Chemistry P41 How was it?

[vivo990]

Well, just to make sure.

[Fe3+] = 2x10^-4
and [I2] = 1x10^-2
(This is given in the question)

I got:
[Fe2+] = 2x10^-2 and
[I-] = 2x10^-4

reaction is:

2Fe3+ + 2I- ------------> 2Fe2+ + I2


so Kc = ( [Fe2+]^2 x [I2]) / ([Fe2+]^2 x [I-]^2)



so Kc = {(2x10^-2)^2 x (1x10^-2)} / { (2x10^-4)^2 x (2x10^-4)^2}
Kc = 2.5x10^9 dm3 mol-1

 

[vivo990]

bro it is the same becuse yo dont see the overall equaton see the individual half equation hence 1:1

Dude, there is a difference between a full equation and equilibrium. There is nothing means half equation when it comes to equilibrium. I mean you have to treat every specie inpendantly.

 

[Omar99]

Yes I got 25 but I was confused , what volume to take after finding the no. Of moles. As moles of fe^3+ were the con. Of fe2+, as I thought that the volume would be 1dm

 

[vivo990]

Yes I got 25 but I was confused , what volume to take after finding the no. Of moles. As moles of fe^3+ were the con. Of fe2+, as I thought that the volume would be 1dm

See, the reactants before equilibrium had the same number of moles, and in the eqn the mol ratio is 1:1. So obviously, I- has to be same as Fe3+. There were nothing of the products in the beginning, so what is produced, has to be in the ratio of 2:1 (between products).

 

[Omar99]

Ok so why did they give us the volume

 

[Omar99]

bro it is the same becuse yo dont see the overall equaton see the individual half equation hence 1:1

Can u show ur steps

 

[vivo990]

Ok so why did they give us the volume

Because, they said equal conc of Fe3+ and I- of 100cm3 each were mixed.
I.e., equal number of moles of Fe3+ and I-,
I think they were just trying to make the question a bit tougher.

 

[vivo990]

Can u show ur steps

I have showed you my steps a little earlier in this thread.

 

[1357913579]

Can u show ur steps

actually i just did it simply see
100cm^3 is used for all means we can use the concentration instead of mole ratio as volume s the same
then
-I wrote(iam not sure if its correct or not) that fe^3+ and fe^2+ have same concentration
-for the second part i multiplied by 2 for iodine
then i just pluged in the values for 25 the answer

 

[Omar99]

And what about the order of hcl

 

[1357913579]

And what about the order of hcl

oder=1

 

[1357913579]

And what about the order of hcl

because when i drew the next curve i got douuble the initial rate that why drawing tangent in the begining of both the graph's.

 

[Omar99]

And what about the last question's formula of polymer was it addition or condensation

 

[1357913579]

And what about the last question's formula of polymer was it addition or condensation

addition for sure your talking about last question second part right?
what was ur order and kc answer?

 

[mady666]

I found the paper a little tough. I hope the threshold is around 55
I got the Kc value as 25.
Was the percentage of Al 6% ?

 

[1357913579]

I found the paper a little tough. I hope the threshold is around 55
I got the Kc value as 25.
Was the percentage of Al 6% ?

yes 6 percent
and i also got 25

 

[Omar99]

I did addition for its structure, order was 1. But I think That I did some thing wrong in kc value because when they asked for conc. of fe^2+ so I took the conc value same for fe^3+ and multiplied it with .1 dmcube of its volume and then divide by 1 for its conc.
Same for I- but the ratio was different, but got the answer as 25. I don't know that's right or wrong

 

[mady666]

And what about the order of hcl

I got a zero order because half life was decreasing

 

[Omar99]

I got it 1st order as both half life's were almost the same

 

[1357913579]

no there were 2 things both were 1st order
one was of the of which the comount dont remmeber ts name it ws on the yaxs of the grph its was of the oder 1 becuase half life was constant.
and for order of hcl also 1 becuase comparing the initial rate of both the graph the one drawn before and the one after( the one which we had to drew)
right??