CHEM 123 SAPLING LEARNING CHAPTER 14

The rate constant for this zero‑order reaction is 0.0270 M·s−1 at 300 ∘C. A⟶products How long (in seconds) would it take for the concentration of A to decrease from 0.950 M to 0.280 M? 𝑡= 24.81s

Solution The relationship between concentration and time for a zero‑order reaction is [A]𝑡=−𝑘𝑡+[A]0 where [A]𝑡 is the concentration of A at time 𝑡, [A]0 is the initial concentration of A, and 𝑘 is the rate constant. Rearranging gives 𝑡=[A]𝑡−[A]0/−𝑘=[A]0−[A]𝑡/𝑘 Finally, substitute the proper values. 𝑡=(0.950 M−0.280 M)/0.0270 M⋅ s−1=24.8 s

After 20 min, a reactant has decomposed to 85% of its original concentration. Which order would the reaction need to be to allow you to solve for the rate constant, 𝑘 , with only this information?

first order

𝑡 (s) [A] (M) 0.0 0.900 25.0 0.609 50.0 0.461 75.0 0.370 For the reaction A⟶products, concentration and time data were collected. Enter these data into the graphing tool to determine the reaction order. What is the reaction order?

2 0.275

For each of the scenarios, identify the order with respect to the reactant, A. A⟶products The half‑life of A decreases as the initial concentration of A decreases. order: A two‑fold increase in the initial concentration of A leads to a four‑fold increase in the initial rate. order: A two‑fold increase in the initial concentration of A leads to a 1.41‑fold increase in the initial rate. order: The time required for [A] to decrease from [A]0 to [A]0/2 is equal to the time required for [A] to decrease from [A]0/2 to [A]0/4. order: The rate of decrease of [A] is a constant. order:

0 2 0.5 1 0 Solution For the scenarios describing the relationship between initial concentration and initial rate, the relevant rate law is rate=𝑘[A]𝑥 where 𝑥 is the order with respect to A. In the scenario where a two‑fold increase in [A] leads to a four‑fold increase in the rate, 4=2𝑥 and 𝑥=2. In the scenario where a two‑fold increase in [A] leads to a 1.41‑fold increase in the rate, 1.41=2𝑥 and 𝑥=1/2. In the scenario where the rate of decrease of [A] is a constant, rate=𝑘, and therefore 𝑥=0. For the scenarios involving half‑life, the relevant formulas are zero-order:𝑡1/2=[A]/2𝑘 first-order:𝑡1/2=0.693/𝑘 second-order:𝑡1/2=1/𝑘[A] In the scenario where half‑life of A decreases as the initial concentration of A decreases, the half‑life is directly proportional to [A], indicating a zero‑order reaction. Finally, saying that the time required for [A] to decrease from [A]0 to [A]0/2 is equal to the time required for [A] to decrease from [A]0/2 to [A]0/4 is equivalent to saying the half‑life is constant, indicating a first‑order reaction.

Calculate the rate constant, 𝑘, for a reaction at 55.0 °C that has an activation energy of 84.0 kJ/mol and a frequency factor of 7.83×1011 s−1. 𝑘=

0.0332 s−1 Solution Use the Arrhenius equation to calculate the rate constant, 𝑘. 𝑘=𝐴𝑒^−𝐸a/𝑅𝑇 In the given equation, 𝐴 is the frequency factor, 𝐸a is the activation energy, 𝑇 is the temperature, and 𝑅 is the ideal gas constant equal to 8.3145 J/(mol·K). For your units to cancel properly, the activation energy must be expressed in joules per mole and the temperature must be expressed in Kelvin. First, convert the activation energy to joules per mole and the temperature to Kelvin. 84.0 kJ/1 mol×1000 J/1 kJ=8.40×104 Jmol 55.0 °C+273.15=328.15 K Input these values along with the frequency factor and 𝑅 into the Arrhenius equation and solve for 𝑘. 𝑘=(7.83×1011 s−1)𝑒(−8.40×104 Jmol)/((8.3145 Jmol⋅K)(328.15 K))=0.0333 s−1

