To speed up the reaction, you need to increase the number of the very energetic particles present at any particular instant - those with energies equal to or greater than the activation energy. Increasing the temperature has exactly that effect - it changes the shape of the graph. In the next diagram, the graph labelled T is at the original temperature. The graph labelled T+t is at a higher temperature.  If you now mark the position of the activation energy, you can see that although the curve hasn't moved very much overall, there has been such a large increase in the number of the very energetic particles that many more now collide with enough energy to react. Remember that the area under a curve gives a count of the number of particles. On the last diagram, the area under the higher temperature curve to the right of the activation energy looks to have at least doubled - therefore at least doubling the rate of the reaction. Summary Increasing the temperature increases reaction rates because of the disproportionately large increase in the number of high energy collisions. It is only these collisions (possessing at least the activation energy for the reaction) which result in a reaction. © Jim Clark 2002 (last modified October 2018) You learned in Chapter 3 "Chemical Reactions" that a catalyst is a substance that participates in a chemical reaction and increases the reaction rate without undergoing a net chemical change itself. Consider, for example, the decomposition of hydrogen peroxide in the presence and absence of different catalysts (Figure 14.4 "The Effect of Catalysts on Reaction Rates"). Because most catalysts are highly selective, they often determine the product of a reaction by accelerating only one of several possible reactions that could occur.
Figure 14.4 The Effect of Catalysts on Reaction Rates A solution of hydrogen peroxide (H2O2) decomposes in water so slowly that the change is not noticeable (left). Iodide ion acts as a catalyst for the decomposition of H2O2, producing oxygen gas. The solution turns brown because of the reaction of H2O2 with I−, which generates small amounts of I3− (center). The enzyme catalase is about 3 billion times more effective than iodide as a catalyst. Even in the presence of very small amounts of enzyme, the decomposition is vigorous (right). Most of the bulk chemicals produced in industry are formed with catalyzed reactions. Recent estimates indicate that about 30% of the gross national product of the United States and other industrialized nations relies either directly or indirectly on the use of catalysts.
Factors that influence the reaction rates of chemical reactions include the concentration of reactants, temperature, the physical state of reactants and their dispersion, the solvent, and the presence of a catalyst.
where M is any molecule, including cyclopropane. Only those cyclopropane molecules with sufficient energy (denoted with an asterisk) can rearrange to propylene. Which step determines the rate constant of the overall reaction?
When we are trying to decide which step is a rate-determining step, you have to look at which step is the slowest of them all, since we have no knowledge of the time scale for either of these steps we have to look elsewhere to determine the rate-determining step. In our specific problem, we do not know what exact molecules we have however we do see that the first step is an equilibrium since it has a double-headed arrow with two k values one that is the reciprocal of the other. Since it is an equilibrium we can reasonably assume that energized propane is rapidly created which would lead us to the conclusion that the second step is the slower of the two steps and is, therefore, the rate-determining step. So, the second step (with k2) determines the rate constant of the overall reaction.
Above approximately 500 K, the reaction between NO2 and CO to produce CO2 and NO follows the second-order rate law Δ[CO2]/Δt = k[NO2][CO]. At lower temperatures, however, the rate law is Δ[CO2]/Δt = k′[NO2]2, for which it is known that NO3 is an intermediate in the mechanism. Propose a complete low-temperature mechanism for the reaction based on this rate law. Which step is the slowest?
