i remember !! thanks
JFraser
  • JFraser
yw
anonymous
  • anonymous
x^2-35x+600=0
anonymous
  • anonymous
sorry it is 46x^2-350x+600=0
JFraser
  • JFraser
then solve for x
anonymous
  • anonymous
which value should i use for eq. conc. x=5 or x=60/23 ??
JFraser
  • JFraser
can the equilibrium concentration of H2 be LARGER thanthe amount of H2 you start with?
anonymous
  • anonymous
no
anonymous
  • anonymous
x=60/23 ? right ?
JFraser
  • JFraser
so which solution for x is the only one that makes sense?
anonymous
  • anonymous
x=60/23
JFraser
  • JFraser
right, now find the equilibrium concentration of H2
anonymous
  • anonymous
0.391
JFraser
  • JFraser
and the concentrations of the rest of the reaction
anonymous
  • anonymous
I2 = 1.391
anonymous
  • anonymous
and 2HI = 5.217
JFraser
  • JFraser
look right to me. to check, plug those values into K and make sure you get 50
anonymous
  • anonymous
YES IT IS CORRECT IT IS COMING 50.04

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