A reactant decomposes with a half-life of 105 s when its initial concentration is 0.279 M. When the initial concentration is 0.666 M, this same reactant decomposes with the same half-life of 105 s. What is the order of the reaction? What is the value and unit of the rate constant for this reaction? 𝑘=

1 0.00660 s−1 Solution Notice that the half-life of the reaction is constant and independent of the initial concentration. Only the first order half-life equation shows an independent relationship between [A]0 and t1/2. 𝑘=ln(2)/𝑡1/2 Rearranging to isolate k gives 𝑡1/2=ln(2)/𝑘 Insert the half-life to solve for k. 𝑘=ln(2)/105 s=0.00660 s−1

The rate constant for this first‑order reaction is 0.550 s−1 at 400 ∘C. A⟶products How long, in seconds, would it take for the concentration of A to decrease from 0.790 M to 0.280 M? 𝑡=

1.89s Solution The relationship between concentration and time for a first‑order reaction is ln[A]𝑡/[A]0=−𝑘𝑡 where [A]𝑡 is the concentration of A at time 𝑡, [A]0 is the initial concentration of A, and 𝑘 is the rate constant. Rearranging to solve for 𝑡 gives 𝑡=ln[A]𝑡/[A]0/−𝑘 Finally, substitute the appropriate values and solve for 𝑡. 𝑡=ln(0.280 M/0.790 M)/−0.550 s−1=1.89 s

The rate constant for this second‑order reaction is 0.160 M−1⋅s−1 at 300 ∘C. A⟶products How long, in seconds, would it take for the concentration of A to decrease from 0.990 M to 0.300 M? 𝑡=

14.38 s Solution The relationship between concentration and time for a second‑order reaction is 1/[A]𝑡=𝑘𝑡+1/[A]0 where [A]𝑡 is the concentration of A at time 𝑡, [A]0 is the initial concentration of A, and 𝑘 is the rate constant. Rearranging to solve for 𝑡 gives 𝑡=(1/[A]𝑡−1/[A]0)/𝑘 Finally, substitute the proper values and solve for 𝑡. 𝑡=(1/0.300 M − 1/0.990 M) / 0.160 M−1⋅ s−1 = 14.5 s

𝑡 (min) [A] (M) 5.0 0.938 10.0 0.583 15.0 0.423 20.0 0.331 25.0 0.273 For the reaction A⟶products , concentration and time data were collected. What is the reaction order? [A]0=

2 2.40M Solution To determine the reaction order, plot the data in three ways. If [A] vs. 𝑡 is linear, the reaction is zeroth order. If ln[A] vs. 𝑡 is linear, the reaction is first order. If 1/[A] vs. 𝑡 is linear, the reaction is second order. Since the plot of 1/[A] vs. 𝑡 is linear, the reaction is second order. In that plot, the equation for the line is 𝑦=0.129994𝑥+0.416008 which is given in 𝑦=𝑚𝑥+𝑏 form. The integrated rate equation for a second order reaction is 1/[A]𝑡=𝑘𝑡+1/[A]0 which means that 𝑏=1[A]0. 1/[A]0=0.416008 [A]0=1/0.416008=2.40 M

The rate constant for this first-order reaction is 0.0203 s−1 at 300 °C. A⟶products If the initial mass of A is 17.93 g, calculate the mass of A remaining after 1.75 min. mass of A:

2.1g Solution The relationship between concentration and time for a first-order reaction is ln[A]𝑡/[A]0=−𝑘𝑡 where [𝐴]𝑡 is the concentration of A at time 𝑡, [𝐴]0 is the initial concentration of A, and 𝑘 is the rate constant. The initial mass, 𝑚0, and the mass at time 𝑡, 𝑚𝑡, of A can be used in place of the concentrations in the integrated rate law. ln𝑚𝑡/𝑚0=−𝑘𝑡 Rearrange the equation to solve for the mass of A after 1.75 min. 𝑚𝑡/𝑚0=e−𝑘𝑡 𝑚𝑡=𝑚0× e−𝑘𝑡 Because the rate constant is given in inverse seconds, the time must first be converted to seconds. 1.75 min×60 s/1 min=105 s Insert the time, rate constant, and initial mass into the equation and solve for 𝑚𝑡. 𝑚𝑡=(17.93 g)× e−(0.0203 s−1)(105 s)=2.1 g

