Affect on volume of solute/solvent on accuracy of Rate Law

  • Thread starter Macroer
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  • #1
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Main Question or Discussion Point

This was part of a lab i preformed to determine rate law for HCl and Mg.

50 mL of 0.5M, 1M, 1.5M and 2M HCl solutions were made using water and this was used with Mg ribbon to find the rate law. The same 0.5M, 1M, 1.5M, 2M HCl solutions were made in 100mL solution that contained water. I was told that the 100mL solution of HCl diluted in water provided more accurate results for the calculation of rate law, than the 50mL solution of HCl dilluted in water. I was wondering why this is the case?

3. The Attempt at a Solution
Don't have that great an idea, but the increased volume of solution provides more surface area for reaction. I know it is not from the accuracy of the equipment because both the 50mL and 100mL diluted HCl was measured in a separate flask. These flasks(forgot the name) had only one tick on it(at 50mL and 100mL respectively),
 
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Answers and Replies

  • #2
Borek
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Hard to tell anything not knowing the procedure you were asked to follow.

While it is quite probable that it is a standard experiment and we could google details, you should ask the question in such a way that all information needed is provided. Could even be trying to organize the data you will realize what the problem is by yourself.

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  • #3
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Procedure as follows:
1.Prepare 50mL of 0.5M,1M,1.5M,2M HCl
2.Place 50mL of 2M HCl into a 250mL Erlenmeyer flask.(volume of flask is not critical)
3.Noting the time, drop a 3 cm piece of Mg into the 2M HCl.
4.Record time when all the ribbon was reacted.
5.Repeat 1-4 with different concentrations of HCl.
6.Repeat 1-5 using a 100mL of 0.5M, 1M,1.5M,2M HCl
7.Find rate law using this information


Data with 50 mL

Concentration Time(s)
0.5 4:22:87
1 0:57:22
1.5 0:19:97
2 0:14;47

Data with 100 mL


Concentration Time(s)
0.5 2:30.84
1 0:32.56
1.5 0:14:31
2 0:9:28

I see that it got faster, but why?Still doesn't ring any bells as to how the procedure/data could have anything to do with it.
 
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  • #4
Borek
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What was the Mg mass? At least approximate?

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  • #5
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Probably around the 0.5g area.
 
  • #6
Borek
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How much HCl was consumed then?

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methods
 
  • #7
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Mg + 2HCl -> MgCl2 + H2
0.5g/24.31g/mol * 2 mol/1 mol * 36.46 g/mol=1.5 g HCl
 
  • #8
Borek
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How much was left?

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  • #9
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Don't know the density of HCl, but assuming it was the same as water.
1.5g=1.5mL

in 50mL
amount left=50mL-1.5mL
=48.5
in 100mL
amount left=100mL-1.5mL
=98.5mL
 
  • #10
Borek
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No no, you got it wrong. Calculate initial moles in solution, calculate how much was consumed, calculate how much was left.
 
  • #11
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Amount of moles used=0.04 mol
For 50 mL
50g/36.46 g/mol=1.37mol-0.04 mol=1.33mol remaining

for 100mL
100g/36.46g/mol=2.74mol-0.04 mol=2.7mol remaining
 
  • #12
Borek
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So, what was the final concentration?
 
  • #13
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50mL
1.33mol/0.05L=26.6mol/L
100mL
2.7mol/0.1L=27mol/L

err, those numbers seem big, must of done a calculation wrong.
 
  • #14
Borek
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I am watching TV so I was not paying attention and missed your mistake. 50 g is mass of solution, not HCl. However, you are given molar concentration and volume...
 
  • #15
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so for 50mL, and 2M solution
2mol/L*(0.05L)=0.1mol
amount used =0.04mol
(.1-0.04mol)/0.05L=1.2 mol/L




and for 100mL and 2M solution
2mol/L*(0.1L)=0.2mol

(0.2mol-0.04mol)/0.1L)=1.6mol/L
 
  • #16
Borek
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So, was the concentration constant during experiment?

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methods
 
  • #17
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Nope, the 50mL one, had a greater change in concentration than the 100mL one, which produces greater error because rate law is only accurate for initial concentrations of reactants?
 
  • #18
Borek
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Not exactly. Rate of the reaction can be constant if the concentration doesn't change. If the concentration changes, rate changes as well (rate law works for any concentrations, but it needs exact actual concentrations). In this case concentration goes down and rate goes down. If you start with larger initial amount of acid change in concentration is smaller, so assuming rate was constant is closer to the reality.

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methods
 
  • #19
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I got a followup question concerning this:

By calculating the volume after we used the same total volume of solution(50mL and 100mL). However, hydrogen gas was lost, doesn't that lower the volume of the solution?
 
  • #20
Borek
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It is not that simple. Hydrogen was lost, but Mg went into the solution. Final volume of the solution depends on its density and is hard to predict. Could be smaller, could be larger, could be almost no change.

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methods
 

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