Newest questions tagged units kinetics - Chemistry Stack Exchange most recent 30 from chemistry.stackexchange.com 2019-09-22T06:12:36Z https://chemistry.stackexchange.com/feeds/tag?tagnames=units+kinetics&sort=newest https://creativecommons.org/licenses/by-sa/4.0/rdf https://chemistry.stackexchange.com/q/104174 -1 Rate constant converion [closed] Pat Srikanth https://chemistry.stackexchange.com/users/70068 2018-11-11T08:39:37Z 2018-11-11T10:45:39Z <p>I need to calculate the half life for a first order reaction for which I need the rate constant <span class="math-container">$k$</span>. From literature I observed that for a very closely simulated experiment I obtained the numerical value of <span class="math-container">$k$</span> but the unit of <span class="math-container">$k$</span> was given in <span class="math-container">$\frac1{\mathrm s}\cdot\frac1{\mathrm{kPa}}$</span>.</p> <p>I am only familiar with <span class="math-container">$k$</span> in <span class="math-container">$\frac1{\mathrm s}$</span>. from which I can easily estimate the <span class="math-container">$t_{1/2}$</span>.</p> <p>Can somebody please explain how I convert the <span class="math-container">$\frac1{\mathrm s}\cdot\frac1{\mathrm{kPa}}$</span> unit to <span class="math-container">$\frac1{\mathrm s}$</span>.</p> <p>Additional informaton: The laboratory experiment was conducted at approx <span class="math-container">$85\ \mathrm{kpa}$</span> and approx <span class="math-container">$100\ \mathrm{kpa}$</span> is <span class="math-container">$1\ \mathrm{atm}$</span>.</p> https://chemistry.stackexchange.com/q/87925 3 Order of the reaction between hemoglobin and oxygen Gennaro Arguzzi https://chemistry.stackexchange.com/users/40502 2017-12-27T12:03:47Z 2017-12-28T11:58:54Z <p>I have the reaction:</p> <p>$$\ce{Hb + O2 &lt;=&gt;[k][k'] HbO2}$$</p> <p>where $\ce{Hb}$ is hemoglobin, $k$ is the rate constant of the forward reaction, $k'$ is the rate constant of the inverse reaction. On my book there is the rate law:</p> <p>$$W = kP(\ce{O2})[\ce{Hb}] - k'[\ce{HbO2}]$$</p> <p>where $W$ is in $\pu{mol m-3 s-1}$, $P(\ce{O2})$ is a partial pressure in Pascal ($\pu{kg m m-2 s-2 = kg m-1 s-2}$).</p> <p>The inverse reaction is a first order reaction because:</p> <p>$$\pu{mol m-3 s-1} = [k'] (\pu{mol m-3})$$</p> <p>thus $[k'] = \pu{s-1}$.</p> <p>I am in trouble with the forward reaction because I got a strange unit of measurement:</p> <p>$$\frac{\pu{mol}}{\pu{m3 s}} = [k] \frac{\pu{kg}}{\pu{m s2}} \frac{\pu{mol}}{\pu{m3}}$$</p> <p>$$[k] = \frac{\pu{m s}}{\pu{kg}}$$</p> <p>Which is the order of the forward reaction?</p> https://chemistry.stackexchange.com/q/83873 3 Could the loss of mass in the kilogram be due to oxidation? The_Sympathizer https://chemistry.stackexchange.com/users/53075 2017-10-08T23:41:19Z 2017-10-08T23:41:19Z <p>I was wondering about this. I have heard, and known for a while, that the famous and celebrated <em>Prototype Kilogram</em> - the lump of metal whose mass is used to define the standard mass scale worldwide - is slowly changing, degrading in some fashion, and this has prompted the search for a new and better definition of the kilogram mass unit on something more stable (the lead is now Planck's constant, but the change has not yet been adopted by the BIPM and looks like it won't be until at least next year.).</p> <p>However what I'm curious about is the <em>reason</em> for the mass change. Apparently it's said it's not clear as to what exactly could be causing it, but I was wondering: is it chemically possible that this mass change could be due to oxidation - tarnishing - of the kilogram and removal of the oxide? I would imagine that the scientists who have studied this must have thought of this, so I'd be more interested in why it's <em>dismissed</em> as an explanation. In particular, it's said that the material it's made of - mostly platinum - has a very low oxidation rate, but "very low" need not equal "zero", and the mass change is reported as up to (unclear, because it's the measuring standard itself!) 