Given that NO3 is an intermediate in the low temperature reaction mechanism, we automatically know two things: 1) NO3 won't show up in the final overall reaction, and thus, 2) NO3 will be in the products of the first reaction of the mechanism and in the reactants of the second reaction. We can also assume that since we're given an intermediate from the problem, this is the only intermediate (so we won't have to dream up any other compounds that might exist in the series of reactions.) These things being said, the lower temperature reaction mechanism will look like this: \[2NO_2(g) \rightarrow NO_3(g) + NO(g) \tag{1}\] \[CO(g) + NO_3(g) \rightarrow CO_2(g) + NO_2(g) \tag{2}\] And the overall reaction will look like this (notice how NO3 is not present): \[CO(g) + NO_2(g) \rightarrow CO_2(g) + NO(g) \tag{overall reaction}\] Now that we have the reaction mechanism written out, we can go about determining which step is the slowest. It would be pretty tricky to do this if we weren't given any further information, however, we know two more things: 1) the high-temperature mechanism's rate = k[NO2][CO] meaning that it took place in one step (given that the overall reaction is also equal to this rate) and 2) the low-temperature mechanism's rate = k′[NO2]2. These two things being said, we've both confirmed that our proposed low-temperature reaction mechanism is in fact two steps, and that we have a means to find which step is slower. By using the "guess-and-check" method we can label each step reaction one at a time as the "slow reaction" and see if the rate matches up with the rate given to us. Let's first try the 2nd reaction. (see above) By using rate laws we can determine that the rate of the reaction must be in terms of its reactants, which follows: \[\textrm{k}\textrm{[CO][NO}_3\textrm{]}\] ... But wait! We can't have the overall reaction rate in terms of an intermediate. By looking at the 1st reaction, we can determine that we can sub in "[NO2] /[NO]" for "[NO3]" since by writing the full reaction rate of the first step and solving for [NO3] this is equivalent. So, we now have the overall rate of the mechanism as the following given the 2nd reaction is the "slow reaction:" \[\textrm{k}\dfrac{\textrm{[CO][NO}_2\textrm{]}^2}{\textrm{[NO}\textrm{]}}\] Note that NO2 is raised to the second power to account for the stoichiometry of the balanced reaction. So clearly the 2nd reaction isn't the slow reaction since the rate is not equivalent to what we were given! Let's check the rate of the 1st reaction now... \[\textrm{k}\textrm{[NO}_2\textrm{]}^2\] What do you know... the rates are equal! We have now confirmed that the 1st reaction is the slow reaction equation, since its rate is equivalent to the overall reaction rate.
Nitramide (O2NNH2) decomposes in aqueous solution to N2O and H2O. What is the experimental rate law (Δ[N2O]/Δt) for the decomposition of nitramide if the mechanism for the decomposition is as follows?
Assume that the rates of the forward and reverse reactions in the first equation are equal.
We know that the slowest step of the reaction is the rate determining step, since it usually has the highest activation energy requirement. As a result, the slowest step of the reaction is the experimental rate law we are looking for. Note: since the slowest step is the rate determining step, that usually means there is some intermediate in between. Intermediates should NEVER be a part of the rate law mechanism. $$rate = rate_2 = {k_2}{[O_2NNH^{-}]}$$ Since Nitramide is an intermediate, we must find some way to substitute it. To solve that problem, we look for where Nitramide is produced and consumed. We see that \[rate_1 = {k_1}{[O_2NNH_2]}\] \[rate_{-1} = {k_{-1}}{[O_2NNH^-]}{[H^+]}\] Since these two rates produce and consume the same amount of \(O_2NNH^{-}\) over the same time period, we can set them equal to each other and solve for the intermediate \[rate_1 = rate_{-1}\] \[{k_1}{[O_2NNH_2]} = {k_{-1}}{[O_2NNH^-]}{[H^+]}\] \[{[O_2NNH^-]} = \frac{k_1{[O_2NNH_2]}}{k_{-1}{[H^+]}}\] Substituting this equation back into our original equation gives us \[rate=rate_2=k_2\frac{k_1[O_2NNH_2]}{k_{-1}[H^+]}\] With all of the rate constants (k), we can clean up our equation a little bit by saying \[k = \frac{k_2{k_1}}{k_{-1}}\] Leaving us with \[rate = \frac{k[O_2NNH_2]}{[H^+]}\]
\(\textrm{rate}=k_2\dfrac{k_1[\mathrm{O_2NNH_2}]}{k_{-1}[\mathrm{H^+}]}=k\dfrac{[\mathrm{O_2NNH_2}]}{[\mathrm{H^+}]}\)
The following reactions are given: \[\mathrm{A+B}\overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}}\mathrm{C+D}\] \[\mathrm{D+E}\xrightarrow{k_2}\mathrm F\] What is the relationship between the relative magnitudes of \(k_{−1}\) and \(k_2\) if these reactions have the following rate law? \[\dfrac{Δ[F]}{Δt} = k\dfrac{[A][B][E]}{[C]}\] How does the magnitude of \(k_1\) compare to that of \(k_2\)? Under what conditions would you expect the rate law to be \[\dfrac{Δ[F]}{Δt} =k′[A][B]?\] Assume that the rates of the forward and reverse reactions in the first equation are equal.
