Intermediate Chemistry - Mono Mole

November 4, 2019September 25, 2024

Rate laws – integral form

An integral rate law mathematically expresses the rate of a reaction in terms of the initial concentration and the measured concentration of one or more reactants over a particular time.

Such a rate law can be derived from its differential form via simple calculus. Consider the decomposition of hydrogen peroxide to oxygen:

$H_2O_2(aq)\rightleftharpoons H_2O(l)+\frac{1}{2}O_2(g)$

The rate law is experimentally determined to be first order: rate = k[H₂O₂]. If we are monitoring the progress of the reaction by measuring the change in concentration of the peroxide, the differential form of the rate law is:

$-\frac{d[H_2O_2]}{dt}=k[H_2O_2]\; \; \; \; \; \; \; \; 12$

Rearranging and integrating eq12,

$\int_{[H_2O_2]_0}^{[H_2O_2]}\frac{1}{[H_2O_2]}d[H_2O_2]=-k\int_{0}^{t}dt$

where [H₂O₂]₀ is the concentration of the peroxide at t = 0, i.e. its initial concentration. We then get:

$ln\frac{[H_2O_2]}{[H_2O_2]_0}=-kt\; \; \; or\; \; \; ln[H_2O_2]=-kt+ln[H_2O_2]_0\; \; \; \; \; \; \; \; 13$

Eq13 is the integral form of the first order rate law for the decomposition of hydrogen peroxide. The general equation for a species, A, that participates in a first order reaction of vA → B is:

$ln\frac{[A]}{[A]_0}=-vkt\; \; \; or\; \; \; ln[A]=-vkt+ln[A]_0\; \; \; \; \; \; \; \; 14$

For a zero order reaction, e.g. the decomposition of excess N₂O on hot platinum, $N_2O\rightarrow N_2+\frac{1}{2}O_2$ , the rate equation is $-\frac{d[N_2O]}{dt}=k$ , and by integrating both sides of the differential rate equation, its integral form is:

$[N_2O]=-kt+[N_2O]_0\; \; \; \; \; \; \; \; 15$

In general, a species, A, that participates in a zero order reaction of vA → B has the equation:

$[A]=-vkt+[A]_0\; \; \; \; \; \; \; \; 16$

For a second order reaction, e.g. NO₂ + CO → NO + CO₂ , the rate equation is:

$-\frac{d[NO_2]}{dt}=k[NO_2]^2$

and its integral form is:

$\frac{1}{[NO_2]}=kt+\frac{1}{[NO_2]_0}\; \; \; \; \; \; \; \; 17$

Once again, the generic second order rate equation that involves only one species, A, in a reaction, vA → B, is:

$\frac{1}{[A]}=vkt+\frac{1}{[A]_0}\; \; \; \; \; \; \; \; 18$

Finally, the diagram below shows the combined concentration-time plot of eq14, eq16 and eq18 for a chemical species A.

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November 4, 2019October 27, 2024

Integral method (chemical kinetics)

The integral method of determining the rate of a reaction makes use of the integral form of a rate law.

Just like the half-life method and the differential method, the integral method is useful for reactions involving a single reactant, e.g.

$N_2O_5(g)\rightleftharpoons 2NO_2(g)+\frac{1}{2}O_2(g)$

If the order of the reaction is known, we can determine the value of the rate constant by plotting the appropriate integral rate equations below using experiment data and finding the vertical intercept of the respective linear functions.

Order	Rate law	Integral rate equation
0	$rate=k$	$[A]=-kt+[A]_0$
1	$rate=k[A]$	$ln[A]=-kt+[A]_0$
2	$rate=k[A]^2$	$\frac{1}{[A]}=kt+\frac{1}{[A]_0}$

If the order of the reaction is not known, the integral rate equation that gives a straight line with the experiment data is the one that describes the order of the reaction. The integral method can also be used together with the initial rate method and the isolation method in determining the unknowns of a rate equation with multiple reactants.

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November 4, 2019September 25, 2024

Initial rate method (chemical kinetics)

The rate of a reaction decreases with time because the concentrations of reactants decrease as a reaction progresses. This poses a problem when we want to compare reactions under different conditions, e.g. to compare the reaction rates for different concentrations of HCl on CaCO₃.

