14.2 Rate Laws | General Chemistry

Chad provides a comprehensive lesson on Rate Laws and how to calculate a rate law from a table of kinetic data. The lesson begins with an introduction to what a rate law looks like including the rate constant, k, the reactant orders, and the overall order of reaction. A description of reactant order and the overall order of reaction is provided. The rate of reaction is not affected by the concentration of a zero order reactant. The rate of reaction is proportional to the concentration of a first order reactant. The rate of reaction is proportional to the square of the concentration of a second order reactant. Finally, the overall order of reaction is equal to the sum of the individual reactant orders.
Chad then shows how to calculate the rate law from a table of experimental data. He explains how to recognize which trials to compare to determine a particular reactant's order and then how to perform the calculation. Once all the individual reactant orders have been calculated, he then shows how to calculate the rate constant. Depending upon the overall order of reaction, the unit of the rate constant vary. Chad also shows how to find rate constant units and an easy way to remember the units for each order of reaction.
I've embedded this playlist as a course on my website with all the lessons organized by chapter in a collapsible menu and much of the content from the study guide included on the page. Check this lesson out at www.chadsprep.com/chads-gener...
If you want all my study guides, quizzes, final exam reviews, and practice exams, check out my General Chemistry Master Course (free trial available) at www.chadsprep.com/genchem-you...
00:00 Lesson Introduction
00:42 Rate Laws, Rate Constants, and Reaction Orders
03:21 Zero Order Reactants, 1st Order Reactants, 2nd Order Reactants
09:21 How to Calculate a Rate Law from a Table of Experimental Data
18:30 How to Calculate the Rate Constant
21:20 How to Find Rate Constant Units
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Glad the video is helping, Andre - Happy Studying!

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Glad the channel is helping you out so much - Happy Studying!

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[@ChadsPrep]

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Thank you and glad the channel is helping you - Happy Studying!

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Radical! I haven't heard 'tubular' since the 80's! You're welcome and Thank You!

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[@ChadsPrep]

Happy to hear the content is helping you! Happy studying :)

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Thank you for the video. How do you determine the initial rate?

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[@zacharypritchett2206]

Hey Chad, what if the concentration from trial one to trial two in NO decreased, but stayed constant in Cl2? would the exponent be a fraction?

[@ChadsPrep]

Hey Zachary - yes we would end up with a fractional exponent in that case. BUT! if the NO concentration halved and the rate halved also it would still be first order :)

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[@leobi1243]

I have a question here, so the rate we calculated is the rate of change in concentration of Cl2 or NO? As I understand it, the rate for NO is double that of Cl2 because of NO's coefficient, so what is the reaction rate representing here?

[@ChadsPrep]

Hi Leo! If you look back at the previous video in the playlist, I introduce rate expressions and explain that the rate of reaction can be expressed in terms of any reactant or product. The absolute value of the rate of reaction will be the same as the rate of change for a reactant or product with a coefficient of 1, half the absolute value of the rate of change for a reactant/product with a coefficient of 2, one-third of the absolute value of the rate of change for a reactant/product with a coefficient of 3, etc. This way, regardless of which reactant or product you consider, the overall rate of reaction will have the same value. Hope this helps!

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[@user-ju6kf1tt6q]

How do you know the trial to pick what concept do you use

[@ChadsPrep]

It is always easier to pick a pair of trials to compare where only the concentration of one reactant is changed if possible.

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[@krgod3644]

The rate we get in rate laws is inital rate?

[@peybak]

Yes, because if you let a reaction go, it might reach equilibrium. So they usually just measure the initial rate.

[@emily191]

I get everything but I’m super confused when it comes to scientific notation. Every time I try to divide it gives me a scientific notation number, not a regular whole number, then I don’t know how to put that decimal to equal a power. Can you please help?

[@ChadsPrep]

Hey Jayooni - is this on your calculator? Can you give me an example of what you are trying to convert?

[@emily191]

@@ChadsPrep yes it is-- it’s a little hard to send the table over. Is it fine if I email it to you?

[@emily191]

@@ChadsPrep but I’m trying to divide 7.0x10^-4 and 2.8x10^-3 Thanks so much for your response I appreciate it a lot!

[@josphellihsilak4588]

@@emily191 This is late, but maybe for someone else needing the same help. Remember an exponent in the denominator can be raised to the numerator by flipping the sign. So I always tell people, in the example you gave, divide the front numbers alone 7/2.8 = 2.5 then flip the exponent in the denominator which pushes it to the numerator (10^-4/10^-3) becomes (10^-4 × 10^3). Multiplying exponents means you add the exponents which now gives you 10^(-4+3) = 10^-1. Now combine it all..2.5 x 10^-1 or .25. Hope that's useful for whomever may need it.

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@@josphellihsilak4588 thank you for your response! I ended up asking my brother and he figured it out in less than a minute lol. You can imagine the look on my face haha

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[@ChadsPrep]

Happy Studying.

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How do I find those list of numbers you have? The numbers: 0.1 0.1 0.18 0.1 0.2 0.36 0.2 0.2 1.44

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From experiments. They’re experimental values.

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I don't know how he got the m-² s-¹. can someone explain pls.

[@ChadsPrep]

So we have (0.18M.s^-1)/[(0.1M^2)(0.1M)]..... just looking at the unit that gives us: (M.s^-1)/[(M^2)(M)].... anything that is on the top and bottom we can cross out.... (-M-.s^-1)/[(M^2)(-M-)] which just leaves us with s^-1 on the top and M^2 on the bottom which is M^-2 s^-1

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@@Jerlogan1191 Most welcome.

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Glad the video helped!