My pleasure. Hope you crush your studies of chemical kinetics!! Best of luck :)

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.

Glad that your persistence has paid off and that my video played a role in your success!!

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.

​@@EricZuckerman1thank you sm proff...

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.

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.

@@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.

@@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!!

Glad you found the video helpful!

Well, your post made my day. Thanks a ton and I am so happy something I made is helping so many.

Glad it helped!!

You are welcome!!

@@EricZuckerman1 or do I minus each [C] at each time from the initial conc [A]0 to get the new [A] for each time?

@@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?

@@EricZuckerman1 yes it does. Thank you

@@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!

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.

@@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

Glad you find the video helpful.

My pleasure!! Good luck on your chemistry journey

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.

@@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.

​@@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.

@@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?

@@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!!

Thank you for the kind words. Best of luck in your chemical kinetics journey!

@@EricZuckerman1 Tnx sir.

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!!

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.

@@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?