Nelson chem reaction rates
Which characteristic of the steric factor is correctly stated?
As the complexity of molecules increases, the steric factor p decreases
The decay constant for mercury-197, a radioisotope used medically in kidney scans, is 1.08×10−2 h−1. What is the half-life of mercury-197?
Decay constant (\lambda) = 1.08 x 10^-2 h-1 half-life (t1/2) = ? radioactive decay follows first order kinetics relation between decay constant and half-life t1/2 = 0.693 / \lambda = 0.693 / 1.08 x 10^-2 =64.17 hours half-life of Hg-197 = 64.17 hour t1/2 = 64.2
What percentage of 146C (t1/2 = 5715 years) remains in a sample estimated to be 18730 years old?
Decay constant k = ln(2)/half life = ln(2)/5715 = 0.000121286 year-1 C/Co = exp(-kt) where C is amount of C-14 at time t, Co is initial amount For t = 18730 years C/Co = exp(-0.000121285 x 18730) = 0.1031 Percent remaining = C/Co x 100% = 0.1031 x 100% = 10.31% N/N0 = 10.3 %
Rate constants for the decomposition of gaseous dinitrogen pentoxide are 3.7×10−5 s−1 at 25 ∘C and 1.7×10−3 s−1 at 55 ∘C. 2N2O5(g)→4NO2(g)+O2(g) What is the activation energy for this reaction in kJ/mol?
Ea = 104 kJ/mol
Rate constants for the reaction NO2(g)+CO(g)→NO(g)+CO2(g) are 1.3M−1s−1 at 700 K and 23.0M−1s−1 at 800 K. What is the value of the activation energy in kJ/mol?
Ea = 134 kJ/mol Use the relation: ln (k'/k) = Ea/R ( 1/T- 1/T' ) Constants: R = 8.314 J/ K.mol (gas constant in joules) R=8.314 X 10^(-3) kJ/ K mol (gas constant in kJ) k' = 23.0 11/M s, T' = 800K k= 1.3/M s, T= 700K Solve for Ea ln (k'/k) = Ea/R ( 1/T- 1/T' ) (R X ln (k'/k)) / (1/T - T') = Ea Subsitute Ea=(R X ln (k'/k)) / (1/T - T') Ea=(8.314 X 10^(-3) X ln (23.0/1.3)) / (1/700-1/800) Ea= 133.77 kJ/mol
What is the activation energy of the reaction?
Ea= ((ln(k1/k2))/((1/T2)-(1/T1))∙-R Ea= (3.3/(−1.51×10−4))∙ -(8.3145) Ea = 182 kJ/mol
The half-life of a reaction, t1/2, is the time it takes for the reactant concentration [A] to decrease by half. For example, after one half-life the concentration falls from the initial concentration [A]0 to [A]0/2, after a second half-life to [A]0/4, after a third half-life to [A]0/8, and so on. on.
For a first-order reaction, the half-life is constant. It depends only on the rate constant k and not on the reactant concentration. It is expressed as t1/2=0.693k For a second-order reaction, the half-life depends on the rate constant and the concentration of the reactant and so is expressed as t1/2=1k[A]0
A certain second-order reaction (B→products) has a rate constant of 2.00×10−3 M−1⋅s−1 at 27 ∘C and an initial half-life of 278 s . What is the concentration of the reactant B after one half-life?
For a second order kinetics, t1/2 = 1/k[Bo] Given, t1/2 = 278 sec k = 2.00E-3/Ms So, [Bo] = 1/kt1/2 = 1/[278 x 2.00E-3] = 1.79856 M After 1 half life, concentration will be half the original concentration. So, 1.79856/2 = .899 M = conc after 1 half-life 0.899 M
In the hydrogenation of ethylene using a nickel catalyst, the initial concentration of ethylene is 1.95 mol⋅L−1 and its rate constant (k) is 0.0018 mol⋅L−1⋅s−1 . Determine the rate of reaction if it follows a zero-order reaction mechanism.
For a zero-order reaction, rate = k So, the rate is 0.0018 mol*L-1*s-1 rate of reaction = 1.8×10−3 mol⋅L−1⋅s−1 The integrated rate law for a chemical reaction is a relationship between the concentrations of the reactants and time. Consider a single reactant A decomposing into products: A→products. For a zero-order reaction, rate=k Because the rate of the reaction is −Δ[A]/Δt, the above rate law can be written as −Δ[A]/Δt=k On integrating this differential equation, the integrated rate law for a zero-order reaction is obtained: [A]t=−kt+[A]0 where [A]t is the concentration at time t, k is the rate constant, and [A]0 is the initial concentration. The variation in concentration with time provides a detailed description of how fast the reaction is occurring.