The rate constant for the reaction is 0.420 M−1⋅s−1 at 200 ∘C. A⟶products If the initial concentration of A is 0.00280 M, what will be the concentration after 215 s? [A]=

2.23×10−3 M Solution The reaction is second order because the units of the rate constant, 𝑘, are M−1⋅s−1. The relationship between concentration and time for a second‑order reaction is 1/[A]𝑡=𝑘𝑡+1/[A]0 where [A]𝑡 is the concentration of A at time 𝑡, [A]0 is the initial concentration of A, and 𝑘 is the rate constant. Rearranging the equation to solve for [A]𝑡 gives [A]𝑡=1/𝑘𝑡+1/[A]0 Finally, substitute the proper values. [A]𝑡=1/(0.420 M−1⋅s−1)(215 s)+1/0.00280 M=0.00223 M

After 44.0 min, 14.0% of a compound has decomposed. What is the half‑life of this reaction assuming first‑order kinetics? 𝑡1/2=

202 min Solution The integrated rate equation for a first‑order reaction is ln[A]𝑡/[A]0=−𝑘𝑡 If 14.0% decomposed, then 86.0% remains. In other words, [A]𝑡=0.860[A]0. ln0.860[A]0/[A]0=−𝑘(44.0 min) Notice that [A]0 cancels out. Solve for 𝑘. ln(0.860)/44.0 min=−𝑘 𝑘=0.00343 min−1 The half‑life, 𝑡1/2, for a first‑order reaction is 𝑡1/2=ln(2)/𝑘 Insert the value of 𝑘 and solve for 𝑡1/2. 𝑡1/2=ln(2)/0.00343 min−1=202 min

Which of the reaction mechanisms is consistent with the energy diagram? Energy is plotted on the y axis versus the reaction progress on the x axis. The reaction starts at some energy and increases to before decreasing slightly. The energy then increases again to a maxima before decreasing to a final energy that is less than the initial energy.

2A−→−fastB B−→−−slowC Solution The energy diagram contains two maxima, which indicates that the reaction contains two elementary steps. The first maximum corresponds to the transition state of the first step and the second maximum corresponds to the transition state of the second step. The activation energy of the first step is less than the activation energy of the second step, meaning the first step proceeds at a faster rate than the second step. Therefore, the most probable reaction mechanism is 2A−→−fastB B−→−−slowC

Time and concentration data were collected for the reaction A⟶products t (s) [A] (M) 0 0.52 20 0.43 40 0.35 60 0.29 80 0.23 100 0.19 The blue curve is the plot of the data. The straight orange line is tangent to the blue curve at 𝑡=40 s. A plot has the concentration of A in molar on the y axis and time in seconds on the x axis. A curve contains the points (0, 0.52), (20, 0.43), (40, 0.35), (60, 0.29), (80, 0.24), and (100, 0.20). A line touches the curve at (40, 0.35) and has a y intercept of (0, 0.48). Approximate the instantaneous rate of this reaction at time 𝑡=40 s. instantaneous rate:

3.25×10−3M/s Solution The instantaneous rate of change of [A] at a particular moment is the slope of the tangent line at that point. Slope is the rise divided by the run. To find the slope of the orange tangent line, identify two sets of coordinates on that line. One point through which it is known that the orange line crosses is 𝑡=40 s and [A]=0.35 M. Another reference point that is relatively easy to read is 𝑡=0 and [A]≈0.48 M. The rise between the two reference points is the difference between the [A] values. rise=(0.35 M)−(0.48 M)=−0.13 M The run between the two reference points is the differnce between the 𝑡 values. run=(40 s)−(0 s)=40 s The slope is rise over run. slope=rise/run =−0.13 M/40 s ≈−0.003 M/s The value is negative because it expresses the rate of change of [A], which is a reactant. The rate of the reaction, however, should be expressed as a poitive value, 0.003 M/s.

