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Topic: Computing the equilibrium concentration of reactants and products  (Read 403 times)

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Offline Win,odd Dhamnekar

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Offline chenbeier

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Re: Computing the equilibrium concentration of reactants and products
« Reply #1 on: November 17, 2020, 12:44:40 PM »
Do you know the mass Action Law
Write it for the given reaction.
Then build the subtraction between initial concentration to the target x .
The Second question :
Values obtained by experiments and measurment of all components.

Offline Win,odd Dhamnekar

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Re: Computing the equilibrium concentration of reactants and products
« Reply #2 on: November 18, 2020, 01:11:54 AM »
Law of Mass Action
  For a general reaction of the form [itex]aA +bB \rightarrow cC +dD [/itex], the law of mass actions states that the equilibrium condition is expressed by the equation,[itex] \frac {C^c \cdot D^d}{A^a\cdot B^b}[/itex]

 Now, for the reaction [itex] H_2(g) + I_2(g) \rightleftharpoons 2HI (g)[/itex], initial concentrations of Hydrogen molecule and Iodine molecule as per experiment (1) 0.024M and 0.0138M respectively. Let X mole/liter of each of the product  be formed.
 At equilibrium, the concentrations would be [itex]H_2=[0.024M-X], I_2=[0.0138-X], 2HI=[X \cdot M] [/itex]

So, [itex]K_c=\frac{X^2}{(0.024-X)(0.0138-X)}=46.42 \Rightarrow X^2=46.42(X^2-0.0378 \cdot X +0.0003312[/itex]

 Now,  solving for X, we get X=1.3432205555e-2 or X=2.52000269414e-2.

 Now X=1.34322e-2  do not match with 0.0252 M of 2HI. So X=0.0252 M of 2HI.
 So, equilibrium concentration of [itex]H_2[/itex]  is 0.024M - 0.0252 =-0.0012 M which is  actual  equilibrium concentration of [itex]I_2[/itex] with minus sign. Does this mean it is the equilibrium centration of Iodine molecules?

and,

equilibrium of concentration of [itex]I_2[/itex] is 0.0138 M- 0.0252 M=-0.0114M which is actual equilibrium concentration [itex]H_2[/itex] with minus sign. Does this mean it is the equilibrium concentration of Hydrogen  molecules?

 Why we should neglect the first value of X?
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Offline Borek

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Re: Computing the equilibrium concentration of reactants and products
« Reply #3 on: November 18, 2020, 03:29:37 AM »
Is the negative concentration something physically possible?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline mjc123

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Re: Computing the equilibrium concentration of reactants and products
« Reply #4 on: November 18, 2020, 05:35:23 AM »
There is a mistake in your calculation. If the concentrations of H2 and I2 each decrease by X M, by how much does the concentration of HI increase? (Note, there is no such compound as 2HI; "2HI" in the equation simply means two equivalents of HI.)

Offline Win,odd Dhamnekar

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Re: Computing the equilibrium concentration of reactants and products
« Reply #5 on: November 18, 2020, 08:15:42 AM »
 So, X=0.0252 is for HI. But in other 3 cases of experiments 2,3,4, using the known initial concentrations of [itex]H_2,I_2[/itex], we could not get the correct concentration of HI. That means This formula is not universal. It is invalid in most of the cases.

  Taking the initial concentrations of HI from the experiment 5,6 also, we could not get the correct initial concentrations of [itex]H_2, I_2[/itex]. For experiment 5, i used the following formula [tex]46.4=\frac{(X-0.0304)\cdot(X-0.0304)}{(0.0304)^2}[/tex]
Any science consists of the following process.
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Offline mjc123

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Re: Computing the equilibrium concentration of reactants and products
« Reply #6 on: November 18, 2020, 08:48:13 AM »
Quote
So, X=0.0252 is for HI.
No it isn't. [HI] ≠ X
Quote
That means This formula is not universal. It is invalid in most of the cases.
No, it means you've got it wrong. Look at the stoichiometry of the equation. Answer the question I asked you. If [H2] = 0.024 - X, what is [HI]?
Write the correct expression for Kc and solve for X.
You're getting it wrong for 5 and 6 for the same reason.

Offline Win,odd Dhamnekar

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Re: Computing the equilibrium concentration of reactants and products
« Reply #7 on: November 19, 2020, 01:04:03 AM »
A)Using experiment (1) data, [tex]\frac{(2x)^2}{(0.024-x)*(0.0138-x)}=46.42[/tex]
If we solve for x, we get x=0.012599979 or x=0.028764377. The value of 0.028764 should be neglected because it is the  concentration of reactants  which is more than the initial concentrations of  [itex] H_2, I_2[/itex] Hence the equilibrium concentrations are [itex] HI=x=0.0252 M,H_2=0.024-x=0.024-0.0126=0.0114,I_2=0.0138-0.0126=0.0012[/itex]

B)Using experiment (2) data, [tex]\frac{(2x)^2}{(0.024-x)(0.0168-x)}=46.42[/tex]
If we solve for x, we get x=0.0147655623466 or x=0.02988167952 . For the same reason mentioned in A), we should neglect the value x=0.0298816.  Hence the equilibrium concentrations are [itex]HI=x=0.029531, H_2=0.024-0.0147655=0.009234, I_2=0.0168-0.0147655623=0.0020344[/itex]

 Likewise, for experiment 3 and 4 also, we get correct answers.
 But i don't understand how to formulate the correct expression for [itex] K_c[/itex] in experiment 5 and 6 where reaction is reversed?

Any science consists of the following process.
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Offline Win,odd Dhamnekar

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Re: Computing the equilibrium concentration of reactants and products
« Reply #8 on: November 19, 2020, 02:49:39 AM »
 I got the correct final concentrations of reactants and products in experiment 5 and 6 after forming correct expression for [itex]K_c[/itex]
Any science consists of the following process.
 1) See2)Hear 3) Smell if needed4) Taste if needed
5) Think 6) Understand 7) Inference 8) take decision [Believe or disbelieve, useful or useless, healthy or unhealthy, cause or effect, favorable or unfavorable, practical or theoretical, practically possible or practically impossible or any other required criteria]

Offline mjc123

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Re: Computing the equilibrium concentration of reactants and products
« Reply #9 on: November 19, 2020, 05:38:44 AM »
Well done.

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