Excellent video! One simple question, you divide 0.1908 by 2 because it is second Order right? if third order we need to divide m by 3?
@saintvendetta20222 ай бұрын
Thanks so much! You did a fantastic job explaining this and making it easy. :)
@EricZuckerman12 ай бұрын
My pleasure. Hope you crush your studies of chemical kinetics!! Best of luck :)
@noblevenom28589 ай бұрын
ms gougle
@ahasanhabibrafy80929 ай бұрын
Thats a freaky good video
@Onlysahar9 ай бұрын
Thank you so muchhhh professor 🫡🫡🫡🫡🫡🫡
@DaivyDjimbi9 ай бұрын
Hello 2M is the initial quantity of products right?
@EricZuckerman19 ай бұрын
In the example, the reaction is 2M --> P. That is, two moles of the substance M react to form a single mole of substance P. Initially, there is no substance P in the container.
@chessasheppard57689 ай бұрын
I've watched about 10+ videos and finally understand it now!
@EricZuckerman19 ай бұрын
Glad that your persistence has paid off and that my video played a role in your success!!
@tranquoclan504311 ай бұрын
What if P data column vs time instead of M data column vs time? Thanks alot!
@sugaryeuphoria248111 ай бұрын
In the assignment... A, B and C is clear to me but what bout D and E. In D and E potential energy of products is lower than that of reactants.
@EricZuckerman111 ай бұрын
The rate of the reaction is related to two variables: T and Ea. It does not matter if the reaction is exothermic or endothermic. Since the reactions are at the same T, that means that the larger the Ea, the slower the reaction.
@sugaryeuphoria248111 ай бұрын
@@EricZuckerman1thank you sm proff...
@user-ft4fs3qm8r11 ай бұрын
Thank you!
@harshjaiswal7821 Жыл бұрын
Concept booster
@zarrkez Жыл бұрын
amazing
@wallomaie1752 Жыл бұрын
Soooooo…the lighter the lid on a pot on a flame, the quicker the water will boil.
@EricZuckerman1 Жыл бұрын
The answer to this is "yes and no". Assuming that the same amount of energy flows in per unit time, then we can discuss how fast the boiling will occur. The real takeaway is the temperature of the water when it boils will be different. So, it's not a race to boiling at a specific temperature, but rather the temperature increases until the criteria for boiling is met. With a lower external pressure, the criteria is met at a lower T.
@stephanyecastelblanco7316 Жыл бұрын
Thanks you!
@rakeshpradeep6575 Жыл бұрын
that was amazing. Thank you
@haengbok7901 Жыл бұрын
Thank you so much sir for this video!!
@nzuzongidi6808 Жыл бұрын
Grade 11
@you2tooyou2too2 жыл бұрын
It seems that the liquid is also dissolving into the gas.
@EricZuckerman12 жыл бұрын
While it may seem so, that would not be technically correct. The solvent is the substance in a mixture that exists in the overwhelmingly larger mole ratio. For dissolved gasses, the % liquid is routinely greater than 99% of the mixture. Also, the liquid in this case stays in the condensed phase and is mainly acting like a pure liquid (with slightly altered physical properties). Meanwhile, the properties of the gas are greatly altered to the extent that it is no longer in the gaseous state at all and its properties are more in line with the solvent.
@you2tooyou2too2 жыл бұрын
@@EricZuckerman1 I'm not sure you followed my thinking; that two substances in contact have a symmetrical process of dissolving some of their mass into the other: some liquid into the gas, and some gas into the liquid. This might be more obvious in different situations, like soldering or especially brazing where the boundary transition between the two substances is more gradual, including a brane of 50/50 solution, allowing the gradual alloying of structural characteristics to provide a stronger bond between the two masses after the transition solution freezes.
@EricZuckerman12 жыл бұрын
@@you2tooyou2too As the video is specific to Henry's Law, the concept is centered around how the gas pressure is related to the dissolved gas within the solvent. The video is also an introductory level discussion, so I definitely don't go to more complicated models. To deal with two partially miscible liquids, a different analysis would need to be done. Certainly, there is quite a difference when the chemical potential of the substances are closer in value and have more similar dependencies on T and P. I must admit, I am not well versed in that area. Thanks for the Interesting insights!!
@bluberikeju2 жыл бұрын
Thanku mr, i hope will be excellent and succes aamiin
@Soumyajit21212 жыл бұрын
SARVESH SIR OPPP
@thepowerofinfoknowlege2 жыл бұрын
Thank you very much, you really made this great and simple. Thank You!
@tqaquotes93792 жыл бұрын
05:14 Vapor Pressure
@Aminulchem2 жыл бұрын
Great.
@noahway132 жыл бұрын
How would a pressure cooker change the cycle ?
@chaseduke92242 жыл бұрын
I am learning about cavitation in hydraulic pumps in school and could not wrap my head around WHY it happens. Thanks to you I now have a MUCH better understanding of this. This video helped me so much! Thank you!