The diagram above shows the pressure of CO₂ liberated over time for the reaction of CaCO₃ with HCl at three different concentrations: A, B and C. Instead of arbitrarily choosing points on each curve to determine the gradients and hence the respective reaction rates (upper left graph), a basis of comparison must be established before finding the gradients. The initial rate of reaction (i.e. the gradient at t ≈ 0, just after the start of the reaction) is eventually chosen as the reference point (upper right graph), because the concentrations of HCl are approximately the same as that before the start of the respective reactions, especially for slow reactions. It is also chosen partly to minimise the effects of temperature (from exothermic or endothermic reactions) on the system when the reaction progresses. Furthermore, the initial rate of reaction method is useful for studying the forward reaction rate of a reversible reaction where the reverse reaction is negligible at the start. The initial rate method can be used to determine rate equations involving either a single or multiple reactants.

To illustrate the method, consider a constant volume reaction with the rate law:

$R=k[A]^i\; \; \; \; \; \; \; \; 25$

The experiment is carried out with four different initial concentrations of the reactant, A, with the data plotted in a concentration versus time graph (see diagram below).

Since,

$R=\frac{change\, in\, concentration\, of \, reactants\, or\, products}{\, change\, in\, time}$

the four gradients of the curves a, b, c and d are the rates of the respective reactions at t = 0, i.e. the initial rates. Taking the natural logarithm of both sides of eq25, we have

$lnR=lnk+iln[A]\; \; or\; \; lnR_0=lnk+iln[A]_0\; \; \; \; \; \; \; \; 26$

where R₀ and [A]₀ are the initial rate of reaction and initial concentration of A respectively. If we plot the natural logarithm of the initial reaction rates of the four curves against the natural logarithm of the respective initial concentrations of A, we get a straight line with gradient i and vertical intercept lnk. Therefore, the rate constant and the order of eq25, and hence the entire rate law, can be found.

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November 4, 2019October 14, 2024

Isolation method (chemical kinetics)

Isolation methods are used for reactions with multiple reactants, e.g.

$A+B\rightarrow C+D$

$rate=k\left [ A \right ]^{i} \left [ B \right ]^{j}\; \; \; \; \; \; \; \; 35$

There are two isolation methods to determine the rate of a reaction. The first, which is used together with the initial rate method, involves varying the concentration of one of the reactants A, while keeping the concentration of the other reactant B constant. The initial rates of these reactions are then determined. Since the concentration of Bis constant for all the experiments and that initial rates are used, eq35 becomes

$initial\: rate=k\, '\left [ A \right ]_0^{\: i}\; \; \; \; \; \; \; \; 36$

where $k\, '=k\left [ B \right ]_0^{\: j}$

By isolating the reactant of interest, A, eq35 has changed from an (i + j) order equation to a pseudo-i order equation. Taking the natural logarithm of both sides of eq36 and plotting the experimental results in a ln(initial rate) versus ln[A]₀graph, we can find the gradient of the curve, which is the order with respect to A, and the vertical intercept of the curve, which corresponds to ln(k[B]₀ ^j). To find j and hence the value of k, we repeat the series of experiments by varying concentrations of B while keeping the concentration of A constant, and carrying out the same graphical analysis as before.

The second isolation method involves using an excess amount of one reactant, e.g. B, such that its concentration remains unchanged during the prolonged course of the reaction. Eq35 becomes

$rate=k''\left [ A \right ]^{i}\; \; \; \; \; \; \; \; 37$

$ln\left ( rate \right )=iln\left [ A \right ]+lnk''\; \; \; \; \; \; \; \; 38$

where $k''=k\left [ B \right ]^{j}$

With just a single experiment, the gradients at various points on the curve of a [A] versus time graph correspond to the reaction rates at different [A]. These data are then used to construct another graph with eq38 to determine i and lnk”. Similarly, to find k and j, we repeat the experiment, this time with [A] in excess and carry out the same graphical analysis as before.

Question

Can the first isolation method be used without the initial rate method, i.e. can we, with the first isolation method, let the reaction run its course as per the situation in eq37?

Answer

No. This is because the concentration of the reactant that is kept constant is not in excess. The concentration of that reactant will change during the course of the reaction, rendering k” a variable. In other words, the concentration of the reactant that is kept constant in the first method is only constant during the duration necessary for the initial rate of the reaction to be taken.

