Any chemistry experts out there?
24 March 2009, 06:26 PM
#1
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Any chemistry experts out there?
My youngest has just sent me this, he's supposed to have it done by tomorrow, but has left it. He can do all the rest of his questions but this one stumps him.
Anyone help?
TIA
Experiment [A] (mol L-1) [B] (mol L -1) Initial rate ( mol L-1s-1)
1,,,,,,,,,,,,,,, 1.6,,,,,,,,,,,, 3.4,,,,,,,,,,,, 3*10-4
2,,,,,,,,,,,,,,, 1.6,,,,,,,,,,,, 6.8,,,,,,,,,,,, 9*10-4
3,,,,,,,,,,,,,,, 3.2,,,,,,,,,,,, 3.4,,,,,,,,,,,,, 3*10-4
Determine the order of the reaction with respect to compound B, explain how you have reached your conclusion.
Sorry about the commas, I'm trying to get it to line up with the columns
Determine the order of the reaction with respect to compound B, explain how you have reached your conclusion.
[/COLOR]
Anyone help?
TIA
Experiment [A] (mol L-1) [B] (mol L -1) Initial rate ( mol L-1s-1)
1,,,,,,,,,,,,,,, 1.6,,,,,,,,,,,, 3.4,,,,,,,,,,,, 3*10-4
2,,,,,,,,,,,,,,, 1.6,,,,,,,,,,,, 6.8,,,,,,,,,,,, 9*10-4
3,,,,,,,,,,,,,,, 3.2,,,,,,,,,,,, 3.4,,,,,,,,,,,,, 3*10-4
Determine the order of the reaction with respect to compound B, explain how you have reached your conclusion.
Sorry about the commas, I'm trying to get it to line up with the columns
Determine the order of the reaction with respect to compound B, explain how you have reached your conclusion.
[/COLOR]
Last edited by tanyatriangles; 24 March 2009 at 06:30 PM.
24 March 2009, 07:40 PM
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According to Mrs PG (Chemistry teacher):
The order with respect to B is 1.5
The following rules are followed:
When the concentration is doubled and the rate of reaction stays the same the order is 0 (as with reactant A).
When the concentration is double and the rate of reaction doubles the order is 1.
When the concentration is doubled and the rate of reaction increases fourfold then the order is 2.
In this reaction the concentration of B is doubled and the rate of reaction increases threefold therefore the rate of reaction must be 1.5.
She doesn't condone cheating though
as long as he reads it, takes it in and understands it she's happy to help 
Paul & LJ
The order with respect to B is 1.5
The following rules are followed:
When the concentration is doubled and the rate of reaction stays the same the order is 0 (as with reactant A).
When the concentration is double and the rate of reaction doubles the order is 1.
When the concentration is doubled and the rate of reaction increases fourfold then the order is 2.
In this reaction the concentration of B is doubled and the rate of reaction increases threefold therefore the rate of reaction must be 1.5.
She doesn't condone cheating though
as long as he reads it, takes it in and understands it she's happy to help Paul & LJ
25 March 2009, 01:46 AM
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May I guess 1.585.
The concentration of A does not affect the reaction. A^n is a constant, so n must = 0.
A^0 is always =1 so ignore.
Therefore rate = k B^m
Rate triples as B doubles.
r = k B^m and 3r = k (2B)^m
multiply left expression by 3
3r = 3 k B^m and 3r = k 2^m B^m
so 3 k B^m = k 2^m B^m
divide both sides by k B^m
3 = 2^m
so m = 1.585
QED
Happy to be proved wrong as IANAC.
The concentration of A does not affect the reaction. A^n is a constant, so n must = 0.
A^0 is always =1 so ignore.
Therefore rate = k B^m
Rate triples as B doubles.
r = k B^m and 3r = k (2B)^m
multiply left expression by 3
3r = 3 k B^m and 3r = k 2^m B^m
so 3 k B^m = k 2^m B^m
divide both sides by k B^m
3 = 2^m
so m = 1.585
QED
Happy to be proved wrong as IANAC.
25 March 2009, 09:58 AM
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Thanks for all replies: we got 1.5, but he's going to get it explained this morning.
Amazing what you can find out on this site, isn't it
?
Amazing what you can find out on this site, isn't it
?
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