Determine the average rate of change of B from 𝑡=0 s to 𝑡=212 s. A⟶2B Time (s) Concentration of A (M) 0 0.600 106 0.390 212 0.180 rateB=

3.96×10−3M/s Solution The reactant, A, changes at a rate of rateA=Δ[A]/Δ𝑡 =(0.180 M−0.600 M)/(212 s−0 s) =−0.00198 M/s The product, B, appears twice as fast as A disappears. −rateA= 1/2 rate rateB= −2× rateA =−2(−0.00198 M/s) =0.00396 M/s

A certain reaction has an enthalpy of Δ𝐻=24 kJ and an activation energy of 𝐸a=56 kJ. What is the activation energy of the reverse reaction? 𝐸a(reverse)=

32 KJ Solution In the diagrams shown, the forward reaction goes from left to right (reactants to products) and the reverse reaction goes from right to left (products to reactants). In the exothermic reaction, the energy of the reactants is greater than the energy of the products. In the endothermic reaction, the energy of the products is greater than the energy of the reactants. In both reactions, the energy reaches a maxima between the reactants and products labeled the transition state. In this case, the reaction is endothermic, as indicated by the sign of Δ𝐻. The activation energy can be described in terms of the transition state, sometimes called the activated complex, and the starting materials. 𝐸a(forward)=𝐸transition−𝐸reactants 𝐸a(reverse)=𝐸transition−𝐸products Taking the difference of the two 𝐸a values for the forward and reverse reactions gives the enthalpy, Δ𝐻. 𝐸a(forward)−𝐸a(reverse)=𝐸products−𝐸reactants=Δ𝐻 Solving for 𝐸a(reverse) gives 𝐸a(reverse)=𝐸a(forward)−Δ𝐻=(56 kJ)−(24 kJ)=32 kJ

Consider the following generic reaction. X+Y⟶Z Each reaction container contains a mixture of X and Y.

A container contains 8 red spheres and 8 blue spheres that are evenly dispersed throughout the container. Solution The collision theory is based on the assumption that for a reaction to occur, the reacting compounds have to come in contact or collide with one another. The rate at which a chemical reaction proceeds is proportional to the frequency of effective collisions. As the concentration of the reactants increase, the frequency with which they collide increases, leading to an increase in the reaction rate. Therefore, the reaction in the container with eight molecules of each reactant will be faster than the reaction in the container with only two molecules of each reactant. In the third container, the reactants are not in contact with each other at all, meaning the reaction in this container will be the slowest, if it happens at all.

Classify the actions based on how they could affect a reaction rate. Rate increases Rate decreases Rate is unaffected

INCREASE THE TEMP. AND ADD A CATALYST DECREASE THE CONCENTRATION OF A REACTANT AND DECREASE THE SURFACE AREA OF A SOLID REACTION Solution A catalyst is a substance that increases a reaction rate by decreasing the activation energy without being consumed in the reaction. Increasing the reaction temperature increases the reaction rate. The effect of temperature on reaction rates is explained by collision theory. Increasing the temperature of a system increases the average speed (kinetic energy) of the reacting molecules. The increased molecular speed causes more collisions to take place within a given time. Decreasing the concentration of a reactant can cause a decrease in the rate of the reaction, depending on the rate order. Because fewer reactant particles are present, the possibility for collisions to occur between the reactant particles decreases. In the solid state, reactions occur at the boundary surface between reactants. The reaction rate decreases as the boundary surface area decreases.

