Sciencegeeknm Posted January 29, 2019 Share Posted January 29, 2019 In a lecture I recently watched there was an example of using the Eyring equation. I have attached a picture showing the values plugged into the equation. The activation energy is 27.2 and the temperature used is 298 K. The answer given in the lecture is 7.0 x 10-8 for the reaction rate. I have tried to work this out on a calculator but can't get the answer given? Maybe I am multiplying the wrong values together. I found an online calculator tool which came to the same answer(2nd screenshot attached) Any help would be appreciated. Link to comment Share on other sites More sharing options...

hypervalent_iodine Posted January 29, 2019 Share Posted January 29, 2019 I did the calculation using your values and get the correct answer, so I have to assume you aren't entering it correctly into your calculator. Could you possibly post a photo of how you are entering it? Link to comment Share on other sites More sharing options...

Sciencegeeknm Posted January 29, 2019 Author Share Posted January 29, 2019 Hi hypervalant_iodine, i am entering 1.38 (EE button) -23 x 298 and then divide by planks constant. But I get a really large number. Can you send a breakdown of each calculation step you are doing to arrive at the answer? Many thanks. Link to comment Share on other sites More sharing options...

hypervalent_iodine Posted January 30, 2019 Share Posted January 30, 2019 What about the rest of the calculation? Link to comment Share on other sites More sharing options...

Sciencegeeknm Posted January 30, 2019 Author Share Posted January 30, 2019 I am entering 1.38 x (EE) -23 x 298= 4.1124e-21 divide by 6.63 x 10-34= 6,202,714,932,126.697 Then -27.2 divide by 0.001987 x 298k ?? Then do you multiply one side of the equation with the other? My answers are way off. Link to comment Share on other sites More sharing options...

Sensei Posted January 30, 2019 Share Posted January 30, 2019 (edited) 23 minutes ago, Sciencegeeknm said: I am entering 1.38 x (EE) -23 x 298= 4.1124e-21 divide by 6.63 x 10-34= 6,202,714,932,126.697 Then -27.2 divide by 0.001987 x 298k ?? Ok, so far. 23 minutes ago, Sciencegeeknm said: Then do you multiply one side of the equation with the other? Not exactly. It's exponential function. https://en.wikipedia.org/wiki/Exponential_function with e constant as a base. https://en.wikipedia.org/wiki/Natural_logarithm Edited January 30, 2019 by Sensei Link to comment Share on other sites More sharing options...

Sciencegeeknm Posted February 3, 2019 Author Share Posted February 3, 2019 Thank you. I'm not sure of what steps to use on a calculator. Can you provide more detail? This is not a homework question just a lecture I watched on YouTube. Thanks. Link to comment Share on other sites More sharing options...

hypervalent_iodine Posted February 3, 2019 Share Posted February 3, 2019 Your calculator should have an ln / e^{x} function. That's what you're looking for. Link to comment Share on other sites More sharing options...

Sciencegeeknm Posted February 3, 2019 Author Share Posted February 3, 2019 Yes it does. I am entering 1.38 ^-23 x 298 = 4.1124 ^-21 divided by 6.63 ^-34 = 6.2027^12 then pressed ex button and entered -27.2 divided by 1.987^-3 x 298 = 2.30x10-7 ?? The answer should be in per seconds. The joules cancel out so is it a kelvin to s-1 conversion? Thanks for your help so far. Link to comment Share on other sites More sharing options...

hypervalent_iodine Posted February 3, 2019 Share Posted February 3, 2019 2 hours ago, Sciencegeeknm said: 1.38 ^-23 x 298 = 4.1124 ^-21 divided by 6.63 ^-34 = 6.2027^12 then pressed ex button and entered -27.2 divided by 1.987^-3 x 298 Are you confusing the rate constant with the Boltzmann constant? The LHS of the equation is k, which is the reaction rate constant (i.e. the value you're trying to calculate). The value you enter there is the Boltzmann constant, which belongs on the RHS of the equation (it is symbolised by k_{B}). I would also suggest using brackets just to be safe you aren't getting caught by order of operations issues. 2 hours ago, Sciencegeeknm said: The answer should be in per seconds. The joules cancel out so is it a kelvin to s-1 conversion? I get s^{-1}, so your math is out again. The Boltzmann constant is in J/K, temperature is in K, and Plank constant is J.s, so it looks like this: ((J/K)*K)/(J*s). The J and K units cancel, leaving 1/s. Link to comment Share on other sites More sharing options...