First, because we have broken the equations down into elementary steps we can write the rate laws for each step. Step1: \[A+B\xrightarrow[]{k_{1}} C+D\] \[rate=k_{1}[A][B]\] Step 2: \[C+D \xrightarrow[]{k_{-1}} A+B\] \[rate=k_{-1}[C][D]\] Step 3: \[D+E \xrightarrow[]{k_{2}} F\] \[rate=k_{2}[D][E]\] If we add a these steps together we see that we get overall reaction \[A+B+E \rightarrow C+F\] we can see that [D] is an intermediate and \[k_{1}=k_{-1}\]. Since we are not told which steps are fast or slow we need to use Steady State Approximation. If the second step is the slower step (k-1>>k2) then our rate determining step would be \[rate=k_{2}[D][E]\] Since we can only write rate laws in terms of products and reactants we have to rewrite this so that we are not including an intermediate. Assume: rate of [D] formation = rate of its disappearance \[k_{1}[A][B]=k_{-1}[C][D]+k_{2}[D][E]\] \[k_{1}[A][B]=[D](k_{-1}[C]+k_{2}[E])\] Solving for [D] we find that \[[D]= \frac{k_{1}[A][B]}{(k_{-1}[C]+k_{2}[E])}\] now we can use this to substitute the intermediate [D] in the rate law to get an appropriate rate law. \[rate=\frac{k_{2} k_{1}[A][B][E]}{(k_{-1}[C]+k_{2}[E])}\] because we had already established k-1>>k2 we can assume that \[k_{-1}[C]+k_{2}[E]\approx k_{-1}[C]\] this would give us the observed rate law \[\frac{\Delta [F]}{\Delta t}=\frac{k_{2}k_{1}[A][B][E]}{k_{-1}[C]}\] to make this clearer we can set \[k=\frac{(k_{2})(k_{1})}{(k_{-1})}\] and we can then simplify it down to \[\dfrac{Δ[F]}{Δt} = k\dfrac{[A][B][E]}{[C]}\] we can see that all of these rate constants are related by this ratio k2k1/k-1. Since k2 is our rate determining step k-1>>k2 and since k1=k-1 then we can see that k1>>k2. We would expect the rate law to be \[\dfrac{Δ[F]}{Δt} =k′[A][B]?\] if the rate determining step aka the slowest step is that corresponding to (step 1) since the rate law for this step is k1[A][B] and this is the exact the same as the rate law that we were given. 14.7: Catalysis
What effect does a catalyst have on the activation energy of a reaction? What effect does it have on the frequency factor (A)? What effect does it have on the change in potential energy for the reaction?
A catalyst lowers the activation energy of a reaction. Some catalysts can also orient the reactants and thereby increase the frequency factor. Catalysts have no effect on the change in potential energy for a reaction.
How is it possible to affect the product distribution of a reaction by using a catalyst?
A heterogeneous catalyst works by interacting with a reactant in a process called adsorption. What occurs during this process? Explain how this can lower the activation energy.
In adsorption, a reactant binds tightly to a surface. Because intermolecular interactions between the surface and the reactant weaken or break bonds in the reactant, its reactivity is increased, and the activation energy for a reaction is often decreased.
What effect does increasing the surface area of a heterogeneous catalyst have on a reaction? Does increasing the surface area affect the activation energy? Explain your answer.
Identify the differences between a heterogeneous catalyst and a homogeneous catalyst in terms of the following.
S14.7.4
An area of intensive chemical research involves the development of homogeneous catalysts, even though homogeneous catalysts generally have a number of operational difficulties. Propose one or two reasons why a homogenous catalyst may be preferred.
Catalysts are compounds that, when added to chemical reactions, reduce the activation energy and increase the reaction rate. The amount of a catalyst does not change during a reaction, as it is not consumed as part of the reaction process. Catalysts lower the energy required to reach the transition state of the reaction, allowing more molecular interactions to achieve that state. However, catalysts do not affect the degree to which a reaction progresses. In other words, though catalysts affect reaction kinetics, the equilibrium state remains unaffected. Catalysts can be classified into two types: homogenous and heterogeneous. Homogenous catalysts are those which exist in the same phase (gas or liquid) as the reactants, while heterogeneous catalysts are not in the same phase as the reactants. Typically, heterogeneous catalysis involves the use of solid catalysts placed in a liquid reaction mixture. For this question, we will be discussing homogenous catalysts. Most of the times, homogeneous catalysis involves the introduction of an aqueous phase catalyst into an aqueous solution of reactants. One reason why homogeneous catalysts are preferred over heterogeneous catalysts because homogeneous catalysts mix well in chemical reactions in comparison to heterogeneous catalysts. However, homogeneous catalyst is often irrecoverable after the reaction has run to completion.