For A⟶products , time and concentration data were collected and plotted as shown. [𝐀] (𝐌) 𝐭 (𝐬) 0.900 00.0 0.474 30.0 0.321 60.0 0.243 90.0 A plot of the concentration of A versus time is concave up and decreases with time. A plot of the natural logarithm of the concentration of A versus time is concave up and decreases with time. A plot of one over the concentration of A versus time increases linearly with time. Determine the reaction order, the rate constant, and the units of the rate constant. order= 2 𝑘= 0.033 [Math Processing Error] (units) Solved

Linear plot Order [A] versus 𝑡 0 ln[A] versus 𝑡 1 1[A] versus 𝑡 2 Use the shape of the given plots to determine the reaction order. In this case, 1[A] versus 𝑡 is a straight line, so the reaction order is 2. The appropriate integrated rate equation is 1/[A]=𝑘𝑡+1/[A]0 where [A]0=0.900 M. [A] can be any of the other concentration values, so long as the value of 𝑡 is taken from that same row. Solving this equation for the rate constant gives 𝑘=(1/[A]−1/[A]0)/𝑡=(1/0.474 M−1/0.900 M)/30 s=0.0333 M−1⋅s−1

The reaction A+B⟶C+Drate=𝑘[A][B]2 has an initial rate of 0.0890 M/s. What will the initial rate be if [A] is halved and [B] is tripled? initial rate: 0.401 M/s What will the initial rate be if [A] is tripled and [B] is halved? initial rate: 0.0668 M/s

Solution According to the rate law, the reaction is first‑order in A and second‑order in B. If the concentration of a first‑order reactant is halved, then the reaction rate changes by a factor of (12)1=12. If the concentration of a second‑order reactant is tripled, then the reaction rate changes by a factor of (3)2=9. The overall reaction rate changes by a factor of 1/2×9. 1/2×9×0.0890 M/s=0.401 M/s If the concentration of a first‑order reactant is tripled, then the reaction rate changes by a factor of (3)1=3. If the concentration of a second‑order reactant is halved, then the reaction rate changes by a factor of (12)2=14. The overall reaction rate changes by a factor of 3×(14). 3×1/4×0.0890 M/s=0.0668 M/s

The reaction described by H2(g)+I2(g)⟶2HI(g) has an experimentally determined rate law of rate=𝑘[H2][I2] Some proposed mechanisms for this reaction are: Mechanism A (1) H2(g)+I2(g)−→𝑘12HI(g) (one-step reaction) Mechanism B (1) I2(g)⥫⥬=𝑘−1𝑘12I(g) (fast, equilibrium) (2) H2(g)+2I(g)−→𝑘22HI(g) (slow) Mechanism C (1) I2(g)⥫⥬=𝑘−1𝑘12I(g) (fast, equilibrium) (2) I(g)+H2(g)−→𝑘2HI(g)+H(g) (slow) (3) H(g)+I(g)−→𝑘3HI(g) (fast) Which of these mechanisms are consistent with the observed rate law? mechanism A mechanism B In 1967, J. H. Sullivan showed that this reaction was dramatically catalyzed by light when the energy of the light was sufficient to break the I−I bond in an I2 molecule. Which mechanism or mechanisms are consistent with both the rate law and this additional observation? mechanism B

Solution For mechanism A, the rate law is rate=𝑘1[H2][I2]=𝑘[H2][I2] This is consistent with the experimentally determined rate law. For mechanism B, the rate law is given by the rate-determining (slow) step rate=𝑘2[H2][I]2 A rate law cannot contain an intermediate. Step 1 can be used to eliminate the intermediate [I] from the rate law. In step 1, the forward and reverse reactions are at equilibrium, therefore 𝑘1[I2]=𝑘−1[I]2 Rearrnge this expression to solve for [I]2. The rate law can be written as rate=𝑘1𝑘2𝑘−1[H2][I2]=𝑘[H2][I2] which is also consistent with the experimentally determined rate law. For mechanism C, the rate law is given by step 2, the rate-determining step. Step 3 is a fast reaction after the rate-determining step (step 2), and it will not influence the rate of reaction. rate=𝑘2[H2][I] As with mechanism B, the intermediate must be removed from the rate law. Eliminate the intermediate [I] from this expression using step 1. In step 1, the forward and reverse reactions are at equilibrium, therefore 𝑘1[I2]=𝑘−1[I]2 Rearrnge this expression to solve for [I]. The rate law can be written as rate=𝑘2(𝑘1/𝑘−1)1/2[H2][I2]1/2=𝑘[H2][I2]1/2 Therefore, only the first two mechanisms are consistent with the observed rate law. The second and third mechanisms involve an I(g) intermediate, meaning both are consistent with the additional observation. However, only the second mechanism is consistent with the rate law. Therefore, only mechanism B is consistent with the experimental observations.

