Also how would you find the average constant?

In summary: From the data it seems like line 1 and 3 have the same concentration of [B], but line 2 has a much higher concentration of [B]. This means that the rate for line 2 is much higher than for line 1 and 3.Based on this, you can conclude that the rate law for line 2 is k[B]^{2}[C]^{z}=108 mol/L*s.
  • #1
1,236
2
Hello

If we have a number of trials, and are given the concentration of 3 reactants is this how you would find the rate law? Also how would you find the average constant?

[tex] 1 : 0.001 \ \ 0.1 \ \ 0.2 \ \ 1.2 \times 10^{-6} [/tex]
[tex] 2 : 0.001 \ \ 0.4 \ \ 0.2 \ \ 0.48\times 10^{-5} [/tex]
[tex] 3 : 0.003 \ \ 0.1 \ \ 0.2 \ \ 108\times 10^{-7} [/tex]
[tex] 4 : 0.003 \ \ 0.4 \ \ 0.4 \ \ 86.4 \times 10^{-6} [/tex]


The 4 numbers are the # of trials, and the three numbers beside them are the concentration of the three reactants. The last number is the rate in [tex] mol/L*s [/tex]


So using the rate law [tex] k[A]^{m}^{n}[C]^{z} [/tex] I know that in trial 1 the rate goes down by a factor of [tex] 1/4 [/tex]. A and C are the same but B changes. Does that mean B's reaction order is 1/4?
Any help in trying to find the average value of the rate constant and the rate law would be appreciated.

Thanks :smile:
 
Last edited:
Physics news on Phys.org
  • #2
would i have to compare rates from three experiments?
 
  • #3
You were close, just one manipulation, and one final step to get the answer.

When trying to find the rate law pick the two trials where the concentrations of two of the substances stay the same and only one changes then solve for the rate law by taking the ln of the solution.

[tex]\frac{rate2}{rate1}=\frac{k[A]^x ^y [C]^z}{k[A]^x ^y [C]^z}[/tex]

Say we take runs 1 and 2 where the 1st (A) and 3rd (C) reactant concentrations stay the same so we can just cancel out the A's, C's and k's: the reaction constant is the same at constant temperature which we are assuming with the experimental data:

[tex]\frac{1.2\times{10^-6}}{0.48\times{10^-5}}=\frac{(0.1)^y}{(0.4)^y}[/tex]

One of the rules of exponents allows us to simplify the right side:

[tex]0.25=(0.25)^y[/tex]

Finally take the ln of both sides to bring down the unknown y and solve for the rate order value for the "B" concentration.

[tex]ln(0.25)=yln(0.25)[/tex]
[tex]y = 1[/tex]
 
  • #4
I believe that because from Trial 1 to Trial 2, goes down 1/4th while A and C remain constant, and the rate also changes by a factor of 1/4th, this means that is zeroth order in relation to the rate constant, so I believe that you can strike it from your rate law; meaning that the rate law will now only depend on A and C.

If I am on the right track, let me advise you to use that knowledge to find the exponents for A and C. Hint: for [C] look at Trials 3 and 4 and apply your knowledge of the rate laws.

Correct me if I'm wrong, it's been a while since I did rate of reactions, so I may be a little fuzzy.

-Art

EDIT: Hrmph never mind then, I must be incorrect. bross, good solution
 
  • #5
so would the reaction equation be [tex] {k[A]^0 ^1 [C]^0} [/tex] making it an overall order of 1? How do you find k? k would be 1 because it cancels out and has no effect on the reaction?
 
Last edited:
  • #6
To find the rate order of A and C you have to repeat the same process as you did for B using the rates where B and C stay the same, to find A, and A and B stay the same, to find C.

Then to find the rate constant k, just take anyone of the three trials and sub the data into the equation you have found as the only unknown is k.
 
  • #7
Artermis said:
Correct me if I'm wrong, it's been a while since I did rate of reactions, so I may be a little fuzzy.
You're right that you're wrong ! :wink:

courtigrad : you have 4 equations in 4 unknowns, so you should be able to solve them.

There's a simple way, by inspection, that will work. You just have to choose the right lines to compare. From bross' work, you know that the rate is linear (first order) in ; meaning that when you double or triple , the rate will correspondingly double or triple.

Next compare lines 1 and 3. What concentrations change between these trials ? By what factor ? And how does this change the rate ? So, what can you tell from this ?
 

1. What is the definition of an average constant?

An average constant is a numerical value that represents the central tendency of a set of data. It is also known as the arithmetic mean and is calculated by adding all the values in the data set and dividing by the total number of values.

2. How do you calculate the average constant?

To calculate the average constant, you need to add up all the values in the data set and then divide the sum by the total number of values. This will give you the arithmetic mean or average constant.

3. Can the average constant be negative?

Yes, the average constant can be negative if the data set contains negative values. It is important to consider the context of the data to determine whether a negative average constant is meaningful.

4. What is an example of finding the average constant?

For example, if you have a data set of test scores: 75, 80, 85, 90, and 95. To find the average constant, you would add all the scores (75+80+85+90+95) and then divide by the total number of scores (5). The average constant in this case would be 85.

5. How is the average constant used in data analysis?

The average constant is commonly used in data analysis to summarize and compare data sets. It provides a single numerical value that represents the central tendency of the data, making it useful for making comparisons and identifying trends.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
483
  • Introductory Physics Homework Help
Replies
32
Views
941
  • Introductory Physics Homework Help
Replies
4
Views
883
  • Introductory Physics Homework Help
Replies
2
Views
3K
Replies
8
Views
726
  • Introductory Physics Homework Help
Replies
2
Views
612
  • Introductory Physics Homework Help
Replies
11
Views
723
  • Introductory Physics Homework Help
Replies
3
Views
958
  • Introductory Physics Homework Help
Replies
4
Views
5K
  • Introductory Physics Homework Help
Replies
3
Views
3K
Back
Top