Altering the order by method of excess reagent

Watch
Ellie.Ginger
Badges: 10
Rep:
?
#1
Report Thread starter 1 year ago
#1
I don't understand what the last paragraph of the following means (which I came across on the textbook) PLSSS EXPLAINNN


"Consider a reaction: A+2B-->C+D
The rate equarion is of the form rate=k[A]^p[B]^q

However if the initial concentration of B is made at least 10 times that of A, the change in [B] during the reaction will be negligible. This means that [B] is constant within experimental error, so the rate equation becomes rate=constant×[A]^p where the constant=k×the approximately constant value of [B]^q

The value of p in order with respect to A can be found by the usual methods of initial rate or half life.

If the experiment is repeated with the same initial concentration of A but 20 times as much B(doubling the concentration of B from the first experiment) the way in which initial rate alters will depend on the order with respect to B. If the rate doubles, the reaction is 1st order with respect to B. If the rate quadruples, the reaction is 2nd order in B."
0
reply
username1445490
Badges: 9
Rep:
?
#2
Report 1 year ago
#2
What it is saying is that if you are trying to find the order of reaction for A (by half lives) then you need to have B in very large excess so that as the reaction proceeds the concentration of A decreases but B pretty much remains constant. This allows you to determine the order of reaction for A by plotting a concentration - time graph and working out successive half lives without the concentration of B changing and affecting the results.
You now know the order for A and need to find the order for B.

So now you repeat the experiment keeping the concentration of A the same but double the concentration of B and use the initial rates method. If the Initial rate in the second experiment is double the first one then the order of B is 1, whereas if the initial rate is quadrupled then the order of B is 2.
Last edited by username1445490; 1 year ago
1
reply
Ellie.Ginger
Badges: 10
Rep:
?
#3
Report Thread starter 1 year ago
#3
(Original post by Madasahatter)
What it is saying is that if you are trying to find the order of reaction for A (by half lives) then you need to have B in very large excess so that as the reaction proceeds the concentration of A decreases but B pretty much remains constant. This allows you to determine the order of reaction for A by plotting a concentration - time graph and working out successive half lives with the concentration of B changing and affecting the results.
You now know the order for A and need to find the order for B.

So now you repeat the experiment keeping the concentration of A the same but double the concentration of B and use the initial rates method. If the Initial rate in the second experiment is double the first one then the order of B is 1, whereas if the initial rate is quadrupled then the order of B is 2.
omg thanks alottt
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Do you have the space and resources you need to succeed in home learning?

Yes I have everything I need (455)
56.38%
I don't have everything I need (352)
43.62%

Watched Threads

View All
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise