New answers tagged

3

As said by M. Farooq in the comments, The Origin and Status of the Arrhenius Equation[1] clearly describes the method by which Arrhenius came with the Arrhenius equation. An excerpt from the article: Arrhenius considered eight sets of published data on the effect of temperature on reaction rates and showed (using more currently conventional symbols) that ...


5

First, be aware that the order of a chemical reaction depends on its mechanism and a great deal of chemical reactions have more than one mechanism, depending on the conditions. The traditional example for zeroth order reaction is a reaction fully dependent on a catalyst (e.g. enzymes in biochemistry). When you have enough reactants and a scarce amount of the ...


13

A standard example in biochemistry are enzyme-catalyzed reactions with a single substrate (e.g. ATPase, where we ignore the second substrate, water, because it is the solvent). At low substrate concentration (low reactant concentration), the reaction is first order (actually, pseudo-first order because of the water). At high susbtrate concentration (high ...


15

I think to some extent, you can think about all zero-order reactions as "pseudo-zero-order" reactions. This is because it's not possible for a reaction to be true zero-order. How can it be that the rate of a reaction does not depend on any one of the reactants? So, anything that is zero-order means that there is some artifact in the system, or that ...


9

Zero-order kinetics can also appear in certain industrial settings. In some steel strip annealing processes where improved bendability of the product is required, steam is applied to decarburize the steel near the surfaces of the strip according to the reaction $\ce{C(s) + H2O(g) -> CO(g) +H2(g)}$ The decarburization process requires carbon to diffuse out ...


7

The best example of a zero-order reaction is the combustion of a candle. If the candle weighs $m_o$ grams at the beginning, and $m$ at any time afterwards, and if $a$ grams of it are burned per minute, the reaction rate $r$ is constant from the very beginning to the end of the candle, and it is : $r = dm/dt = a$ in grams per minute. The integrated rate law ...


4

To complement Nisarg's answer, note that we have the following system of polynomial ODEs $$\begin{bmatrix} \dot a \\ \dot b \\ \dot c\end{bmatrix} = \left( 3 a b - \frac32 c \right) \begin{bmatrix} -1\\ -1\\ +1\end{bmatrix}$$ Thus, given an initial state, the state will flow along a given line. Since concentrations must be non-negative, given an initial ...


0

The equilibrium constant is derived from the rate law and not vice versa. For an elementary reaction, the rate law is written as $$K={[A]}^a{[B]}^b$$ But for a complex reaction, it is not dependent on the stoichiometric coefficient as the rate would be dependent on the multiple elementary reactions taking place. The equilibrium constant which we use is for ...


2

When you are implying reaction quotient, you should use $Q$ instead of $K$. As $Q$ is defined at any concentrations other than equilibrium. When you are changing $[\ce{A}]$ and $[\ce{B}]$, $[\ce{A2B}]$ may change also, which you have neglected. From the given data, the change in rate of reaction cannot be calculated for the reason you have pointed out in ...


1

Who says it's exponential? That the reaction is first-order? If the rate is "too high" early in the reaction when concentration is high, but "too low" later on when concentration is low, that indicates you have a higher concentration dependence than the first-order you assumed when you expected an exponential curve. If we were to assume ...


0

Try examining the Arrhenius Equation and see if that gives you more information about the rate constant. In general, the rate of reaction (in terms of the reactants) is a function of how many successful collisions occur per unit time. There are also unsuccessful collisions. You can also write the rate of reaction in terms of the products. Since the products ...


4

Truly collisionless unimolecular reaction are rare, as even highly unstable molecules must be triggered somehow to gain the reaction activation energy. This is frequently obtained as an energy gain by thermal collision with other molecule, or by a photon absorption. Collisions are usually much more frequent, but photons are usually much more energetic than ...


2

It is reasonable to be confused by this. The behaviour of these sorts of reactions is counter intuitive if you don't have a chemistry background, and it doesn't help that it is sort of obscured by the particular choice of numbers in the assignment. Looking at it naively, you might expect that because the forward rate is twice the backward rate, the final ...


8

From the Eyring equation, we can simply calculate the $k$ value for it. \begin{align} k &= \frac{k_\mathrm{b} T}{h}\exp\left(\frac{-\Delta G^\ddagger}{RT}\right)\\ k_\mathrm{b} &= \pu{1.38E-9 J K^-1}\\ T &= \pu{355 K}\\ h &= \pu{6.626E-34 J s}\\ R &= \pu{8.3145 J K^-1 mol^-1} \end{align} Here, I am assuming the average of $\pu{29 kcal/mol}...


8

You could convert the rate constant($k$) to half-life($t_{1/2}$) which would give you an idea of the time scale required for the reaction to finish at a certain temperature. The equation to obtain half-life from $k$ is different depending on the order of the reaction. For an unimolecular reaction($\ce{A->P}$), it is: $$t_{1/2}=\frac{\text{ln 2}}{k}$$ For ...


Top 50 recent answers are included