The integrated rate law allows chemists to predict the reactant concentration after a certain amount of time, or the time it would take for a certain concentration to be reached. The integrated rate law for a first-order reaction is: [A]=[A]0e−kt Now say we are particularly interested in the time it would take for the concentration to become one-half of its initial value. Then we could substitute [A]02 for [A] and rearrange the equation to: t1/2=0.693k This equation calculates the time required for the reactant concentration to drop to half its initial value. In other words, it calculates the half-life.
Half-life equation for first-order reactions: t1/2=0.693k where t1/2 is the half-life in seconds (s), and k is the rate constant in inverse seconds (s−1).
To calculate average and relative reaction rates. You can measure the rate of a reaction, just like you can measure the speed a jogger runs. While a jogger would be reported to run a specific number of miles in an hour, miles/hour, a reaction is reported to form product or consume reagent in molar concentration per second, M/s. Reaction rate can be defined either as the increase in the concentration of a product per unit time or as the decrease of the concentration of a reactant per unit time. By definition, reaction rate is a positive quantity.
In the reaction X→2Y, for example, Y is being produced twice as fast as X is consumed and thus rate of X=12(rate of Y) Each rate can be expressed as the change in concentration over the change in time, Δt: −Δ[X]/Δt=12(Δ[Y]/Δt)
The oxidation of iodide ion by hydrogen peroxide in an acidic solution is described by the balanced equation H2O2(aq)+3I−(aq)+2H+(aq)→I−3(aq)+2H2O(l) The rate of formation of the red triiodide ion, Δ[I−3]/Δt, can be determined by measuring the rate of appearance of the color (see the figure).
Initial rate data at 25 ∘C are as follows:
The reaction X →products is first-order, and the rate constant is 1.2 × 10−4 min−1. If the original concentration of X is 1.35 M, how long (in hr) will it take the concentration of X to decrease by 85%?
K = 2.303/t log(initial/final) 85% of 1.35 = 1.1475 1.2*10^-4 = 2.303/t log[1.35/(1.35-1.1475)] t = 15812.182min. t = 2.6*10^2 hr. 2.6 × 10² hr
At a certain temperature, initial rate data for the decomposition of gaseous N2O5 are as follows: What is the initial rate of decomposition of N2O5 when its initial concentration is 3.5×10−2 M ?
K=2.4x10^-5 /0.014=171.43x10^-5 s^-1 C - [R] =3x10^-2M _dR / dT =Kx[R] 171.43∙10⁻⁵∙3.5x10⁻²=6∙10⁻⁵ M/s
A sample of 35S undergoes 5260 disintegrations/min initially but undergoes 4860 disintegrations/min after 10.0 days. What is the half-life of 35S in days?
Let the half-life (in days) to be found be x. So, 4860 / 5260 = (1/2)^(10.0/x) Taking log on both the sides: -0.03435 = (10.0/x) (-0.30103) 10.0/x = 0.114108 x = 10.0 / 0.114108 Hence, the half life of 35S in days is, x = 87.636 days t1/2 = 87.6 d
What is the half-life of iron-59, a radioisotope used medically in the diagnosis of anemia, if a sample with an initial decay rate of 16,800 disintegrations/min decays at a rate of 10,860 disintegrations/min after 28.0 days?
Let z be the half-life in days: 10860 / 16800 = (1/2)^(28 days / z) Solve for z algebraically: log 10860 - log 16800 = (28 days / z) x log (1/2) (log 10860 - log 16800) / log (1/2) = 28 days / z z / (28 days) = log (1/2) / (log 10860 - log 16800) z = ((28 days) x log (1/2)) / (log 10860 - log 16800) z = 44.5 days t1/2 = 44.5 d
Polonium-209, an α emitter, has a half-life of 102 years. How many α particles are emitted in 1.1 s from a 1.6 ng sample of 209Po?
N = 1090
The decay constant of 35S is 7.95×10−3day−1. What percentage of an 35S sample remains after 175 days?
N = 24.9 % Decay constant=r=7.95*10^-3 /days Let initial sample be So S=So e^-rt After 175 days S/So =e^(-175*7.95*10^-3) =0.24876=24.9%
To understand how elementary steps make up a mechanism and how the rate law for an elementary step can be determined. Very often, a reaction does not tell us the whole story. For instance, the reaction NO2(g)+CO(g)→NO(g)+CO2(g) does not involve a collision between an NO2 molecule and a CO molecule. Based on experimental data at moderate tempertures, this reaction is thought to occur in the following two steps: NO2(g)+NO2(g)→NO3(g)+NO(g) NO3(g)+CO(g)→CO2(g)+NO2(g) Each individual step is called an elementary step. Together, these elementary steps are called the reaction mechanism.
