9 votes
Accepted

Deriving kinetic equations for reversible reactions

Actually you need concept of coupled differential equations but I suppose you can do a 'trick' to avoid it: $$ \frac{dA}{dt} =- a A + b B$$ $$ - \frac{dB}{dt} = - aA + bB$$ Are the form of the two ...
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7 votes

Can anyone explain, why there are coefficients 2 in the equation for rate of change [A]?

$$\ce{A_n ->[$k$] nA}$$ For the reaction given above, rate of formation of $\ce{A}$ is defined as: $$\frac{1}{n}\frac{\mathrm d\ce{[A]}}{\mathrm dt} = k\ce{[A_n]}$$ In this case, we have three ...
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6 votes
Accepted

What is the rate law of a reaction: 2A + 3B → products?

If such a reaction "$\ce{2A + 3B -> }$ products" occurs, it is the sum of several more elementary reactions, which are taking place successively. Some may be fast, some slower. The rate ...
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  • 20.7k
6 votes

Why is it that first order reactions never end?

Note that basics of reaction kinetics were not created with the quantized nature of matter in mind. In this context, reactions of the 1st order do not end for the same reason why the function $\exp{(-...
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  • 27.3k
6 votes
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Does 'k' in the rate law change with a change in volume

(Disclaimer: It's been a long time since I did this.) The collision frequency (or collision density, to use the term in Atkins' Physical Chemistry 9ed) is indeed proportional to concentration. We ...
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5 votes

What is the rate law of a reaction: 2A + 3B → products?

If $\ce{2A + 3B -> \text{product(s)}}$ describes the stoichiometry of a reaction, than you describe the balance of starting materials against products; this however does not state much about the ...
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5 votes

Why do alkenes (olefins) burn hotter than alkanes (paraffins), even though alkanes are denser and more hydrogen-rich?

Flame temperatures are not primarily determined by the thermodynamics of the reactions in the flame Intuitively it might seem that the temperature of a flame is dependent on the thermodynamics of the ...
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5 votes

Finding the rate of a reaction given experimental data for reaction kinetics

As I said in the comments, your image is quite low resolution. But, if I read it correctly, you have got the exponents in the rate law wrong. Solving this type of problem is simple, you just need to ...
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  • 5,249
4 votes
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Determine vmax and enzyme concentration (Michaelis–Menten)

Michaelis-Menten kinetics is given by the equation: $$V = V_\mathrm{max}\frac{S}{K_M+S} \tag1$$ Where $V_\mathrm{max} = k_\mathrm{cat}\cdot [\ce{E_T}]$ and $K_M = \dfrac{k_\mathrm{on} + k_\mathrm{cat}...
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4 votes

Determine vmax and enzyme concentration (Michaelis–Menten)

Michaelis-Menten kinetics is given by $$V = V_{\max} \frac{S}{K_M + S}$$ Taking the inverse of both sides to linearise the equation, $$\frac{1}{V} = \frac{1}{V_\max} +\frac{K_M}{V_{\max} S}$$ Taking $...
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4 votes

Does 'k' in the rate law change with a change in volume

The original formulation gives the temperature-dependence of the rates: Source: https://zenodo.org/record/1749766#.YhWZ85ZOklw The given formula is showing the relationship between two rates (same ...
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4 votes

Are there any genuine, elementary ternary reactions?

We cannot really distinguish a ternary collision from two binary ones, because the binary collision takes time for the interaction to run its course and therefore any system has a nonzero chance that ...
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3 votes

Calculate the concentration of the reactant from the absorbance of the product without molar absorptivity

Reactant $\ce{A}$ does not absorb the light, so the measured absorbance is only due to $\ce{B}$. Let $[\ce{B}]_t$ and $A_t$ be the measured values of $[\ce{B}]$ and $A$ at any time during the ...
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3 votes

Deriving kinetic equations for reversible reactions

I'll just post what I did after @Buraian's answer for a general case - Let the reaction be $\ce{A <=>[$k_\mathrm f$][$k_\mathrm b$] B}$ Let the concentration of $\ce{A}$, $\ce{[A]} = a$, and $\...
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  • 1,773
3 votes
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Are the orders of reactants with respect to a reaction different for different starting concentrations?