@EricZuckerman12 жыл бұрын
Glad you found the video helpful!
@fawazalfawaz95383 жыл бұрын
Thank you
@davidswalec35983 жыл бұрын
That was just amazing! I was having difficulty with psychometric charts, even thou I understood, I thought refrigeration pressures and temperatures, now it is so easy. I am going to recommend technicians learn this before they get into all the other HVAC concepts.
@EricZuckerman13 жыл бұрын
Well, your post made my day. Thanks a ton and I am so happy something I made is helping so many.
@tlotlisogladwinmore36033 жыл бұрын
why have you changed the reaction order on [H] form 2 > 1/2 > 1??
@amin92453 жыл бұрын
Life saver
@EricZuckerman13 жыл бұрын
Glad it helped!!
@got7ahgasse3373 жыл бұрын
how can you solve the rate order reaction when the data only has 1 set of similar concentrations?
@ketpinnara47203 жыл бұрын
That is so good explanation.
@kimtaehyungspreciouslaugh16143 жыл бұрын
Thank you for this!!
@EricZuckerman13 жыл бұрын
You are welcome!!
@sunitarawat90123 жыл бұрын
Very helpful...
@naviitaramro34743 жыл бұрын
If it were a first Order would finding the slope be m/2?
@sofiadeumacansandoval33523 жыл бұрын
Thank you so much! I struggled to make my graphs until I watched your video!
@Random-sm5gi3 жыл бұрын
Thank you soooo much boss love ya
@tanzirahmed633 жыл бұрын
if we were given concentration of the product vs time, how could we use this to find the order and value of rate constant? someone said to me you need to convert the concentrations of the product to reactant concentrations but i dont know how.
@tanzirahmed633 жыл бұрын
@@EricZuckerman1 or do I minus each [C] at each time from the initial conc [A]0 to get the new [A] for each time?
@EricZuckerman13 жыл бұрын
@@tanzirahmed63 You are looking at this as math. You need to understand the concept, so let's look at it that way. Imagine a stack of A's, where each stack is 1 mole of A. Every time an A is used, it creates a B. So, the A stack gets smaller and the B stack grows. This happens almost instantaneously (it's fast). Now continue that to B. B is used to create a C, sot he B stack goes down by 1 and C stack goes up by 1. The overall picture is here is that A decreased by 1 and C increased by 1, and there is no B remaining. Thus, the [A] = [A]0 - 1 and the [C] = 1. At any time, the concentration of A can be indirectly measured by measuring the concentration of C. In other words, for every x M of C created, x moles of A has been used and we have [A]0 - x concentration of A remaining. Does this clear it up?
@tanzirahmed633 жыл бұрын
@@EricZuckerman1 yes it does. Thank you
@Vunami3 жыл бұрын
not all heroes wear cape... this guy explained this too well
@omoralsuny57553 жыл бұрын
Dear Professor, Hope you are doing well. I am working on Soil Phosphorus Sorption Kinetics. After calculation , I have found both adsorption and desorption simultaneously and the equilibrium point has not followed any definite pattern and the equilibrium point is not obvious too. While plotting them on pseudo-first-order, pseudo-second-order, elovich equation, I can't take each value as that shows error, eliminationg those values I tried to plot, this time R squared value is too low(even 0.008). For first order, I have plotted -log(qe-qt) vs t where as t/qt vs t for second order and qt vs ln(t) for Elovich equation. I can't understand what should I do? Will you please help me?
@omoralsuny57553 жыл бұрын
@@EricZuckerman1 Thanks for reading my comment! I appreciate your generous attitude! I am searching for the solution. Hopefully I will find a solution! And again thanks for wishing me luck. Stay safe and sound!
@nardaparama99533 жыл бұрын
So to find the rate constant, what we need is to know the temperature and concentration?
@EricZuckerman13 жыл бұрын
The common method of determining the rate constant is as follows. Measure k directly by running multiple trials of the reaction at the desired temperature. For each trial, you can use the initial concentrations and measured initial rate to compute k from the rate law of the reaction. Average the k from multiple all the trials to get the bet k. So, you need a rate law for the reaction (in the form rate = k [A]^x [B]^y...). You need initial concentrations. And you need to measure initial rates of reaction. In the example reaction, we know the generic rate law is rate = k [N2O5]^x. But, the rate constants given have units of s^-1 or 1/s. That is the units of k for a reaction that is first order overall, so x must equal 1. Thus, we know the rate law is rate = k [N2O5]. The k values would then be measured at each temperature by using an initial concentration of N2O5 and measuring the rate. k is computed using the results of that experiment.
@nardaparama99533 жыл бұрын
@@EricZuckerman1 ok so I understand how to find the rate constant. Now the data I have got includes gas collected (mL)/time and temperature. How do I measure the initial rate of that? Experiment is the decomposition of H2O2 with the use of MnO2 catalyst
@nizettedugue96213 жыл бұрын
I am not able to select the separate numbers on excel and its driving me crazy
@thembelanimiya89643 жыл бұрын
Explicit explanation 👌🏾...many thanks Sir.