Overall, the resulting reaction is NO2(g)+CO(g)→NO(g)+CO2(g) Notice that in the elementary steps NO3 appears both as a product and then as a reactant; therefore it cancels out of the final chemical equation. NO3 is called a reaction intermediate. Also notice that 2 molecules of NO2 appear in the reactants of the first step and 1 molecule of NO2 appears as product of the second step, the net effect leaves only 1 molecule of NO2 as a reactant in the net equation. Molecularity is the proper term for "how the molecules collide" in a reaction. For example, step 1 is bimolecular because it involves the collision of two molecules. Step 2 is also bimolecular for the same reason. Unimolecular reactions involve only one molecule in the reactants. Though rare, collisions among three molecules can occur. Such a reaction would be called termolecular.
What is the initial rate of formation of I3− when the initial concentrations are [H2O2] = 0.100 M and [I−] = 0.200 M ?(see slide 6)
Rate = 2.30×10−4 M/s (see picture for example)
What is the average rate of consumption of Br− during the same time interval?
Rate = 9.2×10−6 M/s
At a certain temperature, initial rate data for the decomposition of gaseous N2O5 are as follows: What is the rate law?
Rate = k[N2O5
What is the rate law for the hypothetical reaction X + 2Y →2Z? The table below gives the initial concentrations and rates for several experiments.
Rate = k[X]2
What is the average rate of consumption of H+ during the same time interval?
Rate of consumption of H+ = 1.87×10−4 Ms Since we know the rate of consumption of Br− (given) and the rate of formation of Br2 (calculated), we can use any one of them to find the rate of consumption of H+: rate of consumption of H+=65(rate of consumption of Br−)=2(rate of formation of Br2) Significant Figures Feedback: Your answer 1.9⋅10−4Ms was either rounded differently or used a different number of significant figures than required for this part.
Determine the average rate of decomposition of H3PO4 between 10.0 and 40.0 s.
Rate of decomposition of H3PO4 = 4.00×10−4 Ms For every two moles of H3PO4 that disappear, only one mole of P2O5 is formed. The rate of decomposition of H3PO4 is twice the rate of formation of P2O5.
Consider the reaction 5Br−(aq)+BrO−3(aq)+6H+(aq)→3Br2(aq)+3H2O(l) The average rate of consumption of Br− is 1.56×10−4 M/s over the first two minutes. What is the average rate of formation of Br2 during the same time interval?Rate of formation of Br2
Rate of formation of Br2 = 9.36×10−5 M/s
Consider the reaction 2H3PO4→P2O5+3H2O Using the information in the following table, calculate the average rate of formation of P2O5 between 10.0 and 40.0 s.
Rate of formation of P2O5 = 2.00×10−4 Ms
What is the rate law.
Rate=k[B]2
Bromomethane is converted to methanol in an alkaline solution. The reaction is first order in each reactant. CH3Br(aq)+OH−(aq)→CH3OH(aq)+Br−(aq) Write the rate law. How does the reaction rate change if the OH− concentration is decreased by a factor of 7?
Rate=k[CH3Br][OH−] Rate2/Rate1 = 0.143 Rate2/Rate1 = 4 (If OH- concentration is reduced by a factor or 7 => [OH-(2)]/[OH-(1)] = 1/7 If conc of OH- decreases by 7, the rate decreases by 7. If the concentation of BOTH reactants doubles, the rate goes up by 2 times 2, or four.)
What is the rate law for the formation of I−3? (see slide 6)
Rate=k[H2O2][I−]
The gas-phase reaction of hydrogen and iodine monochloride, is first order in H2 and first order in ICl. H2(g)+2ICl(g)→2HCl(g)+I2(g) What is the rate law? What are the units of the rate constant?
Rate=k[H2][ICl] 1/(M⋅s) (rate = k[A] [B] rate = k[H2] [HCl] M/t in sec = k M *M k units: M-1 s-1 or M-1 min-1or M -1 hr-1, etc.)
Hydrogen iodide gas decomposes at 410 ∘C: 2HI(g)→H2(g)+I2(g) The following data describe this decomposition: Is the reaction first order or second order?
Second
To understand how to use integrated rate laws to solve for concentration. A car starts at mile marker 145 on a highway and drives at 55 mi/hr in the direction of decreasing marker numbers. What mile marker will the car reach after 2 hours? This problem can easily be solved by calculating how far the car travels and subtracting that distance from the starting marker of 145. 55 mi/hr×2 hr=110 miles traveled milemarker 145−110 miles=milemarker 35 If we were to write a formula for this calculation, we might express it as follows: milemarker=milemarker0−(speed×time) where milemarker is the current milemarker and milemarker0 is the initial milemarker.