Consider the rate law. rate=𝑘[A]𝑥 Determine the value of 𝑥 if the rate doubles when [A] is doubled. 𝑥= 1 Determine the value of 𝑥 if the rate quadruples when [A] is doubled. 𝑥= 2

Solution In this rate law, 𝑥 represents the order of the reaction with respect to A. A zero‑order relationship means that the rate is independent of concentration. A first‑order relationship means that the rate is directly proportional to concentration. A second‑order relationship means that the rate is directly proportional to the square of the concentration. If the rate doubles when [A] is doubled rate2/rate1=𝑘[A]𝑥2/𝑘[A]𝑥1 2=2𝑥 𝑥=1 If the rate quadruples when [A] is doubled rate2rate1=𝑘[A]𝑥2/𝑘[A]𝑥1 4=2𝑥 𝑥=2

For the reaction A⟶products, concentration and time data were collected. Enter these data into the graphing tool and determine the value and units of the rate constant, 𝑘, for this reaction. 𝑡 (min) [A] (M) 0.0 1.00 3.0 0.76 6.0 0.52 9.0 0.28 𝑘= 0.08 Units of 𝑘: M⋅min−1

Solution Linear plot Order Slope Units of 𝑘 [A] vs. 𝑡 0 −𝑘 M⋅min−1 ln[A] vs. 𝑡 1 −𝑘 min−1 1/[A] vs. 𝑡 2 𝑘 M−1⋅min−1 Start by entering the given data into the graphing tool. Then, click the tabs above the graph to see the three different plots and determine the reaction order. The linear plot is [A] vs. 𝑡 because the points fall most closely along the linear "best fit" line, as indicated by the highest value of 𝑟2. Thus, the reaction is zeroth order. The value of 𝑘 is the absolute value of the slope of the line. To determine the slope, 𝑚, look at the equation of the line. The equation is in the form 𝑦=𝑚𝑥+𝑏, so the slope is the coefficient of 𝑥. In this case, 𝑘=0.080 M⋅min−1 . The units of 𝑘 depend on the reaction order as shown in the table.

Using the data in the table, determine the rate constant of the reaction and select the appropriate units. A+2B⟶C+D Trial [𝐀] (𝐌) [𝐁] (𝐌) Rate (M/s) 1 0.360 0.250 0.0183 2 0.360 0.500 0.0183 3 0.720 0.250 0.0732 𝑘=0.141 Units M−1s−1

Solution Looking at trials 1 and 3, [A] doubles while [B] remains constant. The rate increases by a factor of 4, indicating a second‑order relationship between [A] and the rate. Looking at trials 1 and 2, [B] doubles while [A] remains constant. The rate is not affected, indicating a zero‑order relationship between [B] and the rate. Thus, the rate law is rate=𝑘[A]2 and the rate constant is 𝑘=rate/[A]2=0.0183 M/s/(0.360 M)2=0.141 M−1⋅s−1

A particular reactant decomposes with a half‑life of 161 s when its initial concentration is 0.398 M. The same reactant decomposes with a half‑life of 205 s when its initial concentration is 0.313 M. Determine the reaction order. 2 What is the value and units of the rate constant for this reaction? 𝑘= 0.0156 Units M−1⋅s−1

