Arrhenius equations; A2 Chemistry coursework

In summary: But when I try to use the equation: Ea = -ln(k)I get an error message saying "y = mx + c is undefined when k is a negative number". Am I doing something wrong?In summary, the person is trying to use the 'arrhenius equation' to ascertain the activation energy/rate of reaction for a certain acid/particular molarity at a range of different temperatures. They have the range of data (taken from 'inverted titrations') for 5 different temperatures- aka the volume of gas produced every 10 seconds over a two minute period
  • #1
Synkronised
4
0
Hello everyone,
I'm in my last year of A-level chemistry here in the UK, and am currently writing up my chemistry coursework. Its a project I was given on the kinetics of the reactions between magnesium ribbon and several different acids (I wanted to explore some much more exciting fields, but sadly couldn't!)

Anyway, I need some help on something...

i'm trying to use the 'arrhenius equation' to ascertain the activation energy/rate of reaction for a certain acid/particular molarity at a range of different temperatures, as an extension to the original project. I have the range of data (taken from 'inverted titrations' if you will!) for 5 different temperatures- aka the volume of gas produced every 10 seconds over a two minute period. (for example 8.8cm^2 after 10 seconds, 20.2cm^2 after 20 seconds)

my problem is I was visiting universities in the last lesson before the holidays started, and we were shown how to use k=Ae (-EA / RT), and how this produces a graph of results... involving logarithms and... things. ahem. My maths isn't outstanding sadly, and I've hit a huge dead end with this. I also can't ask any staff as it's due in the day we return- about a weeks time.

So, here's my desparate plea. If anyone knows what I'm talking about and feels particulary kind, could you please either show me what to do, or point me in the right direction? If this is in the wrong forum or section of the site i apologise, I haven't been active here very long at all!

Thankyou in advance to any help :smile:

(p.s. does anyone know any websites that tell you the percentage error of simple scientific equipment such as burettes/ conical flasks/ measuring cylinders etc? or am i wishing on a star? heh)

<edited for a spelling mistake, there's probably more, its late!>
 
Chemistry news on Phys.org
  • #2
Synkronised said:
Hello everyone,
I'm in my last year of A-level chemistry here in the UK, and am currently writing up my chemistry coursework. Its a project I was given on the kinetics of the reactions between magnesium ribbon and several different acids (I wanted to explore some much more exciting fields, but sadly couldn't!)

Anyway, I need some help on something...

i'm trying to use the 'arrhenius equation' to ascertain the activation energy/rate of reaction for a certain acid/particular molarity at a range of different temperatures, as an extension to the original project. I have the range of data (taken from 'inverted titrations' if you will!) for 5 different temperatures- aka the volume of gas produced every 10 seconds over a two minute period. (for example 8.8cm^2 after 10 seconds, 20.2cm^2 after 20 seconds)

my problem is I was visiting universities in the last lesson before the holidays started, and we were shown how to use k=Ae (-EA / RT), and how this produces a graph of results... involving logarithms and... things. ahem. My maths isn't outstanding sadly, and I've hit a huge dead end with this. I also can't ask any staff as it's due in the day we return- about a weeks time.

So, here's my desparate plea. If anyone knows what I'm talking about and feels particulary kind, could you please either show me what to do, or point me in the right direction? If this is in the wrong forum or section of the site i apologise, I haven't been active here very long at all!
Fear not; you've come to the right place. If this were math homework, I would not be doing the following for you :

[tex]k = Ae^{-E_a/RT} [/tex]

Take natural logarithms on both sides of the above equation. Remember the following properties of logarithms :

(i) ln(AB) = lnA + lnB
(ii)[itex]ln(e^x) = x[/itex]

Applying these above gives you :
[tex]ln(k) = ln(A) - \frac {E_a}{RT} [/tex]

This has the form :

[tex]y = mx + c [/tex] where y = ln(k), c = ln(A), x = 1/T, m = -Ea/R

In short, if you plot ln(k) vs. 1/T, the slope (m) will tell you the value of Ea.

(p.s. does anyone know any websites that tell you the percentage error of simple scientific equipment such as burettes/ conical flasks/ measuring cylinders etc? or am i wishing on a star? heh)
You should be able to find out the accuracy in your glassware by calling the manufacturer or looking in their website.
 
