Lab Report Determining the Activation Energy of a Reaction

[TwoSaints]

I have to do a Lab Report on the topic of kinetics involving the Activation Energy of a reaction by measuring the rate of a reaction at different temperatures. What I cannot get my head around is how can I find calculate the activation energy (presumably using the Arrhenius equation) without the rate constant and the Arrhenius constant of the reaction? I am quite "confuzzled" (yes, that is the extend of my confusion) and any suggestions whatsoever would be greatly appreciated.

[nuka]

it depends on your practical... which lab did you do? the activation energy of what?

because i could help if your practical was similar to what we did...

our practical's aim was:

The aim of the experiment was to calculate the activation energy in the reduction of peroxodisulphate (VI) ions by iodine ions. This was done by measuring the time taken for the reaction to occur, tested through when a colour change was displayed.

so if it's something like that, then i can help

[Keel]

To calculate the activation energy you have to use Arrhenius' equation, k=Ae^(-Ea/RT), where A is collision frequency (which you can ignore), Ea is activation energy, R is the gas constant (8.314)and T is temperature in Kelvin.

You can rearrange this equation into something similar to y = mx +c

ln k= ln A- Ea/RT

ln k= (-Ea/R)(1/T)+ln A

Let y = ln k, m= (-Ea/R), x = (1/T) and c=lnA

Plot a graph of ln k on the y axis and (1/T) on the x axis

The activation energy Ea therefore = -(Gradient of graph*8.314)

[TwoSaints]

To calculate the activation energy you have to use Arrhenius' equation, k=Ae^(-Ea/RT), where A is collision frequency (which you can ignore), Ea is activation energy, R is the gas constant (8.314)and T is temperature in Kelvin.

You can rearrange this equation into something similar to y = mx +c

ln k= ln A- Ea/RT

ln k= (-Ea/R)(1/T)+ln A

Let y = ln k, m= (-Ea/R), x = (1/T) and c=lnA

Plot a graph of ln k on the y axis and (1/T) on the x axis

The activation energy Ea therefore = -(Gradient of graph*8.314)

I can't do that because I don't have "k" so I cant calculate ln(k). And to the previous poster that is actually the exact same lab, but I figured it out how to do it; I just had to replace ln(k) with ln(rate) because they're both proportional to each other. But thanks for the answers anyway

[di27]

Okay so I have the same problem as well. I don't get what k is! Like basically my experiment is the Determination of the Activation Energy for the Reaction between Bromide and Bromate (V) Ions. I have all the temperatures and time taken for all of them with averages and calculations of ln(1/t) but I don't get what 'k' is! HELP?

TABLE of results attached!

Please help!

[Drake Glau]

To calculate the activation energy you have to use Arrhenius' equation, k=Ae^(-Ea/RT), where A is collision frequency (which you can ignore), Ea is activation energy, R is the gas constant (8.314)and T is temperature in Kelvin.

You can rearrange this equation into something similar to y = mx +c

ln k= ln A- Ea/RT

ln k= (-Ea/R)(1/T)+ln A

Let y = ln k, m= (-Ea/R), x = (1/T) and c=lnA

Plot a graph of ln k on the y axis and (1/T) on the x axis

The activation energy Ea therefore = -(Gradient of graph*8.314)

I can't do that because I don't have "k" so I cant calculate ln(k). And to the previous poster that is actually the exact same lab, but I figured it out how to do it; I just had to replace ln(k) with ln(rate) because they're both proportional to each other. But thanks for the answers anyway

rate=k[D]^mn[C]^o

k=rate/([D]^mn[C]^o)

I really didn't read what data you had, it was just the way of finding k that I thought of

Edited June 25, 2011 by Drake Glau