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November 4, 2019December 12, 2024

Why is initial rate proportional to 1/t?

The ‘initial rate proportional to 1/t’ concept is applicable to a chemical kinetics experiment that has the initial concentration of a reactant varied but the measured amount of a product fixed. One such experiment is the decomposition of hydrogen peroxide over a platinum catalyst, while another is the iodine clock. For the decomposition experiment, we have:

$2H_2O_2(aq)\rightleftharpoons2H_2O(l)+O_2(g)$

Using various initial concentrations of H₂O₂, different times taken to collect a fixed volume of oxygen are recorded. The diagram below illustrates the setup, where the fixed volume of oxygen produced under constant temperature is measured by a fixed change in pressure of the constant-volume system.

Since the reaction is very slow, we assume that the initial concentration of H₂O₂ for each experiment remains relatively unchanged for the volume of oxygen generated (the change in pressure of the system can be fixed at an amount that does not exceed three minutes to occur). Using eq10, and assuming gases behave ideally, we can write

$initial\, rate=\frac{d[O_2]}{dt}=\frac{\Delta (n_{O_2}/V)}{\Delta t}=\frac{\Delta p}{RT\Delta t}\; \; \; \; \; \; \;\; 27$

To produce a fixed volume of oxygen, the change in amount of oxygen, and hence the change in pressure of the system, is constant regardless of the concentration of H₂O₂ used. Thus,

$initial\, rate=C\frac{1}{\Delta t}\; \; \; \; \; \; \; \; 28$

where C is a constant and is equal to $\frac{\Delta p}{RT}$ .

This answers the question of why initial rate is proportional to 1/t.

To develop an equation that that allows us to further analyse the decomposition reaction, we propose the following rate law:

$initial\, rate=k[H_2O_2]_0^{\;\: i}\; \; \; \; \; \; \; \; 29$

where [H₂O₂]₀is the initial concentration of H₂O₂.

Taking the natural logarithm for both sides of eq28 and eq29, we have

$ln(initial\, rate)=ln\frac{1}{\Delta t}+ln\frac{\Delta p}{RT}\; \; \; \; \; \; \; \; 30$

and

$ln(initial\, rate)=iln[H_2O_2]_0+lnk\; \; \; \; \; \; \; \; 31$

Substitute eq31 in eq30

$ln\frac{1}{\Delta t}=iln[H_2O_2]_0+ln\frac{k}{\Delta p/RT}\; \; \; \; \; \; \; \; 32$

Therefore, using the time measurements for the fixed change in pressure to occur for different initial concentrations of H₂O₂, a plot of $ln\frac{1}{\Delta t}$ versus $ln[H_2O_2]_0$ gives a gradient of the order of the reaction, i, and a vertical intercept of $ln\frac{k}{\Delta p/RT}$ where k is the rate constant.

Question

Eq32 is derived using the rate equation $\frac{d[O_2]}{dt}=k[H_2O_2]_0^{\; \, i}$ . What if we define the rate equation as $-\frac{1}{2}\frac{d[H_2O_2]}{dt}=k_1[H_2O_2]_0^{\; \, i}$ ?

Answer

$-\frac{1}{2}\frac{\Delta n_{H_2O_2}/V_{H_2O_2}}{\Delta t}=k_1[H_2O_2]_0^{\: \, i}$

$ln\left ( -\frac{1}{2}\frac{\Delta n_{H_2O_2}}{V_{H_2O_2}} \right )+ln\frac{1}{\Delta t}=lnk_1+iln[H_2O_2]_0$

$ln\frac{1}{\Delta t}=iln[H_2O_2]_0+ln\left ( \frac{k_1}{-\Delta n_{H_2O_2}/2V_{H_2O_2}} \right )\; \; \; \; \; \; \; \; 33$

To produce a fixed volume of oxygen, the change in amount of H₂O₂, $\Delta n_{H_2O_2}$ , is a constant value regardless of the initial amounts of H₂O₂. For example, to produce 1 mole of oxygen, the change in amount of H₂O₂ is always -2 moles irrespective of how many moles of H₂O₂ there are to begin with. Furthermore, the same volume of H₂O₂is used as we vary [H₂O₂]. So, the 2^nd term on RHS of eq33 is a constant, which makes eq33, like eq32, a linear function.