Consider the following reaction between cerium(IV) and thallium(I) ions: \[\ce{2Ce^{4+} + Tl^+ → 2Ce^{3+} + Tl^{3+}}\] This reaction is slow, but Mn2+ catalyzes it, as shown in the following mechanism: \[\ce{Ce^{4+} + Mn^{2+} → Ce^{3+} + Mn^{3+}}\] \[\ce{Ce^{4+} + Mn^{3+} → Ce^{3+} + Mn^{4+}}\] \[\ce{Mn^{4+} + Tl^{+ }→ Tl^{3+} + Mn^{2+}}\] In what way does \(Mn^{2+}\) increase the reaction rate?
The Mn2+ ion donates two electrons to Ce4+, one at a time, and then accepts two electrons from Tl+. Because Mn can exist in three oxidation states separated by one electron, it is able to couple one-electron and two-electron transfer reactions.
The text identifies several factors that limit the industrial applications of enzymes. Still, there is keen interest in understanding how enzymes work for designing catalysts for industrial applications. Why?
Enzymes are expensive to make, denature and fail at certain temperatures, are not that stable in a solution, and are very specific to the reaction it was made for. However, scientists can use the observations from enzymes to create catalysts that are more effective in aiding the reaction and cost less to produce. Overall, catalysts still play a large part in lowering the activation energy for reactions. Creating new catalysts can help in the improvement of areas such as medical, ecological, and even commercial products.
Most enzymes have an optimal pH range; however, care must be taken when determining pH effects on enzyme activity. A decrease in activity could be due to the effects of changes in pH on groups at the catalytic center or to the effects on groups located elsewhere in the enzyme. Both examples are observed in chymotrypsin, a digestive enzyme that is a protease that hydrolyzes polypeptide chains. Explain how a change in pH could affect the catalytic activity due to (a) effects at the catalytic center and (b) effects elsewhere in the enzyme. (Hint: remember that enzymes are composed of functional amino acids.) Q14.7.10 At some point during an enzymatic reaction, the concentration of the activated complex, called an enzyme–substrate complex (\(ES\)), and other intermediates involved in the reaction is nearly constant. When a single substrate is involved, the reaction can be represented by the following sequence of equations: \[\text {enzyme (E) + substrate (S)} \rightleftharpoons \text{enzyme-substrate complex (ES)} \rightleftharpoons \text{enzyme (E) + product (P)}\] This can also be shown as follows: \[E + S \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} ES \underset{k_{-2}}{\stackrel{k_{2}}{\rightleftharpoons}} E+P\] Using molar concentrations and rate constants, write an expression for the rate of disappearance of the enzyme–substrate complex. Typically, enzyme concentrations are small, and substrate concentrations are high. If you were determining the rate law by varying the substrate concentrations under these conditions, what would be your apparent reaction order?
\(\dfrac{Δ[ES]}{Δt}=−(k_2+k_{−1})[ES]+k_1[E][S]+k_{−2}[E][P] \approx 0\); zeroth order in substrate.
A particular reaction was found to proceed via the following mechanism:
What is the overall reaction? Is this reaction catalytic, and if so, what species is the catalyst? Identify the intermediates
A particular reaction has two accessible pathways (A and B), each of which favors conversion of X to a different product (Y and Z, respectively). Under uncatalyzed conditions pathway A is favored, but in the presence of a catalyst pathway B is favored. Pathway B is reversible, whereas pathway A is not. Which product is favored in the presence of a catalyst? without a catalyst? Draw a diagram illustrating what is occurring with and without the catalyst.
In both cases, the product of pathway A is favored. All of the Z produced in the catalyzed reversible pathway B will eventually be converted to X as X is converted irreversibly to Y by pathway A.