  • #3
heys, thanks for trying to help but I'm still rather confused! (call me stupid, I know I am)
I'm still endevouring to find the value of Ea by plotting ln(k) vs. 1/T as you so helpfully suggested, but I am having trouble with actually finding values to plot!
To explain further: 2.0mol dm^-3 sulphuric acid was reacted with magnesium at 30, 40, 50, 60 and 70°C. I drew a graph with the reactions plotted on it, and drew tangents to the curve at t=0 to achieve 5 initial rates. (0.98, 1.52, 1.78, 1.99 and 2.25cm^3 s^-1 respectively). Are these the figures I use to try to find the "ln rate?"
To achieve the ln rate, a formulae in a textbook quoted (simply) "ln rate = constant - Ea/R x 1/t", and i still don't understand sadly.Obviously I'm not asking for you or anyone else to do this work for me, just help me understand in laymans terms what to do if possible.

For the 1/T axis, I take it i convert the temperature into degrees kelvin and divide 1 by it...

Again, any help would be appreciated more then you realize (and when i said my maths was bad, i meant its GCSE standard. I stupidly didn't take it at A-level!)

Thankyou in advance!
 
  • #4
Synkronised said:
heys, thanks for trying to help but I'm still rather confused! (call me stupid, I know I am)
I'm still endevouring to find the value of Ea by plotting ln(k) vs. 1/T as you so helpfully suggested, but I am having trouble with actually finding values to plot!
To explain further: 2.0mol dm^-3 sulphuric acid was reacted with magnesium at 30, 40, 50, 60 and 70°C. I drew a graph with the reactions plotted on it, and drew tangents to the curve at t=0 to achieve 5 initial rates. (0.98, 1.52, 1.78, 1.99 and 2.25cm^3 s^-1 respectively). Are these the figures I use to try to find the "ln rate?"
Yes, they are - if you want to use ln(rate) instead of ln(k). Either way will work; the only difference will be in the c-intercept of the line. The slope of the stright line will be unaffected.

Alternatively (and you can do this for brownie points, if you feel like it) you can determine the rate constant (k) from the rate equation and the measured initial rates and concentrations. In this case, the reaction is a first order reaction (I would be surprised if not), so the rate equation would be :

[tex]\frac{d[A]}{dt}(t=0) = -k[A_0] [/tex] where the LHS = initial rate (slope of curve at t=0), k is the rate constant and Ao is the initial concentration of sulphuric acid. So, from each slope (which will be a negative number as it points downwards), you can determine the corresponding rate constant, k.

To achieve the ln rate, a formulae in a textbook quoted (simply) "ln rate = constant - Ea/R x 1/t", and i still don't understand sadly.Obviously I'm not asking for you or anyone else to do this work for me, just help me understand in laymans terms what to do if possible.
I derived nearly this very same formula for you from the Arrhenius Equation, in my previous post. Go over it again, and in the place of k, simply substitute Ro/Ao, where Ro is the initial rate (the term on the LHS of the above rate equation) . It is simply a matter of taking logarithms and applying the rules for logs.

For the 1/T axis, I take it i convert the temperature into degrees kelvin and divide 1 by it...
Correct.
 

Related to Arrhenius equations; A2 Chemistry coursework

1. What is the Arrhenius equation?

The Arrhenius equation is a mathematical formula that relates the rate of a chemical reaction to its activation energy, temperature, and frequency factor. It was proposed by Swedish scientist Svante Arrhenius in 1889.

2. How is the Arrhenius equation used in A2 Chemistry coursework?

In A2 Chemistry coursework, the Arrhenius equation is used to calculate the activation energy of a reaction by plotting the natural logarithm of the reaction rate against the reciprocal of temperature. This allows students to determine the effect of temperature on the rate of a reaction and make predictions about the behavior of the reaction at different temperatures.

3. What is the significance of the activation energy in the Arrhenius equation?

The activation energy in the Arrhenius equation represents the minimum amount of energy required for a chemical reaction to occur. It is a measure of the difficulty of breaking the bonds between reactant molecules and is indicative of the speed at which a reaction will occur.

4. How does temperature affect the rate of a reaction according to the Arrhenius equation?

The Arrhenius equation states that as the temperature increases, the rate of a reaction also increases. This is because higher temperatures provide more energy to the reactant molecules, allowing them to overcome the activation energy barrier and react more quickly.

5. Are there any limitations to the Arrhenius equation?

While the Arrhenius equation is a useful tool for predicting the effect of temperature on reaction rates, it does have limitations. It assumes that the reaction is a simple, one-step process and does not take into account factors such as catalysts or changes in the reaction mechanism. Additionally, the equation may not accurately predict the behavior of reactions at extremely high or low temperatures.

Similar threads

Replies
1
Views
1K
Replies
2
Views
6K
Replies
2
Views
3K
  • New Member Introductions
Replies
1
Views
156
  • Biology and Chemistry Homework Help
Replies
2
Views
3K
  • Biology and Chemistry Homework Help
Replies
1
Views
2K
Replies
16
Views
3K
Replies
1
Views
2K
Back
Top