Solution Notice that the half‑life of the reaction increases as the initial concentration decreases. Only the second‑order half‑life equation shows an inverse relationship between [A]0 and 𝑡1/2. 𝑡1/2=1/𝑘[A]0 Rearranging to isolate 𝑘 gives 𝑘=1/𝑡1/2[A]0 Plug in either set of values to find the rate constant. Both sets of numbers should give identical rate constants, but the values might be slightly different due to rounding. Using the half‑life of 161 s and initial concentration of 0.398 M, 𝑘 is 𝑘=1/(161 s)(0.398 M)=0.0156 M−1⋅s−1 Using the half‑life of 205 s and initial concentration of 0.313 M, 𝑘 is 𝑘=1/(205 s)(0.313 M)=0.0156 M−1⋅s−1

Consider the reaction and its rate law. 2A+3B⟶productsrate=𝑘[B]2 What is the order with respect to A? order: 0 What is the order with respect to B? order: 2 What is the overall reaction order? overall order: 2

Solution The order of a reactant is indicated by its exponent in the rate law. If a reactant is absent from the rate law, it has an order of zero because anything to the zeroth power is equal to 1. In this case, the order with respect to A is 0, and the order with respect to B is 2. Summing these orders gives the overall reaction order. 0+2=2 Unless you know that the given reaction is an elementary step, you cannot assume that the coefficients are equal to the orders. In contrast, the exponents in the rate law always indicate the correct orders of the reactants.

Initial‑rate data at a certain temperature is given in the table for the reaction C2H5Cl(g)⟶C2H4(g)+HCl(g) [𝐂2𝐇5𝐂𝐥]0(𝐌) Initial rate (M/s) 0.100 0.665×10−30 0.200 1.33×10−30 0.300 2.00×10−30 Determine the value and units of the rate constant. 𝑘= 6.67 ×10 −30 units: s−1

Solution The rate law has the general form rate=𝑘[C2H5Cl]𝑥 To find the order with respect to C2H5Cl, compare the first two trials. (1.33×10−30 M/s)/(6.65×10−31 M/s) = 𝑘(0.200 M)𝑥/𝑘(0.100 M)𝑥 Cancel 𝑘, then solve for 𝑥. 2.00=(0.200 M/0.100 M)^𝑥 2.00=2.00 𝑥=1 Since the reaction is first order, the complete rate law is rate=𝑘[C2H5Cl] To find the rate constant, plug in values from any trial, then solve for 𝑘. The first trial is used in this calculation. 6.65×10−31 M/s=𝑘(0.100 M) 𝑘=6.65×10−31 M/s/(0.100 M)=6.65×10−30 s−1

Consider the reaction between nitric oxide and ozone. The reactants are N O and O 3. In the N O molecule a nitrogen atom is double bonded to an oxygen atom. In the O 3 molecule, an oxygen atom is double bonded to a second oxygen atom and single bonded to a third oxygen atom. The products are N O 2 and O 2. In the N O 2 molecule, the nitrogen atom is double bonded to one oxygen atom and single bonded to a second oxygen atom. In the O 2 molecule, the oxygen atoms are connected by a double bond. NO(g)+O3(g)⟶NO2(g)+O2(g) Select a plausible transition state for the reaction of nitric oxide with ozone.

The nitrogen atom in N O collides with the oxygen atom in O 3 that is single bonded to the central oxygen atom. Solution According to the collision theory, a bimolecular reaction will occur when two molecules collide with the proper orientation and sufficient energy. For a reaction to occur between NO and O3, the molecules must be oriented so that the nitrogen atom in NO collides with an oxygen atom in O3. Energetically, the reaction is most plausible if the nitrogen atom collides with one of the terminal oxygen atoms in ozone instead of the central oxygen atom. Therefore, the most plausible transition state is The nitrogen atom in N O collides with the oxygen atom in O 3 that is single bonded to the central oxygen atom.

The rate of a reaction is studied by measuring the concentration of the reactant and the concentration of the product as a function of time. The results are shown in the graph. A graph with time on the x axis and concentration on the y axis. A red curve labeled X, starts at a concentration of zero at time zero. The concentration of X increases as time increases. The shape of the curve is concave down. A blue curve labeled Y, starts at some concentration above zero at time zero and decreases as time increases. The shape of the curve is concave up. The final concentration of Y is near zero. The final concentration of X is approximately twice the initial concentration of Y. Which chemical equation is consistent with the graph?

