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Topic: Learning the Arrhenius Equation  (Read 8114 times)

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Offline yoyoils

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Learning the Arrhenius Equation
« on: August 20, 2011, 12:51:54 AM »
my book says:

k = 1014e -58.4/1.53 = 3.03 x 10-3 s-1

so far im trying 10 to the 14th power times (58.4 divided by 1.53)
along with -58.4 inplace of 58.4
but the answer is 3.03 x 10-3 and i cant seem to find out how to put it in the calc or if im even doing it right
I'm pretty much asking how to type this in a calculator. Like an example pretty pleeease :3

Offline Fluoroantimonicacid

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Re: Learning the Arrhenius Equation
« Reply #1 on: August 20, 2011, 01:23:12 AM »
Arrhenius equation is
ln A = ln k - Ea/RT
A is frequency factor, Ea is activation energy by J/mol, R is universal gas constant 8,314 J/K mol,and T is temperature by Kelvin.
Try solving it this way.

Offline Fluoroantimonicacid

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Re: Learning the Arrhenius Equation
« Reply #2 on: August 20, 2011, 02:38:42 AM »
Hey, I was trying to modify the other reply, I forgot to add k is speed constant. You can delete that.

Offline yoyoils

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Re: Learning the Arrhenius Equation
« Reply #3 on: August 20, 2011, 05:11:27 AM »
I understand that so far, I'm just having trouble figuring out how to put it in the calculator bcuz I never learned anything past basic algebra.
and i really have to learn it this way


my book says:
A = 1014
Ea = 58.4 kcal mol-1
R = 1.986 cal deg-1 mol-1
and then says calculate k at 500degC for this reaction.

Offline Fluoroantimonicacid

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Re: Learning the Arrhenius Equation
« Reply #4 on: August 20, 2011, 05:50:16 AM »
R is not cal/oC mol. It is cal/K mol. Celsius is an empirical measure unit just based on the water's boiling point.

k = 1014 * e(-58400 cal/mol /(1.986 cal/K mol * 773 K))
= 3.013 * 10-3

You can put it that way.

Offline yoyoils

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Re: Learning the Arrhenius Equation
« Reply #5 on: August 20, 2011, 06:06:12 AM »
first of all thank you so much helping me so much.

k = 1014 * e(-58400 cal/mol /(1.986 cal/K mol * 773 K))

it's the "e" that i'm having a problem with. Once I calculate what's within the (())'s, I don't know what to do with the -38.04119131

ive tried pressing [LN] also labeled ex button on the calculator with that but it goes ERROR so that's where I'm not sure what to do.

Offline yoyoils

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Re: Learning the Arrhenius Equation
« Reply #6 on: August 20, 2011, 06:10:39 AM »
like i'll get to:

k = 1014 * e(-38.04119131)

and think hmm... ok clicks LN/ex with "-38.04119131" up in the calc but it says error lol

Offline Fluoroantimonicacid

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Re: Learning the Arrhenius Equation
« Reply #7 on: August 20, 2011, 06:52:37 AM »
e is 2,7183. so you can try 2,7183 ^ -38,04 = 3.01*10-17

Offline Borek

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Re: Learning the Arrhenius Equation
« Reply #8 on: August 20, 2011, 10:06:42 AM »
I never learned anything past basic algebra.

The sooner you will realize you will be not able to move ahead without learning more math, the better.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline yoyoils

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Re: Learning the Arrhenius Equation
« Reply #9 on: August 20, 2011, 02:30:33 PM »
I just need to know how to literally use the calculator to get the answers to Arrhenius equation.

question:
k = 1014 * e(-58400 cal/mol /(1.986 cal/K mol * 773 K))

like for example:

[1][.][9][8][6] times [7][7][3] = 1535.178
[STO][1]

[5][8][4][0][0][+-] divided by [RCL][1] = -38.04119131

would someone please tell me what i'm supposed to type in next to answer this with my calculator

the answer is supposedly = 3.01*10-17 but i need to know how to get that

Offline yoyoils

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Re: Learning the Arrhenius Equation
« Reply #10 on: August 20, 2011, 02:37:53 PM »
I never learned anything past basic algebra.
The sooner you will realize you will be not able to move ahead without learning more math, the better.

I prefer gaining math through learning chemistry. I'd strongly rather not go studying math in order to just understand a little more chemistry. Chemistry pretty much is math, it's just more specified towards "science", experimentation, and matter. so i really really want to continue gaining grounds through only my chemistry book. "Organic Chemistry Structure and Function FORTH EDITION" When it comes to learning how others do, I'm hindered by extreme attention to detail.. i'm sorry. but i do appreciate the guidance.

Offline yoyoils

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Re: Learning the Arrhenius Equation
« Reply #11 on: August 20, 2011, 11:58:57 PM »
I'm using this calculator now bcuz I found out mines too old http://web2.0calc.com/

I put in: 10^14 * e(-38.04119131) and it = 3.01245472874 which is 3.012 x 10^-3

but I'm still worried if that's not accurate enough because my book states:

k = 10^14 e -58.4/1.53 = 3.03 x 10^-3 s-1
« Last Edit: August 21, 2011, 12:26:09 AM by yoyoils »

Offline Fluoroantimonicacid

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Re: Learning the Arrhenius Equation
« Reply #12 on: August 21, 2011, 07:01:36 AM »
There isn't anything to be added in it. Perhaps they used more round calculation.

Offline Jorriss

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Re: Learning the Arrhenius Equation
« Reply #13 on: August 22, 2011, 12:10:58 AM »
I never learned anything past basic algebra.
The sooner you will realize you will be not able to move ahead without learning more math, the better.

I prefer gaining math through learning chemistry. I'd strongly rather not go studying math in order to just understand a little more chemistry. Chemistry pretty much is math, it's just more specified towards "science", experimentation, and matter. so i really really want to continue gaining grounds through only my chemistry book. "Organic Chemistry Structure and Function FORTH EDITION" When it comes to learning how others do, I'm hindered by extreme attention to detail.. i'm sorry. but i do appreciate the guidance.
That won't work for long.

No chemistry book will teach you calculus. Basically every class past general chemistry will just assume you know calculus.

You need to learn calculus, linear algebra and some basic differential equations. The rest you can learn through your natural chemistry courses but you really do need those.

Seriously, take calculus and linear algebra. It is extremely necessary and worth it.

If you learn math you won't learn a little more chemistry, you'll learn A LOT more. In fact, you won't be capable of learning most chemistry without math - you'll completely severe yourself from physical chemistry and large aspects of inorganic, analytical and organic chemistry.

For example, basically ANY problem that involves rates of change is a differential equation, which can be approximated with some linear algebra (which is just a fraction of its importance).




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