The kinetics of an enzyme-catalyzed reaction can be analyzed by plotting the reaction rate versus the substrate concentration. This type of analysis is referred to as a Michaelis–Menten treatment. At low substrate concentrations, the plot shows behavior characteristic of first-order kinetics, but at very high substrate concentrations, the behavior shows zeroth-order kinetics. Explain this phenomenon. Page 2
Reaction rates are usually expressed as the concentration of reactant consumed or the concentration of product formed per unit time. The units are thus moles per liter per unit time, written as M/s, M/min, or M/h. To measure reaction rates, chemists initiate the reaction, measure the concentration of the reactant or product at different times as the reaction progresses, perhaps plot the concentration as a function of time on a graph, and then calculate the change in the concentration per unit time. The progress of a simple reaction (A → B) is shown in Figure \(\PageIndex{1}\); the beakers are snapshots of the composition of the solution at 10 s intervals. The number of molecules of reactant (A) and product (B) are plotted as a function of time in the graph. Each point in the graph corresponds to one beaker in Figure \(\PageIndex{1}\). The reaction rate is the change in the concentration of either the reactant or the product over a period of time. The concentration of A decreases with time, while the concentration of B increases with time. \[\textrm{rate}=\dfrac{\Delta [\textrm B]}{\Delta t}=-\dfrac{\Delta [\textrm A]}{\Delta t} \label{Eq1} \] Square brackets indicate molar concentrations, and the capital Greek delta (Δ) means “change in.” Because chemists follow the convention of expressing all reaction rates as positive numbers, however, a negative sign is inserted in front of Δ[A]/Δt to convert that expression to a positive number. The reaction rate calculated for the reaction A → B using Equation \(\ref{Eq1}\) is different for each interval (this is not true for every reaction, as shown below). A greater change occurs in [A] and [B] during the first 10 s interval, for example, than during the last, meaning that the reaction rate is greatest at first.
A Video Discussing Average Reaction Rates. Video Link: Introduction to Chemical Reaction Kinetics(opens in new window) [youtu.be] (opens in new window)
We can use Equation \(\ref{Eq1}\) to determine the reaction rate of hydrolysis of aspirin, probably the most commonly used drug in the world (more than 25,000,000 kg are produced annually worldwide). Aspirin (acetylsalicylic acid) reacts with water (such as water in body fluids) to give salicylic acid and acetic acid, as shown in Figure \(\PageIndex{2}\). Because salicylic acid is the actual substance that relieves pain and reduces fever and inflammation, a great deal of research has focused on understanding this reaction and the factors that affect its rate. Data for the hydrolysis of a sample of aspirin are in Table \(\PageIndex{1}\) and are shown in the graph in Figure \(\PageIndex{3}\).
The data in Table \(\PageIndex{1}\) were obtained by removing samples of the reaction mixture at the indicated times and analyzing them for the concentrations of the reactant (aspirin) and one of the products (salicylic acid). The average reaction rate for a given time interval can be calculated from the concentrations of either the reactant or one of the products at the beginning of the interval (time = t0) and at the end of the interval (t1). Using salicylic acid, the reaction rate for the interval between t = 0 h and t = 2.0 h (recall that change is always calculated as final minus initial) is calculated as follows: \[\begin{align*}\textrm{rate}_{(t=0-2.0\textrm{ h})}&=\frac{[\textrm{salicyclic acid}]_2-[\textrm{salicyclic acid}]_0}{\textrm{2.0 h}-\textrm{0 h}} \\&=\frac{0.040\times10^{-3}\textrm{ M}-0\textrm{ M}}{\textrm{2.0 h}}=2.0\times10^{-5}\textrm{ M/h} \end{align*} \nonumber \] The reaction rate can also be calculated from the concentrations of aspirin at the beginning and the end of the same interval, remembering to insert a negative sign, because its concentration decreases: \[\begin{align*}\textrm{rate}_{(t=0-2.0\textrm{ h})}&=-\dfrac{[\textrm{aspirin}]_2-[\textrm{aspirin}]_0}{\mathrm{2.0\,h-0\,h}} \\&=-\dfrac{(5.51\times10^{-3}\textrm{ M})-(5.55\times10^{-3}\textrm{ M})}{\textrm{2.0 h}} \\&=2\times10^{-5}\textrm{ M/h}\end{align*} \nonumber \] If the reaction rate is calculated during the last interval given in Table \(\PageIndex{1}\)(the interval between 200 h and 300 h after the start of the reaction), the reaction rate is significantly slower than it was during the first interval (t = 0–2.0 h): \[\begin{align*}\textrm{rate}_{(t=200-300\textrm{h})}&=\dfrac{[\textrm{salicyclic acid}]_{300}-[\textrm{salicyclic acid}]_{200}}{\mathrm{300\,h-200\,h}} \\&=-\dfrac{(3.73\times10^{-3}\textrm{ M})-(2.91\times10^{-3}\textrm{ M})}{\textrm{100 h}} \\&=8.2\times10^{-6}\textrm{ M/h}\end{align*} \nonumber \]
