New answers tagged

1

Yes, you are right, the reason primarily looking at it from a theoretical point of view is the high bond dissociation energy of $\ce{N#N}$ Let's look at this aspect from every point of view Theoretical- High bond dissociation energy of $\ce{N#N}$ Kinetics- We have the Arrhenius equation given below, which states the relationship between the rate constant ...


0

The mechanism of the gas phase halogenation reactions is discussed in some detail in a paper by Benson and Bus[1]. In the following, I submarised section II ( mechanism and rate laws). The mechanism of the reaction thermal reaction is $\ce{X2 + M -> 2X^. + M \;\; k_1 \quad(1) }$ $\ce{2X^. + M -> X2 + M \;\; k_{-1} \quad(-1) }$ $\ce{...


2

I suspect that you could be forming an aqueous solution of $\ce{(NH4)2PtCl6}$. I would like to make an observation as a person who has done platinum group metal chemistry, the anionic chloro complexes of PGMs like $\ce{PdCl4^2-}$ and $\ce{PtCl4^2-}$ are very toxic. They can induce a nasty allergy to PGMs. I would suggest that you do not work at home with ...


1

In the book that Wikipedia cites (Advanced Organic Chemistry: Reaction Mechanisms By Reinhard Bruckner, ISBN 9780080498805), they have a different set of assumptions. They are saying the formation of chlorine radicals proceeds by a fast equilibrium. Then, they say that the organic radical is at steady state, ignoring reaction (4) given by the OP (the ...


1

Your analysis of the expectation is correct, and we can look at it more quantitatively in this way: Since the hydroxide concentration is much higher than the CV concentration in all of the reactions, let's approximate it as unchanging. Assuming the reactions are first-order in each reactant, we have a rate law for the concentrated reaction of: $$-\frac{d[\...


-1

All the reaction have different concerntration at start . As you have taken so much excess NaOH it would remain constant. We could look it as you have a solution of 50 ml with 0.05 mM violet red and 0.5M NaOH . You dilute the N times and then to have to compare rates. Threotically we could find that but when we say discolouration in the solution does it mean ...


2

To summarize, and to answer the question, it is correct to say that the rate of a chemical reaction can be expressed in % per second if and only if the reaction is first order. This percentage is the first order rate constant $k$.


6

According to IUPAC Recommendations for chemical kinetics [1], rates $\nu$ are defined for reactant $\ce{B}$ and product $\ce{Y}$ as a time $(t)$ derivative of the amount $n$ (in general) or per unit volume $V$ (for the closed system): $$ \begin{array}{lll} \hline \text{System} & \text{Consumption} & \text{Formation} \\ \hline \text{Open} & \...


0

For a reaction like this: $\ce{aA +bB⟶cC +dD}$, the rate is given by Rate=$−a\frac{d[A]}{dt}=−b\frac{d[B]}{dt}=c\frac{d[C]}{dt}=d\frac{d[D]}{dt}$ where the unit is usually $\ce{mol L^{-1}{s^{-1}}}$. The rate of a reaction is different for different orders and has different units of the same. It's unit usually has the terms $\ce{mol, L, and s}$. For a zero ...


2

This answer is mostly a compilation of the excellent comments by Spontification and Chet Miller. Differential rate law The overall reaction order tells you what happens to the rate when you change all concentrations by the same factor at the same time. For example, if you decrease all concentrations by a factor 2 and the overall order is 1, the reaction ...


2

First, note that one does not always measure initial rates. It is fairly common to fit an entire reaction curve (or nearly entire), but it requires a little bit more math. It is also common to measure an "initial" rate that is not quite initial. That is, it is the rate after a burst phase has occurred, and the system has settled into something close to a ...


1

Allosteric inhibitors changes the conformation of enzyme enzyme for preventing the substrate binding, and allosteric activators modify the enzyme conformation so that substrate binding can take place. Competitive inhibitors don't change the enzyme conformation, as these inhibitors compete with the substrate to bind at enzyme's active site. Non-...


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