Comparing eq33 with eq32,

$k_1=-\frac{\Delta n_{H_2O_2}RT}{2V_{H_2O_2}\Delta p}k$

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November 4, 2019September 25, 2024

Arrhenius equation (chemical kinetics)

The Arrhenius equation was conceived by Svante Arrhenius, a Swedish chemist, in 1889.

It relates the rate constant, k, of a reaction with the temperature of the reaction, T, as follows:

$k=Ae^{-\frac{E_a}{RT}}\; \; \; \; \; \; \; \; 39$

where E_a is the activation energy of the reaction, R is the universal gas constant and A is the pre-exponential factor. The values of E_a and A of a reaction are experimentally determined.

The Arrhenius equation is based on the van’t Hoff equation: $\Delta_rH^o=RT^2\frac{dlnK}{dT}$ . Consider a reversible elementary reaction $A+B\; \; \begin{matrix} k_1\\\rightleftharpoons \\ k_2 \end{matrix}\; \; P$ , where k₁and k₂ are the rate constants of the forward and reverse reactions respectively. At equilibrium, $rate_1=rate_2$ and therefore, $k_1[A][B]=k_2[P]$ or $\frac{k_1}{k_2}=\frac{[P]}{[A][B]}=K$ . Substituting $K=\frac{k_1}{k_2}$ in the van’t Hoff equation,

$\Delta_rH^o=RT^2\frac{dlnk_1}{dT}-RT^2\frac{dlnk_2}{dT}$

If we define $RT^2\frac{dlnk}{dT}=E_a$ ,

$\Delta_rH^o=E_{a1}-E_{a2}\; \; \; \; \; \; \; \; 40$

Eq40 is best interpreted using a potential energy graph:

The definition of activation energy, $E_a=RT^2\frac{dlnk}{dT}$ , results in eq39 when integrated. Taking the natural logarithm on both sides of eq39,

$lnk=lnA-\frac{E_a}{RT}\; \; \; \; \; \; \; \; 41$

Eq41 is a linear function with dependent variable $lnk$ and independent variable $\frac{1}{T}$ , so that a plot of $lnk$ versus $\frac{1}{T}$ gives a gradient of $-\frac{E_a}{R}$ and vertical intercept of $lnA$ .

Question

Are the activation energy E_a and pre-exponential factor A of a reaction independent of temperature?

Answer

For the Arrhenius equation to be applicable, a plot of lnk versus 1/T must, with strong linear correlation, produce a straight line when values of k at various T are substituted in eq41. This implies that the activation energy E_a and pre-exponential factor A of a reaction are constants and therefore independent of temperature. In fact, the Arrhenius equation works reasonably well for many reactions over a temperature range of about 100 K. However, deviations from the equation do occur for some other reactions.

A more rigorous approach to analyse the relation between k and T using the transition state theory (TST) reveals the temperature-dependence of the activation energy of a reaction, with $E_a=\Delta^{\ddagger}H^o+xRT$ , where x = 1 for unimolecular gas-phase reactions and x = 2 for bimolecular gas-phase reactions. The TST also shows that the pre-exponential factor is dependent on temperature, where $A=\frac{\kappa e^xkRT}{p^oh}^2e^{\frac{\Delta^{\ddagger}S^o}{R}}$ .

So why does the Arrhenius equation work for so many reactions when both the activation energy and pre-exponential factor of a reaction are temperature-dependent? For E_a, the value $\Delta^{\ddagger}H^o$ is usually much bigger than xRT, and for A, T²is dwarfed by the term $\frac{\kappa e^xkR}{p^oh}e^{\frac{\Delta^{\ddagger}S^o}{R}}$ .

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November 4, 2019September 25, 2024

Iodine clock (chemical kinetics)

The iodine clock is a chemical clock experiment devised by Hans Landolt, a Swiss chemist, in 1886. It illustrates the theories of chemical kinetics and redox chemistry via reactions that generate aqueous iodine as a product, and manifests the concept of a chemical clock through the time taken by one or more parallel reactions to consume the iodine produced.

An example is the oxidation of iodide by hydrogen peroxide to form aqueous iodine:

$H_2O_2+2H^++2I^-\rightarrow 2H_2O+I_2\; \; \; \; \; \; \; \; 46$

The iodine produced immediately reacts with a known amount of reducing agent, e.g. sodium thiosulphate, which is added to the peroxide-iodide mixture together with some starch solution at the start of the reaction.