In the preceding example, the stoichiometric coefficients in the balanced chemical equation are the same for all reactants and products; that is, the reactants and products all have the coefficient 1. Consider a reaction in which the coefficients are not all the same, the fermentation of sucrose to ethanol and carbon dioxide: \[\underset{\textrm{sucrose}}{\mathrm{C_{12}H_{22}O_{11}(aq)}}+\mathrm{H_2O(l)}\rightarrow\mathrm{4C_2H_5OH(aq)}+4\mathrm{CO_2(g)} \label{Eq2} \] The coefficients indicate that the reaction produces four molecules of ethanol and four molecules of carbon dioxide for every one molecule of sucrose consumed. As before, the reaction rate can be found from the change in the concentration of any reactant or product. In this particular case, however, a chemist would probably use the concentration of either sucrose or ethanol because gases are usually measured as volumes and, as explained in Chapter 10, the volume of CO2 gas formed depends on the total volume of the solution being studied and the solubility of the gas in the solution, not just the concentration of sucrose. The coefficients in the balanced chemical equation tell us that the reaction rate at which ethanol is formed is always four times faster than the reaction rate at which sucrose is consumed: \[\dfrac{\Delta[\mathrm{C_2H_5OH}]}{\Delta t}=-\dfrac{4\Delta[\textrm{sucrose}]}{\Delta t} \label{Eq3} \] The concentration of the reactant—in this case sucrose—decreases with time, so the value of Δ[sucrose] is negative. Consequently, a minus sign is inserted in front of Δ[sucrose] in Equation \(\ref{Eq3}\) so the rate of change of the sucrose concentration is expressed as a positive value. Conversely, the ethanol concentration increases with time, so its rate of change is automatically expressed as a positive value. Often the reaction rate is expressed in terms of the reactant or product with the smallest coefficient in the balanced chemical equation. The smallest coefficient in the sucrose fermentation reaction (Equation \(\ref{Eq2}\)) corresponds to sucrose, so the reaction rate is generally defined as follows: \[\textrm{rate}=-\dfrac{\Delta[\textrm{sucrose}]}{\Delta t}=\dfrac{1}{4}\left (\dfrac{\Delta[\mathrm{C_2H_5OH}]}{\Delta t} \right ) \label{Eq4} \]
Consider the thermal decomposition of gaseous N2O5 to NO2 and O2 via the following equation: \[\mathrm{2N_2O_5(g)}\xrightarrow{\,\Delta\,}\mathrm{4NO_2(g)}+\mathrm{O_2(g)} \nonumber \] Write expressions for the reaction rate in terms of the rates of change in the concentrations of the reactant and each product with time. Given: balanced chemical equation Asked for: reaction rate expressions Strategy:
A Because O2 has the smallest coefficient in the balanced chemical equation for the reaction, define the reaction rate as the rate of change in the concentration of O2 and write that expression. B The balanced chemical equation shows that 2 mol of N2O5 must decompose for each 1 mol of O2 produced and that 4 mol of NO2 are produced for every 1 mol of O2 produced. The molar ratios of O2 to N2O5 and to NO2 are thus 1:2 and 1:4, respectively. This means that the rate of change of [N2O5] and [NO2] must be divided by its stoichiometric coefficient to obtain equivalent expressions for the reaction rate. For example, because NO2 is produced at four times the rate of O2, the rate of production of NO2 is divided by 4. The reaction rate expressions are as follows: \(\textrm{rate}=\dfrac{\Delta[\mathrm O_2]}{\Delta t}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t}\)
The contact process is used in the manufacture of sulfuric acid. A key step in this process is the reaction of \(SO_2\) with \(O_2\) to produce \(SO_3\). \[2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)} \nonumber \] Write expressions for the reaction rate in terms of the rate of change of the concentration of each species. Answer \(\textrm{rate}=-\dfrac{\Delta[\mathrm O_2]}{\Delta t}=-\dfrac{\Delta[\mathrm{SO_2}]}{2\Delta t}=\dfrac{\Delta[\mathrm{SO_3}]}{2\Delta t}\)