@EricZuckerman13 жыл бұрын
Glad you find the video helpful.
@abhishekdwivedi42093 жыл бұрын
Excellent way to teach sir. Thanks for sharing your knowledge.
@EricZuckerman13 жыл бұрын
My pleasure!! Good luck on your chemistry journey
@OctobrTea3 жыл бұрын
thank you sooo much
@keshavbassi65013 жыл бұрын
@Eric Zuckerman Is the liquid in equilibrium with vapour at boiling point?
@EricZuckerman13 жыл бұрын
Yes, they are in equilibrium as long as the external pressure does not change. As more energy is added, the equilibrium shifts more to the vapor side until all of the liquid becomes vapor. Until that point, the system is at equilibrium. Now it gets challenging to describe this for open systems, like a pot of water on a stove.
@keshavbassi65013 жыл бұрын
@@EricZuckerman1 I am not able to infer what's my book trying to imply. There are two statements given : 1. "At the saturation temperature, the boiling liquid is in equilibrium with its own vapour." 2. "The steam is called saturated when the molecules escaping from liquid become equal to the molecules returning to it." In case of first statement, I think that definition is not right. The liquid can be in equilibrium with its own vapour at a temperature lower than boiling point too. Same argument holds for second statement too.
@EricZuckerman13 жыл бұрын
@@keshavbassi6501 The book is telling you the story in reverse, as I read your description. A "saturated steam" (#2) is one in which the number of moles of gas are constant. So, for every new evaporated molecule there is a different molecule that is condensing. In other words, the idea of saturation is that no more gas molecules can exist as steam at this temperature. When "saturated", the pressure caused by the vapor is constant....it is the vapor pressure. I've not heard this use of saturated before. That brings us to #1. At some temperature, the vapor pressure caused by the "saturated vapor" will be equal to the external pressure. That temperature is the boiling point. Remember that the liquid to gas equilibrium can occur at temperatures lower than the boiling point. Only at the boiling point does a system have enough pressure to overcome the external pressure. To give an example, consider a pot of water on your stove at room temperature. The external pressure is caused by the air if no lid is present. If you put a lid on the pot, then the external pressure is the pressure needed to lift that lid off the pot. I hope this helps.
@keshavbassi65013 жыл бұрын
@@EricZuckerman1 So basically the vapour generated at boiling point in a closed container is saturated steam and it is in equilibrium with liquid water, i.e. the molecules escaping from liquid become equal to the molecules returning to it. And, as we add more energy the amount of steam generated ( or number of moles) goes on increasing. Is it right?
@EricZuckerman13 жыл бұрын
@@keshavbassi6501 Be careful not to always imagine the liquid is water. In general, in a closed container any liquid will evaporate some and create a vapor pressure in the container. Even when not boiling, the system is at equilibrium. At the boiling point (temperature), it just so happens that the vapor pressure is equal to the external pressure. The boiling point also represents the temperature and pressure combination where both the liquid and gas states are equally stable, which is why a phase transition can happen for the substance. Otherwise, I think you have a good handle on this topic. Best of luck!!
@katrothanil35403 жыл бұрын
What a explanation Sir. No one can beat your teaching.👌👍👈
@EricZuckerman13 жыл бұрын
Thank you for the kind words. Best of luck in your chemical kinetics journey!
@katrothanil35403 жыл бұрын
@@EricZuckerman1 Tnx sir.
@a.g.8383 жыл бұрын
when you minimized the x-axis range, you eliminated one of the points on the graph that had the value less than 0.003
@EricZuckerman13 жыл бұрын
Great catch. I never noticed that before. While that is a mistake that no one should repeat, it will not and does not impact the equation of the line obtained by Excel. Excel uses the data that you select to make the graph and does not adjust for what is scaled onto the axis or not (thank goodness). Appreciate you giving me the heads up...first to notice after several years!!
@tcgsam3 жыл бұрын
Is the temperature used in this the final temperature of the solution?
@EricZuckerman13 жыл бұрын
The activation energy is determined by running the same reaction at multiple different temperatures. Each of those trials would need to be done at a constant temperature if possible. If you are mixing solutions and the temperature changes during the reaction (exothermic or endothermic reaction), then you should use the initial temperature. If this is a real experiment you are running, then be sure to note when you measured the temperature so that the reader of your report knows.
@tcgsam3 жыл бұрын
@@EricZuckerman1 Thanks for your response. I performed the experiment in real life, performing 3 trials each at 5 different temperatures but my k values were very similar to each other, causing all my ln(k) values to all be negative and relatively close to each other. That should still work for the calculations, right?
@samueldeng83913 жыл бұрын
Is the temperature used the final temperature of the solution?