$2S_2O_3^{\; \: 2-}+I_2\rightarrow S_4O_6^{\; \: 2-}+2I^-\; \; \; \; \; \; \; \; 47$

As the thiosulphate-iodine reaction occurs at a much faster rate than the peroxide-iodide reaction and it regenerates the exact stoichiometric amount of iodide ions that is present at the start of the reaction, the presence of iodine is only detected when all of the thiosulphate has been reacted.

When there is no more thiosulphate to remove the iodine, iodine combines with the unreacted iodide to form the triiodide ion, which in turn forms an intense blue-black charge-transfer complex with starch:

$I_2+I^-\rightleftharpoons I_3^{\; \; -}\; \; \begin{matrix} starch\\\rightarrow \end{matrix}\; \; coloured\, complex$

The different times needed for the mixtures to turn blue-black are recorded for a series of experiments using different concentrations of hydrogen peroxide, acid and iodide. For a constant amount of thiosulphate, the longer the time taken for the mixture to turn blue-black, the slower the peroxide-iodide reaction. In other words, the rate of the peroxide-iodide reaction is inversely proportional to the time taken for the mixture to turn blue-black.

Another version of the iodine clock experiment is the reaction between ammonium persulphate and potassium iodide:

$S_2O_8^{\; \; 2-}+2I^-\rightarrow 2SO_4^{\; \; 2-}+I_2$

$I_2+2S_2O_3^{\; \; 2-}\rightarrow2I^-+S_4O_6^{\; \; 2-}$

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November 4, 2019November 30, 2024

Reaction mechanism (chemical kinetics)

A reaction mechanism is a stepwise molecular depiction of the way reactants interact to formed products in a chemical reaction. It is represented by one or more balanced chemical equations depending on whether the type of reaction it is describing is an elementary reaction or a complex reaction, respectively.

An elementary reaction is a chemical reaction where one or more species react in a single step to form products via a transition state. An example is the thermal decomposition of phosphorous (V) chloride:

$PCl_5\rightleftharpoons PCl_3+Cl_2$

Elementary reactions are classified according to their molecularity, which is the number of molecules reacting to form products in an elementary reaction.

Molecularity	Mechanism	Rate law	Order	Example
Unimolecular	$A\rightarrow P$	$rate=k[A]$	1^st	$PCl_5\rightleftharpoons PCl_3+Cl_2$
Bimolecular	$A+A\rightarrow P$	$rate=k[A]^2$	2^nd	$Br+Br\rightarrow Br_2$
Bimolecular	$A+B\rightarrow P$	$rate=k[A][B]$	2^nd	$CH_3Br+OH^-\rightleftharpoons CH_3OH+Br^-$
Trimolecular or Termolecular	$A+A+A\rightarrow P$	$rate=k[A]^3$	3^rd	$Cl+Cl+Cl\rightarrow Cl_2+Cl$
	$A+A+B\rightarrow P$	$rate=k[A]^2[B]$		$NO+NO+O_2\rightarrow 2NO_2$
	$A+B+C\rightarrow P$	$rate=k[A][B][C]$		$H+O_2+M\rightarrow HO_2+M$

A unimolecular reaction involves a species rearranging its bonds or dissociating to form products. Bimolecular and termolecular reactions occur when two and three molecules respectively collide to form products. The probability of elementary reactions involving the simultaneous collision of more than three molecules is very low.

From the table above, the rate law and hence the order of an elementary reaction can be predicted from the molecularity of the reaction, where unimolecular, bimolecular and termolecular reactions are, overall, first, second and third order reactions respectively.

As mentioned above, one or more species in an elementary reaction react via a transition state to form the products. The transition state is a point of highest potential energy on the potential energy profile curve of a reaction. It is where an unstable, transient and non-isolable chemical structure called an activated complex is formed (see diagram below).

For example, the bimolecular reaction between bromomethane and hydroxide ion to give methanol proceeds through the activated complex (denoted by $[\; \; ]^{\ddagger}$ ), where the C-Br bond weakens and the C-OH bond starts to form (see diagram below).