The instantaneous rate of a reaction is the reaction rate at any given point in time. As the period of time used to calculate an average rate of a reaction becomes shorter and shorter, the average rate approaches the instantaneous rate. Comparing this to calculus, the instantaneous rate of a reaction at a given time corresponds to the slope of a line tangent to the concentration-versus-time curve at that point—that is, the derivative of concentration with respect to time. The distinction between the instantaneous and average rates of a reaction is similar to the distinction between the actual speed of a car at any given time on a trip and the average speed of the car for the entire trip. Although the car may travel for an extended period at 65 mph on an interstate highway during a long trip, there may be times when it travels only 25 mph in construction zones or 0 mph if you stop for meals or gas. The average speed on the trip may be only 50 mph, whereas the instantaneous speed on the interstate at a given moment may be 65 mph. Whether the car can be stopped in time to avoid an accident depends on its instantaneous speed, not its average speed. There are important differences between the speed of a car during a trip and the speed of a chemical reaction, however. The speed of a car may vary unpredictably over the length of a trip, and the initial part of a trip is often one of the slowest. In a chemical reaction, the initial interval typically has the fastest rate (though this is not always the case), and the reaction rate generally changes smoothly over time.
Using the reaction shown in Example \(\PageIndex{1}\), calculate the reaction rate from the following data taken at 56°C: \[2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)} \nonumber \]
Given: balanced chemical equation and concentrations at specific times Asked for: reaction rate Strategy:
A Calculate the reaction rate in the interval between t1 = 240 s and t2 = 600 s. From Example \(\PageIndex{1}\), the reaction rate can be evaluated using any of three expressions: \[\textrm{rate}=\dfrac{\Delta[\mathrm O_2]}{\Delta t}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t} \nonumber \] Subtracting the initial concentration from the final concentration of N2O5 and inserting the corresponding time interval into the rate expression for N2O5, \[\textrm{rate}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t}=-\dfrac{[\mathrm{N_2O_5}]_{600}-[\mathrm{N_2O_5}]_{240}}{2(600\textrm{ s}-240\textrm{ s})} \nonumber \] B Substituting actual values into the expression, \(\textrm{rate}=-\dfrac{\mathrm{\mathrm{0.0197\;M-0.0388\;M}}}{2(360\textrm{ s})}=2.65\times10^{-5} \textrm{ M/s}\) Similarly, NO2 can be used to calculate the reaction rate: \[\textrm{rate}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=\dfrac{[\mathrm{NO_2}]_{600}-[\mathrm{NO_2}]_{240}}{4(\mathrm{600\;s-240\;s})}=\dfrac{\mathrm{0.0699\;M-0.0314\;M}}{4(\mathrm{360\;s})}=\mathrm{2.67\times10^{-5}\;M/s} \nonumber \] Allowing for experimental error, this is the same rate obtained using the data for N2O5. The data for O2 can also be used: \[\textrm{rate}=\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}=\dfrac{[\mathrm{O_2}]_{600}-[\mathrm{O_2}]_{240}}{\mathrm{600\;s-240\;s}}=\dfrac{\mathrm{0.0175\;M-0.00792\;M}}{\mathrm{360\;s}}=\mathrm{2.66\times10^{-5}\;M/s} \nonumber \] Again, this is the same value obtained from the N2O5 and NO2 data. Thus, the reaction rate does not depend on which reactant or product is used to measure it.
Using the data in the following table, calculate the reaction rate of \(SO_2(g)\) with \(O_2(g)\) to give \(SO_3(g)\). \[2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)} \nonumber \]
In this Module, the quantitative determination of a reaction rate is demonstrated. Reaction rates can be determined over particular time intervals or at a given point in time. A rate law describes the relationship between reactant rates and reactant concentrations. Reaction rates are reported as either the average rate over a period of time or as the instantaneous rate at a single time. Reaction rates can be determined over particular time intervals or at a given point in time.
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Why does the reaction rate of virtually all reaction increase with an increase in temperature?
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