50 micrograms over 100 years - which is nonetheless a <em>very slow</em> rate of loss - about $10^{-16}$ mol Pt/s, so that it would take $10^{16}$ s (10 Ps) to oxidize just one mole, or about 300 million years. Could this be within range of the very-slow-but-<em>not</em>-zero range of oxidization of platinum in air? In particular I find this:</p> <p><a href="https://www.google.com/url?sa=t&amp;rct=j&amp;q=&amp;esrc=s&amp;source=web&amp;cd=3&amp;cad=rja&amp;uact=8&amp;ved=0ahUKEwjTsN2bleLWAhWB5YMKHf7_A14QFggyMAI&amp;url=http%3A%2F%2Fwww.technology.matthey.com%2Fpdf%2Fpmr-v19-i4-135-140.pdf&amp;usg=AOvVaw2D87XQhqSGonkmrjf0kD_a" rel="nofollow noreferrer">https://www.google.com/url?sa=t&amp;rct=j&amp;q=&amp;esrc=s&amp;source=web&amp;cd=3&amp;cad=rja&amp;uact=8&amp;ved=0ahUKEwjTsN2bleLWAhWB5YMKHf7_A14QFggyMAI&amp;url=http%3A%2F%2Fwww.technology.matthey.com%2Fpdf%2Fpmr-v19-i4-135-140.pdf&amp;usg=AOvVaw2D87XQhqSGonkmrjf0kD_a</a></p> <p>talking about platinum oxidizing in air upon mild heating, and even <em>losing mass</em>. Now it seems that a rule of physics is "whatever isn't forbidden is compulsory", that is, whatever isn't absolutely impossible will eventually happen given a suitably long time period, and so if the oxidation is possible could it still be going on at very reduced rates at room temperature that nonetheless might still be sufficient to degrade the kilogram? Or is the rate then so low as to not contribute at all? (I'd think about Arrhenius rate law, but no idea what the relevant constants would be. Nonetheless exponentials are pretty steep, so this might rule it out.)</p> https://chemistry.stackexchange.com/q/64354 1 Reaction rate unit conversion: how to convert from molarity to ppm Gladius https://chemistry.stackexchange.com/users/38706 2016-12-12T15:32:20Z 2017-04-01T00:20:59Z <p>How can I convert a reaction rate $k$ with units of $\mathrm{1\over M~~s}$ to $\mathrm{1\over ppm~~s}$?</p> <p><strong>For context:</strong> I want to make a numerical simulation of a system of reactions that happens in air (and in water droplets suspended in the air), I have already built a part of the model for which I used $\mathrm{ppm}$ for my concentrations and thus, e.g., $\mathrm{1\over ppm~~s}$ as the units for the $k$ values of second-order reactions in the system. (Reactions of different orders have different units, obviously, but those details are irrelevant to the question.) The article I have on hand lists a $k$ value for the reaction I need in $\mathrm{1\over M~~s}$ and the reaction takes place in the liquid phase.I assume that the PPM is weight-based although the article in question does not actually specify this.</p> <p>I am asking the above question because I am not sure if this conversion is necessarily as straightforward as a simple factor of $10^{-6}$.</p> https://chemistry.stackexchange.com/q/58642 1 Rate constants using Arrhenius equation atreyav https://chemistry.stackexchange.com/users/34607 2016-09-07T22:34:51Z 2016-09-07T23:11:29Z <p>If I'm trying to use the Arrhenius equation in this form:</p> <p>$$\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_\mathrm a}R\cdot\left(\frac1{T_2}-\frac1{T_1}\right)$$</p> <p>And I am given the activation energy ($E_\mathrm a$) in kJ, do I need to convert it into J before I plug in my other givens? </p> https://chemistry.stackexchange.com/q/43008 9 Units of rate constant (cycles per second vs radians per second) in Eyring equation slaw https://chemistry.stackexchange.com/users/24236 2015-12-30T18:22:56Z 2019-05-10T15:49:07Z <p>In the Eyring equation (EE),</p> <p>$$k = \frac{k_\mathrm B T}{h} \exp\left(\frac{-\Delta G_{\mathrm f}}{RT}\right),$$</p> <p>the units of $k$ are $\mathrm{s^{-1}}$. However, in general rate constants are usually expressed in $\frac{\mathrm{rad}}{\mathrm{s}}$. For instance, in expressions for damped oscillations of the form $\exp[(\mathrm{i} \omega - k)t]$, where $\omega$ is by definition in $\frac{\mathrm{rad}}{\mathrm{s}}$ and $k$ has to bear units providing consistent dimensions.