Complex reactions, on the other hand, occur through a series of steps. An example is the acid-catalysed iodination of propanone:

$CH_3COCH_3+I_2\; \; \begin{matrix} H^+\\\rightarrow \end{matrix}\; \; CH_3COCH_2I+HI$

The rate of the reaction depends on the concentration of two species but, surprisingly, does not depend on the concentration of iodine:

$rate=k[CH_3COCH_3][H^+]\; \; \; \; \; \; \; \; 42$

The mechanism of the reaction consists of four steps, each with a particular reaction rate (see diagram below).

The overall rate of reaction is dependent on the slowest step, which is called the rate determining step. As the rate determining step for the above reaction is the rearrangement of the first intermediate CH₃C(OH)CH₃⁺ to the second intermediate CH₃C(OH)CH₂, the proposed rate law is:

$rate=k_2\left [ CH_3C(OH)CH_3^+ \right ]\; \; \; \; \; \; \; \; 43$

However, in chemical kinetics, we are interested in analysing the effect of varying concentrations of reactants on the rate of the reaction. Therefore, we modify eq43 to reflect the concentrations of reactants by noting that

- k₂ « k_-1, such that we assume [CH₃C(OH)CH₃⁺] attains a constant value before the conversion of CH₃C(OH)CH₃⁺ to CH₃C(OH)CH₂ commences.
- The forward and reverse rates for the first step are equal at equilibrium and hence:

$k_1[CH_3COCH_3][H^+]=k_{-1}[CH_3C(OH)CH_3^{\; \: +}]$

$[CH_3C(OH)CH_3^{\; \: +}]=\frac{k_1}{k_{-1}}[CH_3COCH_3][H^+]\; \; \; \; \; \; \; \; 44$

Substitute eq44 in eq43

$rate=k[CH_3COCH_3][H^+]\; \; \; \; \; where\; \; k=\frac{k_1k_2}{k_{-1}}$

which is eq42.

An energy profile diagram of this reaction (see diagram above) shows four activated complexes with energies coinciding with the peaks (denoted by green dots) and three intermediates with potential energies matching the three troughs of the curve (denoted by red dots). It is important to note that an intermediate is a relative stable and isolable chemical species that exists for a finite time, while an activated complex is an unstable, non-isolable chemical structure that has only a transient existence.

Question

Propose a two-step mechanism for the reaction $2NO+O_2\rightarrow 2NO_2$ , which has the experimentally observed rate law: $rate=k[NO]^2[O_2]$ .

Answer

The rate determining step must involve O₂ or NO or both. Let’s suppose it is the final step of the mechanism and has O₂ as a reactant:

$M+O_2\; \; \begin{matrix} k_2\\\rightarrow \end{matrix}\; \; 2NO_2$

where M is an unknown intermediate molecule.

This implies that the preceding step is possibly:

$2NO\; \begin{matrix} k_1\\\rightleftharpoons \\ k_{-1} \end{matrix}\; M$

Balancing the above equations reveals the intermediate as: N₂O₂ . To check whether the proposed mechanism is reasonable, we write the rate law for the rate determining step:

$rate=k_2[N_2O_2][O_2]\; \; \; \; \; \; \; \; 45$

If we assume k₂ « k_-1 such that [N₂O₂] attains a constant value before it reacts with O₂, we can substitute the equilibrium expression of $k_1[NO][NO]=k_{-1}[N_2O_2]$ in eq45 to give:

$rate=k[NO]^2[O_2]\; \; \; \; \; where\; k=\frac{k_1k_2}{k_{-1}}$

To further validate the proposed mechanism, we need to isolate the intermediate.

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November 4, 2019January 25, 2025

Iodine clock experiment (chemical kinetics)

A typical iodine clock experiment seeks to determine the rate law of a reaction.

For the peroxide-iodide reaction $H_2O_2+2H^++2I^-\rightarrow 2H_2O+I_2$ , a proposed rate law is:

$rate=k[H_2O_2]^x[H^+]^y[I^-]^z\; \; \; \; \; \; \; \; 48$

Since the rate of the peroxide-iodide reaction is relatively slow, the method of initial rates can be used to determine the rate law. Let’s suppose a series of peroxide-iodide experiments is conducted at room temperature with the following parameters:

Experiment	Flask 1			Flask 2
Experiment	0.040M H₂O₂/cm³	0.05M H₂SO₄/cm³	Starch/ drops	0.010M KI /cm³	0.001M Na₂S₂O₃/cm³	H₂O/cm³
1	10	10	3	10	10	10
2	10	10	3	20	10	–
3	20	10	3	10	10	–
4	10	20	3	10	10	–

Water is added to experiment 1 so that all four experiments are conducted under the same constant volume. For every experiment, a stopwatch is turned on once the contents of the two flasks are mixed together and turned off when the mixture turns blue-black.