Sciencegeeknm Posted February 5, 2019 Author Share Posted February 5, 2019 Do you have to use the whole equation to reach the reaction rate of 7.0 x10-8? What do You mean when you say the Boltzman constant belongs on the right hand side of the equation? Thanks. Link to comment Share on other sites More sharing options...

hypervalent_iodine Posted February 5, 2019 Share Posted February 5, 2019 48 minutes ago, Sciencegeeknm said: Do you have to use the whole equation to reach the reaction rate of 7.0 x10-8? What do You mean when you say the Boltzman constant belongs on the right hand side of the equation? Thanks. I don’t understand your first question. Or your second one, really. The Boltzmann constant is not k, it is kB (the B should be subscripted). In your calculation, you put the Boltzmann constant where k goes, which is incorrect. The value for k is the reaction rate constant, and this is what you are trying to solve. IOW, you treat k (the term on the LHS) as an unknown and solve the equation. This is what you have written in the picture you uploaded in the OP. From what I can tell, the only thing you were doing wrong there there was not using e correctly. Link to comment Share on other sites More sharing options...

Sciencegeeknm Posted February 5, 2019 Author Share Posted February 5, 2019 I know that k is the reaction rate to be solved and kb is the Boltzman constant. The screenshot attached is from the actual lecture. I really have no idea how to use the values given to reach the answer Can you tell me how you reached the correct answer? Sorry to be a pain Thanks Link to comment Share on other sites More sharing options...

hypervalent_iodine Posted February 5, 2019 Share Posted February 5, 2019 Never mind, I misread the post! I got the right answer by plugging your numbers into the equation. Are you using brackets around the exponential term? And are you then multiplying your values together? My suspicion is that you aren’t accounting for order of operations with the terms in your fractions. I’ve attached a quick scribble for how I got to the correct answer: I changed the units for delta G and R, because I don’t like kcal, but the numbers should still work regardless. Link to comment Share on other sites More sharing options...

Sciencegeeknm Posted February 6, 2019 Author Share Posted February 6, 2019 Thank you for sending the photo showing your workings. It really helped and now I can reach the right answer using your values and the ones quoted in the lecture. My mistake when doing the RHS of the equation was to divide and multiply the values all in one go. You first have to multiply the gas constant by the temperature and then divide top by bottom. 2 separate steps. You then get 45.94. Then ex 45.94 to get 1.12 ^-20 Then 6.203^12 x 1.12^-20= 0.0000000694736 After some rounding up 7.0^-8 Thanks again for your help. I will now have a better understanding when approaching other equations that come up in the lectures. Link to comment Share on other sites More sharing options...

hypervalent_iodine Posted February 6, 2019 Share Posted February 6, 2019 30 minutes ago, Sciencegeeknm said: Thank you for sending the photo showing your workings. It really helped and now I can reach the right answer using your values and the ones quoted in the lecture. My mistake when doing the RHS of the equation was to divide and multiply the values all in one go. You first have to multiply the gas constant by the temperature and then divide top by bottom. 2 separate steps. You then get 45.94. Then ex 45.94 to get 1.12 ^-20 Then 6.203^12 x 1.12^-20= 0.0000000694736 After some rounding up 7.0^-8 Thanks again for your help. I will now have a better understanding when approaching other equations that come up in the lectures. Yes, this is what I was talking about with the order of operations issues and was pretty much what I thought was going on. When you do it in the calculator, you should put your various terms in brackets as I did in my work-through. Operations in brackets are performed before other multiplication steps. Essentially what you were doing was what I have written in the first example (sorry about all the hand-written notes, I am just useless with LaTex): When you put the equation into your calculator, make it a habit to use brackets around everything (as I did in my calculation in my last post). I would be cautious about doing each operation separately and then combining the values, only because you get rounding errors this way and sometimes that can mean you fall outside the range of accepted answers in an exam. Link to comment Share on other sites More sharing options...

Sciencegeeknm Posted February 7, 2019 Author Share Posted February 7, 2019 Thanks. Yes using brackets does work better. Once you find the rate it can be used to find the half life of the reactant. I was just wondering on the screenshot attached where the value of 0.693 comes from? Something to do with log2? If I press log and enter 5 a value near to 0.693 is displayed ? Link to comment Share on other sites More sharing options...

hypervalent_iodine Posted February 7, 2019 Share Posted February 7, 2019 It's the natural log (ln) of 2. You can see it derived here: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Nuclear_Chemistry/Nuclear_Kinetics/Half-Lives_and_Radioactive_Decay_Kinetics. Link to comment Share on other sites More sharing options...

Sciencegeeknm Posted February 7, 2019 Author Share Posted February 7, 2019 Thanks! Now I understand. Link to comment Share on other sites More sharing options...

Sciencegeeknm Posted February 11, 2019 Author Share Posted February 11, 2019 Hi, I have another question. From the attachment you can see that at 298 k the delta G is 27.2 and at 353 it is 26.4 There is a difference of 0.8 Kcal. The lecturer says that 0.8 then gives you 0.016 and he has 16 Kcal as the delta S value. Why does 0.8 then give you 0.016? I don’t understand. Link to comment Share on other sites More sharing options...

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