Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: kurashi on July 03, 2006, 11:09:54 AM
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Hiz all ,
Currently, I am doing a assignment to determine the rate law of this eqn through a experiment using a batch reactor at (15s intervals) and conductivity meter.
0.01M 0.05M
NaOH + CH3COOC2H5---------->CH3COONa + C2H5OH
As from the equation, only NaOH and CHCOONa contribute to the conductivity of the solution. However, I can only obtain the total conductivity of the solution ( conductivity of NaOH + conductivity of CH3COONa ) May I know is it possible to obtain the conductivity of each individual's reactant?
Pls help
Thank you
Beat regards
Desmond
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May I know is it possible to obtain the conductivity of each individual's reactant?
No, but if you assume that conductivities are additive you can find out how the reaction goes measuring total.
Hint: Na+ concentration doesn't change, [OH-] + [CH3COO-] = const.
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But sir,
what do you mean by addictive?
My hypothesis is the conductivity total will decrease with respect to time . Am I right?
Regds
Desmond
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Yes, you would be right about the conductivity dropping because the reaction is replacing the highly mobile hydroxide anions with the less mobile acetate.
Didn't you measure the conductivity of 0.01 M NaOH before adding the ethyl acetate? That should be the solution's conductivity at t=0 sec., right? And when the conductivity finally bottoms out, wouldn't that suggest that the reaction is complete, with the solution containing 0.01 M sodium acetate instead of 0.01M NaOH?
Anyway, since [NaOH] + [CH3COONa] = 0.01 throughout the reaction, you can say the overall conductivity of the solution is equal to ax + b(0.01-x), with a=conductivity of sodium hydroxide, b=cond. of sodium acetate, x=[NaOH] & (0.01-x) being [sodium acetate]. If you measure the conductivity of the NaOH before adding any ethyl acetate, you can easily solve for a. Then using the final data point (assuming you allowed the reaction to completion), you can calculate b. Once you've done that, you can find out the concentrations at any data point.
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Thanks for your reply...it has been very helpful. However, i found out that the experiment conducted for one of the reactants have to be in excess. Hence. the experiment have to be done twice for each reactant(in excess).
What i'm afraid is that the conductivity measured, includes the excess e.g(NaOH) that was not reacted. Therefore, I'm not sure whether the method/equation given by you is applicable. ???
For the experiment, I have to set the concentration of the reactants [NaOH] :[CH3COOC2H5] to the ratio of 1:5 respectively and vice versa. Hence, I have to use the method of excess to calculate the rate law.
If thats the case, how to find the conductivity of each reactant from the conductivity total?
I am supposed to calculate the conductivity of NaOH and conductivity of NaAc using the eqn below,
Conductivity total=conductivity of NaOH + conductivity of NaAc
conductivity of NaOH = 0.195[1+0.0184(T-194)][NaOH]
conductivity of NaAc = 0.07[1+0.0284(T-194)][NaAC]
And since one of the reactants ( NaOH ) is in excess, which reactant (NaOH or CH3COOC2H5) should I put in first so as not to spoil the conductivity meter?
Thank you
Regds
Desmond
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OK, for the reaction involving 0.05M NaOH & 0.01M ethyl acetate, you can do the same thing. Measure the conductivity of the hydroxide like before, only it will be 5x more concentrated. Like you said in the first post, ethyl acetate doesn't contribute to the conductivity so there's really no point measuring it. After a while, the conductivity will bottom out (though at a higher level this time since there's still 0.04M NaOH in solution).
As for calculating the rate law, just realize [NaOH]+[NaAc]= initial [NaOH] and [EtAc]+[NaAc]= initial [EtAc]. It'll apply for both experiments so you don't need to worry about reagents being in excess.
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Sir,
If that was the case,can I say that only 0.01 M if NaOH has been reacted for the 2 experiments , the equation that you have given overall conductivity= ax+b(0.01-x) is vaild?
However, is the a and b constant? for my experiment, a and b is a variable hence is it still vaild?
Regds
Desmond
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overall conductivity= ax+b(0.01-x) is vaild?
I would rather go from the start for
overall conductivity= c + ax + b(0.01-x)
as you may expect some conductivity even before NaOH addition. When there is no NaOH excess, c will be small (perhaps even neglectable). In the case of NaOH excess c will be much larger, but calculations made will be in both cases identical.
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However for my experiment, I will have a list of conductivity total at 15 s intervals, hence, from the eqn, the conductivity of NaOH and conductivity of NaAc will be a variable. However, as far as i known, the a and b is contant throughout?
Please correct me if Im wrong.
Thank you
Best regards
Desmond
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a, b and c are constants.
Note that overall conductivity is linearly dependent on the x:
overall conductivity= c + ax + b(0.01-x) = (c + 0.01b) + (a-b)x = a' + b'x
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Ok. I understood. However, currently, I have to determine the Rate law by using a batch reactor for where at each time intervals, I can only obtain the data for overall conductivity. And at each time intervals, the conductivity will change, hence, I have to calculate the concentration of NaOH from the time intervals and conductivity respectivity.
Hence, does the method still works?
Thank you
Regds
Desmond
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I have to calculate the concentration of NaOH from the time intervals and conductivity respectively.
You may easily calculate x from the conductivity data.
Hence, does the method still works?
Yes. But I am not sure finding NaOH concentration is the simplest approach. I think you should concentrate on x (defined as reaction fraction).
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Ok.. If that is the case, Lets assume X as the reducing factor or so called the conversion factor,
then, for eg,
at t=0, Cond = 1 X0=0 [NaOH] not reacted = y(1-X0)
at t=1, Cond = 0.5 X1=0.5 NaOH] not reacted = y(1-X1)
at t=2, Cond = 0.3 X2=0.7 [NaOH] not reacted = y(1-X2)
is my steps vaild as i assume the conductivity data u have mentioned refers to my data from the experiment.
Thank you
Regds
Desmond
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Looks OK.
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However, from the steps i introduced, i realise that the
overall conductivity = drop in conductivity of NaOH + increase in Conductivity of CH3C00Na,
Hence, the difference of overall conductivity consist part of " some" conductivity of CH3COONa, hence, is it still possible to use it?
Best regds
Desmond
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However, from the steps i introduced, i realise that the
overall conductivity = drop in conductivity of NaOH + increase in Conductivity of CH3C00Na,
Hence, the difference of overall conductivity consist part of " some" conductivity of CH3COONa, hence, is it still possible to use it?
Yes, it is covered by the ax + b(0.01-x) - ax is the CH3C00Na increase, b(0.01-x) is the NaOH drop.
Note that what is really happening is that OH- is replaced by CH3COO-, as Na+ concentration doesn't change.
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Hi i am doing the same experiment as well.
But I can't figure out how am I going to obtain the rate law through the method of excess as i do not have the concentration of ethyl acetate, i would be unable to use the integral method to find the order of reaction with respect to ethyl acetate. And i can't figure out how to find K as well if i use the method of excess as K in the method of excess is different from the K in the rate law.