Y⟶2X Solution When a reaction progresses overtime, the concentrations of reactants are expected to decrease because reactants are consumed in the reaction. The concentration of prooducts are expected to increase with time because products are formed in the reaction. In the graph, the concentration of Y is decreasing and the concentration of X is increasing. Thus, Y is the reactant and X is the product. The graph shows that the final concentration of the reactant Y is approximately zero, and that the final concentration of the product X is approximately twice the initial concentration of the reactant Y. This indicates that 2 mol of X are formed for every 1 mole of Y consumed. Thus, the chemical equation that is consistent with the graph is Y⟶2X

The concentration of X over time is shown in the graph. A graph with time on the x axis and the concentration of X on the y axis. The curve is concave down. There are three points labeled on the curve 1, 2, and 3, where point 1 is closest to time zero. Is X a reactant or product of the reaction? How is the reaction rate changing as time progresses? What conclusion can be drawn about the average rate of the reaction between points 1 and 2 and between points 2 and 3?

product slowing down The rate of the reaction between points 1 and 2 is faster than the rate of reaction between points 2 and 3 because the concentration of reactants decreases. Solution In a chemical reaction, the reactant concentration decreases with time as the reactant is consumed and the product concentration increases as the product is formed. The graph shows that the concentration of X increases with time. Thus, X is a product. The rate of the reaction can be expressed in terms of the concentration of X reaction rate=+1/𝑎 * Δ[X]/Δ𝑡 where Δ[X] is the change in concentration, Δ𝑡 is the change in time and 𝑎 is the coefficient of X in the balanced equation of the reaction. Based on the shape of the curve, the change in the concentration of X is decreasing (getting smaller) over time. Since X is a product, that means that the reaction is slowing down. This can be observed by comparing the rate of the reaction between points 1 and 2 and points 2 and 3 on the graph. The rate at which X is formed is greater between points 1 and 2 than between points 2 and 3, meaning the reaction is faster between points 1 and 2 than between points 2 and 3. This decrease in rate is most likely due to a decrease in the concentration of the reactants (not shown on the graph).

Consider the following balanced chemical equation. 2Fe+3H2O⟶Fe2O3+3H2 How is the rate of appearance of H2, Δ[H2]Δ𝑡, related to the rate of disappearance of Fe?

−3/2(Δ[Fe]/Δ𝑡) Solution The rate of the reaction can be expressed as the decrease in concentration of a reactant over a certain length of time or as the increase in concentration of a product over a certain length of time. For the reaction 𝑎A+𝑏B⟶𝑐C+𝑑D where A and B are reactants, C and D are products, and 𝑎, 𝑏, 𝑐, and 𝑑 are stoichiometric coefficients, the rate of reaction can be expressed as rate=−1/𝑎(Δ[A]/Δ𝑡)=−1/𝑏(Δ[B]/Δ𝑡)=1/𝑐(Δ[C]/Δ𝑡)=1/𝑑(Δ[D]/Δ𝑡) Note that the rate of disappearance of each reactant and rate of appearance of each product are multiplied by the inverse of their coefficients in the balanced equation. Also, the rate of appearance of the products has the same sign as the overall reaction rate, whereas the rate of disappearance of the reactants has the opposite sign. For the reaction 2Fe+3H2O⟶Fe2O3+3H2 the rate of reaction can therefore be expressed as rate=−1/2(Δ[Fe]/Δ𝑡)=−1/3(Δ[H2O]/Δ𝑡)=1/1(Δ[Fe2O3]/Δ𝑡)=1/3(Δ[H2]/Δ𝑡) Simplify the expression to only focus on Fe and H2. rate=−1/2(Δ[Fe]/Δ𝑡)=1/3(Δ[H2]/Δ𝑡) Finally, to isolate the rate of appearance of H2, multiply both the Fe and H2 rate expressions by 3. (Δ[H2]/Δ𝑡)=−3/2(Δ[Fe]/Δ𝑡)