</p> <p>By expressing in the EE the energy $k_\mathrm B T$ as $\nu h$, where frequency $\nu$ is in cycles per second $\left(\mathrm{Hz}=\frac{\mathrm{cyc}}{\mathrm{s}} = \mathrm{cps}\right)$, it appears that $k$ is also given in $\mathrm{cps}$ in this equation. The desired units of $\frac{\mathrm{rad}}{\mathrm{s}}$ would be obtained if $h$ were replaced by $\hbar$. However, I have never seen the EE formulated that way. Has anybody an idea in this matter?</p> https://chemistry.stackexchange.com/q/40908 -2 Reaction rate constant conversion [closed] unterm- https://chemistry.stackexchange.com/users/23031 2015-11-18T09:06:56Z 2015-11-18T13:24:03Z <p>I'm flustered, I'm not into chemistry and need to convert $\mathrm{(nM\cdot s)^{-1}}$ to $\mathrm{(M\cdot s)^{-1}}$. Is it correct to multiply with $10^9$?</p> https://chemistry.stackexchange.com/q/34045 7 Units in modified Arrhenius equation? J-S https://chemistry.stackexchange.com/users/7317 2015-07-16T00:57:57Z 2015-07-31T17:25:13Z <p>The modified Arrhenius equation is used to express the rate constant in a chemical mechanism model I'm working with. The equations is as follows: $$k_\mathrm{f} = A\times T^b\times\exp\left(-\frac{E_\mathrm{a}}{RT}\right)$$ The paper states that "Units are Moles, cm3, Seconds, K, and Calories/Mole" so what would be the final units of the rate constant? I'm a bit confused due to the presence of the exponential function. When I simply plug the values in as given in the paper the rate becomes huge when multiplied by the molar concentration, is this because it gives molecules per second? To give you and idea for the following reaction: $$\ce{C2H5OH + OH &lt;=&gt;C2H4OH + H2O}$$ The Arrhenius constants are as follows: $$A = 1.74E+11$$ $$b = 0.27$$ $$E_a = 600.0$$ I am yet to calculate the reverse reaction using Gibbs Free Energy, is it just equally as large and thus it all cancels out or are the final units really not in $\mathrm{mol\ s^{-1}}$?</p> https://chemistry.stackexchange.com/q/30823 2 Unit of time used in rate law equations Caesium-133 https://chemistry.stackexchange.com/users/7689 2015-05-04T02:42:17Z 2016-11-12T21:14:45Z <p>If doing a problem where you use the integrated rate law for a second order reaction</p> <p>($\frac{1}{[\mathrm{A}]}-\frac{1}{[\mathrm{A_0}]}=kt$) to first find $k$, then use that to find the half-life. Does it matter seconds are used, or minutes, or whatever?</p> https://chemistry.stackexchange.com/q/22410 3 Why is the units of kcat 1/s? CiaranWelsh https://chemistry.stackexchange.com/users/9434 2015-01-05T14:22:59Z 2015-11-16T19:55:54Z <p>I understand that $k_\text{cat}$ measures the turnover number of an enzyme. This measure is therefore a quantity of molecule conversions per unit of time. I suspect that my problem is more that of a lack of maths but why is the unit expressed in $\mathrm s^{-1}$? Why not just seconds?</p> https://chemistry.stackexchange.com/q/6462 4 How to find the units of the rate constant? Ayush https://chemistry.stackexchange.com/users/2450 2013-10-09T21:13:58Z 2017-10-14T09:10:45Z <blockquote> <p>Consider the reaction $$\ce{2A + B -&gt; Products}$$ When the concentration of $\ce{B}$ alone was doubled, the half-life did not change. When the concentration of $\ce{A}$ alone was doubled, the rate increased by two times. The units of the rate constant is:</p> <p>a) $\mathrm{s^{-1}}$</p> <p>b) $\mathrm{L\ mol^{-1}\ s^{-1}}$</p> <p>c) Unitless</p> <p>d) $\mathrm{mol\ L^{-1}\ s^{-1}}$</p> </blockquote> <p>I have tried it solving through this method:</p> <p>According to me, when the concentation of A alone was doubled, reaction rate also increased by two time, implies that Reaction is first order according to A. In same way, the reaction should be 0 order according to B. So net order would be 1. So this gives me answer $\mathrm{s^{-1}}$ but the answer is $\mathrm{L\ mol^{-1}\ s^{-1}}$. Where am I going wrong?</p>