New answers tagged

1

To understand the behavior described by the equation $$\rho(T)=A\mathrm e^{-E_\mathrm a/(RT)}$$ a little better, consider a few properties (where it is assumed that $A$ and $E_a$ are independent of $T$): The value of $\rho(0)=0$ (as $T$ goes to zero $1/T\rightarrow \infty$ so that $\exp(-\infty)=1/\exp(\infty)=1/\infty=0$) The value of $\mathrm d \rho (T)/...


4

This is how the graph between the rate constant and temperature looks like: This might not seem intuitive looking at the Arrhenius equation which says: $$k=Ae^{\frac{-E_a}{RT}}$$ After using some mathematical prowess someone might say that as $T\rightarrow \infty, ~k\rightarrow A$ and so the graph should look something more like: Which on first glance ...


1

If all you want to find is the final concentration of all the species at equilibrium than you don't need to solve these differential equations. To solve this types of questions easily in chemistry we define a quantity $K_{eq}$. Where $K_{eq}=\frac{K_f}{K_b}$ and $K_{eq}=\frac{\text{product of concentrations of products at equilibrium to the power of their ...


2

$1$) 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 not change if the amount of reactant is increased. But the rate of reaction ${r = - dn/dt}$ does increase when the amount of reactant ${n}$ increases, whatever ...


4

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 $K_M = \dfrac{k_\mathrm{on} + k_\mathrm{cat}}{k_\mathrm{off}}$ (using the conventional description of Briggs and Haldane's derivation of the Michaelis-Menten equation; Ref.1): $$\ce{E + S <=>[$k_\...


4

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 $y = 1/V$ and $x = 1/S$ in $y = A + Bx$ leads by linear regression to $A = 0.4767$, $B = 4.4754$ (not including units). Therefore, $$V_\max = \frac{1}{A} = \pu{2....


4

This is an exercise in applying the Hammond postulate (multiple times). To accurately compare reaction rates, we need to have the activation energy. We don't have this, and the actual energy of the transition state is unknowable without experimental data. Hammond's postulate states that the structure of the transition state is more similar to the ...


13

The ambiguity doesn't lie so much in the definition of the rate, because as long as the constants $a$ and $b$ in $ap_\ce{X}$ and $b[\ce{X}]$ are chosen appropriately, the two expressions are entirely equivalent. So you can define the rate either way, and there is no physical difference, not even a mathematical difference. It is somewhat akin to asking ...


-2

Why be qualitative when with just a little more effort you can be quantitative? Acetylsalicylic acid has a pK$_a$ of 2.97, i.e., it is fairly acidic (Ref 1). Salicylic acid, one of the hydrolysis products, is comparably acidic (pK$_a$ = 2.79 Ref 2) and its sodium salt is fairly neutral (pH = 6.0 - 8.5 for a commercial material Ref 3; calculations suggest ~7....


0

First I was thinking that maybe a chemical indicator could be used, however the pKa values of all three components are relatively close to each other (especially aspyrin and salicylic acid). It might still be a bit of a help to stick a pH electrode in the system and observe the change - however, this is not exactly something you can see easily. Another thing ...


0

When you look at the overall reaction, it is if you have an equal concentration of each reactant, then by the reaction progressing or adding equal amounts of both, the reaction rate won't change. For example, if you have 2M of both A and B, $[A]^{-1}[B]^1 = 1/2*2=1$, and increasing the concentration to 3M for both won't change the rate since $1/3*3=1$. The ...


3

TL;DR Catalysts by definition do not alter the magnitude of changes in thermodynamic properties that accompany conversion of reactants to products, they only alter the conversion (reaction) rates. The reason a catalyst does not alter the entropy of a reaction is that this is one part of the definition of a catalyst! Ideally none of the changes in ...


-1

The simplest and most direct explanation is that isotopically heavier molecules have higher boiling points. Regular water (R) boils at 100.0$^o$ C. Heavy water (H) boils at 101.4$^o$ C. A mixture of the two (R + H) can be separated by a distillation column. The heavier molecules require more energy to evaporate, to push away one atmosphere. Any similarity to ...


1

As very simple model to find what properties are important in evaporation consider a molecule that when close to the surface moves around in a cell of side $a$ and collides with frequency $f$ into its neighbouring molecules. There is an area $a^2$ through which it can escape the surface and in a second it moves on average a distance $v= 2af$ back and forth ...


1

For this we can first write the equation for the reaction again: $$MH_2 => He_{(g)} + Noble gas + H_{2(g)}$$ Next we can find the moles of gas at 80 mins (I am assuming it is 80 minutes and not 82). You already did this and you got 0.045 moles. Since each mole of $MH_2$ forms 3 moles of gas, $\frac{0.045}{3}=0.015$ moles of $MH_2$ used in 80 minutes. Then ...


3

Evaporation depends on molecular velocity not on the energy The simple mistake in your assumption is that evaporation is driven by molecules having enough energy to escape the liquid phase. This is wrong. Evaporation depends on whether the molecule is travelling fast enough to escape the liquid. The molecules in the liquid will have, on average, the same ...


2

[OP] The catalytic efficiency of an enzyme is given by $k_\mathrm{cat}/K_\mathrm{M}$ where $k_\mathrm{cat}$ is the turnover number, or the number of molecules that can be produced per second per active site of an enzyme. The last part is not quite accurate. $k_\mathrm{cat}$ is the rate of the reaction under saturating conditions divided by the enzyme ...


6

The turnover number or catalytic constant $ k_{\mathrm{cat}}$ in the Michaelis-Menten model is the rate constant for the productive dissociation of intermediate $\ce{ES}$: $$\nu = k_{\mathrm{cat}}[\ce{ES}]$$ The constant $k_{\mathrm{cat}}$ says how much product forms from intermediate but does not say how much intermediate forms in the first place. It is ...


6

The most basic kinetic scheme for enzymes is represented as $$\ce{E + S <=>[K_m] ES ->[k_{cat}] E + P}$$ As should be clear, the $k_{cat}$ is the rate constant for the reaction that occurs after substrate is bound to the enzyme. The resulting rate (kcat[E]tot) is only achieved when every molecule of enzyme essentially always is in the act of ...


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