A reaction's rate does depend upon the consumption of its reactants, and the manner in which the reactants interact. Let's consider the following reaction: $$\ce{A + B -> final products}$$ The rate ...
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  • 558
3 votes

Why do alkenes (olefins) burn hotter than alkanes (paraffins), even though alkanes are denser and more hydrogen-rich?

Double-bonds are less stable, i.e., their formation is less exothermic, so it takes less energy to break one double bond than two singles. Or one triple vs. three single bonds. Consider triple-bonded ...
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3 votes

Steady-state approach - how to find rate equation for intermediate?

Split the scheme into three parts and write down each step. Start by definition with $$\displaystyle -\frac{1}{2}\frac{d[\ce{NO}]}{dt}=+\frac{d[\ce{N2O2}]}{dt}$$ then the first step $\displaystyle -\...
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2 votes

Steady-state approach - how to find rate equation for intermediate?

If you apply Goldberg & Waage's Law of Mass Action to the rate determining (slowest) step, we will get: $$\frac {d[\ce{N2O2}]}{dt}=-k_2 [\ce{N2O2}][\ce{O2}]$$ Moreover, we have: $$\frac {k_1}{k'...
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2 votes
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Does mass affect rate of reaction?

$1$) Let's admit first that the reaction rate is proportional to the amount of reactants, in the usual way. Whatever the order $1$ or $2$ of the reaction, the rate constant ${k_{1} }$ or ${k_2}$ will ...
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2 votes

First or second order reaction

Use the 2nd order scheme $2A \to B$ with initial amount of $A=A_0$ so $B=A_0-A$ and B is the product you observe. The rate of change is $\displaystyle \frac{1}{2}\frac{dA}{dt}=-k_2A^2$ and I'm not ...
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2 votes

Problem understanding the rate of reaction of reversible reactions

Let say the kinetic rate of neutralization $$\ce{CH3COOH + OH- -> CH3COO- + H2O}$$ is $k_\mathrm{f}[\ce{CH3COOH}][\ce{OH-}]$ And the kinetic rate of hydrolysis is $$\ce{CH3COO- + H2O -> CH3COOH +...
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2 votes
Accepted

Calculate concentration of product after time passed

The kinetics is given by : $\pu{d[A]/dt}$ = $\pu{- k_{1}[A] - k_{2}[A] = - k_t[A]}$ Integration gives : $\pu{[A] = [A]_{0} e^{-k_tt}}$ At any time, the proportion of $\pu{P_{1}}$ and $\pu{P_{2}}$ are $...
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  • 20.7k
2 votes

Deriving kinetic equations for reversible reactions

This is a classic example of the kinetics of reversible reactions. From the mechanism $\ce{A <=>[$k_\mathrm f$][$k_\mathrm b$] B}$ you can write the kinetic equations \begin{align} \frac{\...
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  • 451
2 votes

How to calculate rate of carbon dioxide from rate equation?

Suppose we start with $n$ moles of $\ce{CO2}$ for simplicity. Then, at any point during the reaction, we will have $n-V_\text{gas}[\ce{CO2}]$ moles of each of the products. This means we could write ...
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1 vote

Calculate rate constant of second order reaction and pseudo first order reaction

The general rate law for a second order kinetics if $\pu{A_o < B_o}$, is $$\pu{\frac{1}{B_o - A_o} \ln\frac{A_o B}{A B_o} = kt}$$ If you only want to check that it is second oder, you can forget ...
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1 vote

pH for optimal kinetic for enzyme reaction

I don't think the rate equation is $v= \frac{k_2 [E]_0 [S]}{K_M(1+ [H+]/K_1 + K_2/[H+]) + [S]}$ I suspect it is rather: $$v= \frac{k_2 [E]_0 [S]}{(K_M + [S]) (1+ [\ce{H+}]/K_1 + K_2/[\ce{H+}])}$$ ...
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1 vote

Writing rate of disappearance and rate of appearance using rate law

You overthink it. It is just matter of eventual multiplication or division with respective stoichiometric coefficients. If there is an elementary reaction $$\ce{a A -> b B}$$ then for respective ...
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  • 27.3k
1 vote

Deriving kinetic equations for reversible reactions

Using SymPy: ...
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1 vote

Can the RDS be different in different direction in a multi-step reversible reaction?

After discussion with OP in chat, the issue in logic can be rectified as follows. Simply put: $\ce{A<=>B<=>[slow]C <=> D}$ doesn't imply that $\ce{B <=> C}$ is the RDS in both ...
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