$H_2O_2+2H^++2I^-\rightarrow 2H_2O+I_2\; \; \; \; \; \; \; \; 49$

$2S_2O_3^{\; \: 2-}+I_2\rightarrow S_4O_6^{\; \: 2-}+2I^-\; \; \; \; \; \; \; \; 50$

When the mixture turns blue-black, the initial amount of 1.0×10^-5 moles of thiosulphate in each experiment has reacted with 5.0×10^-6 moles of iodine (eq50), which in turn required 1.0×10^-5 moles of iodide to produce (eq49). This means that we are measuring the time for a constant amount of 5.0×10^-6 moles of iodine to form in each experiment. Hence, we can rewrite eq48 as:

$\frac{d[I_2]}{dt}=\frac{\frac{n_{I_{2,f}}}{V}-\frac{n_{I_2,i}}{V}}{t_f-t_i}=k[H_2O_2]^x[H^+]^y[I^-]^z$

Since $n_{I_2,f}=5.0\times10^{-6}\: moles$ , $n_{I_2,i}=0$ , $V\approx0.05\, dm^3$ , and $t_i=0$

$\frac{\frac{5.0\times10^{-6}}{0.05}}{t_f}=k[H_2O_2]^x[H^+]^y[I^-]^z\; \; \; \; \; \; \; \; 51$

Comparing eq51 and eq48, $rate\propto\frac{1}{t_f}$ . Let’s assume that the experiment is completed with the different times $t_f$ recorded in the table below:

Experiment	Flask 1			Flask 2			Time (t_f)/s	(1/t_f)/10^-2 s^-1
Experiment	0.040M H₂O₂/cm³	0.05M H₂SO₄/cm³	Starch/ drops	0.010M KI /cm³	0.001M Na₂S₂O₃/cm³	H₂O/cm³	Time (t_f)/s	(1/t_f)/10^-2 s^-1
1	10	10	3	10	10	10	21.5	4.65
2	10	10	3	20	10	–	11.2	8.93
3	20	10	3	10	10	–	10.4	9.62
4	10	20	3	10	10	–	22.3	9.18

Comparing experiments 1 and 2, doubling the concentration of KI increases the reaction’s initial rate by $\frac{8.93}{4.65}=1.92$ or approximately two fold. From experiments 1 and 3, doubling the concentration of H₂O₂ increases the reaction’s initial rate by $\frac{9.62}{4.65}=2.07$ or approximately two fold. With reference to experiments 1 and 4, doubling the concentration of the acid increases the initial rate of the reaction by $\frac{9.17}{4.65}=1.97$ or approximately two fold. Notice that the factor, $\frac{5.0\times10^{-6}}{0.05}$ , disappears when we divide $\frac{\frac{5.0\times10^{-6}}{0.05}}{t_f}$ from different experiments to compare the initial reaction rates. Therefore, we use $\frac{1}{t_f}$ instead of $\frac{\frac{5.0\times10^{-6}}{0.05}}{t_f}$ to represent the initial reaction rates and conclude that the rate law is first order with respect to each reactant:

$rate=k[H_2O_2][H^+][I^-]$

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October 30, 2019September 25, 2024

Real gases: overview

Real gases deviate from ideal behavior under high pressure and low temperature, exhibiting complex interactions and properties that are essential for understanding their behavior in various scientific and industrial applications.

The ideal gas law functions well under conditions when gas particles are regarded as:

1. in constant random motion.
2. point masses with zero volume.
3. very far apart from one another
4. devoid of intermolecular forces of attraction or repulsion.
5. perfectly elastic.

Real gases, however, contain molecules that occupy a certain volume that is not negligible compared to the volume of the reaction vessel. Furthermore, intermolecular forces of attraction and repulsion between real gas molecules can be significant under certain conditions. Therefore, real gases deviate from ideality and require other equations of state to describe.

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