Similarly, the integrated rate law for a zero-order reaction is expressed as follows: [A]=[A]0−rate×time or [A]=[A]0−kt since rate=k[A]0=k A zero-order reaction (Figure 1) proceeds uniformly over time. In other words, the rate does not change as the reactant concentration changes. In contrast, first-order reaction rates (Figure 2) do change over time as the reactant concentration changes. Because the rate of a first-order reaction is nonuniform, its integrated rate law is slightly more complicated than that of a zero-order reaction. The integrated rate law for a first-order reaction is expressed as follows: [A]=[A]0e−kt where k is the rate constant for this reaction. The integrated rate law for a second-order reaction is expressed as follows: 1[A]=kt+1[A0] where k is the rate constant for this reaction.
What is the mechanism that is consistent with the rate law rate = k[D]2 for the reaction D + E →F + G?
Step 1. D + D →F + H, slow Step 2. H + E →G + D, fast.
A certain first-order reaction (A→products) has a rate constant of 7.50×10−3 s−1 at 45 ∘C. How many minutes does it take for the concentration of the reactant, [A], to drop to 6.25% of the original concentration?
Suppose [A0] = 1 M then [At] = 0.0625 M from the formula t = (2.303 / k)* log[A0]/[At] = ( 2.303 / 7.50×10−3 s−1) * log ( 1 / 0.0625 ) = 307.067 s x 1.204 = 369.75 seconds
The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as k=Ae−Ea/RT where R is the gas constant (8.314 J/mol⋅K), A is a constant called the frequency factor, and Ea is the activation energy for the reaction. However, a more practical form of this equation is lnk2k1=EaR(1T1−1T2) which is mathmatically equivalent to lnk1k2=EaR(1T2−1T1) where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1 and T2). The activation energy of a certain reaction is 33.6 kJ/mol . At 22 ∘C , the rate constant is 0.0180s−1. At what temperature in degrees Celsius would this reaction go twice as fast?
T2 = 38 ∘C see pic for example: The activation energy of a certain reaction is 42.3 kJ/mol . At 22 ∘C , the rate constant is 0.0190s−1. At what temperature in degrees Celsius would this reaction go twice as fast?)
In the collision model it is assumed that reactions occur as a result of collisions between molecules with the correct kinetic energy and orientation. The minimum energy required for a reaction to occur when molecules collide is called the activation energy, Ea. The number of collisions that are favorably oriented for a reaction is described by the frequency factor, A. The Arrhenius equation relates both of these factors to the rate constant, k: k=A⋅e−Ea/(R⋅T) where R=8.314 J/(mol⋅K) and T is the temperature in kelvins. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, assuming they are all at the same temperature and that each starts with the same initial concentration.
The activation energy, Ea, for a reaction is like a barrier that must be overcome for the reaction to occur. The frequency factor, A, is the fraction of collisions that are favorably oriented.
The overall order of an elementary step directly corresponds to its molecularity. Both steps in this example are second order because they are each bimolecular. Furthermore, the rate law can be determined directly from the number of each type of molecule in an elementary step. For example, the rate law for step 1 is rate=k[NO2]2
The exponent "2" is used because the reaction involves two NO2 molecules. The rate law for step 2 is rate=k[NO3]1[CO]1=k[NO3][CO] because the reaction involves only one molecule of each reactant the exponents are ommitted.
In a certain hypothetical experiment, the initial concentration of the reactant R is 1.00 mol⋅L−1 , and its rate constant is 0.0150 mol⋅L−1⋅s−1. It follows a zero-order reaction mechanism for the consumption of reactant R. Plot the graph of concentration versus time. Consider the time intervals as 0, 10, 20, 30, 40, and 50 s.
The graph of concentration versus time for a typical zero-order reaction is a straight line with a negative slope. The slope of the line is equal to −k and the y intercept is [R]0. There is a steady decrease in concentration until the reaction stops.
At 500 ∘C, cyclopropane (C3H6) rearranges to propene (CH3−CH=CH2). The reaction is first order, and the rate constant is 6.7×10−4s−1. If the initial concentration of C3H6 is 5.50×10−2 M . What is the half-life (in minutes) of this reaction?
The half-life (in minutes) of this reaction is (see pic)
The rate of reaction in terms of the "rate law expression" includes the rate constant (k), the concentration of the reactants, and the orders of the reaction with respect to the different reactants. Consider the following reaction: A+B→C+D The initial concentrations of the reactants A and B are 0.280 M and 0.350 M, respectively. The rate of reaction is 0.060 M⋅s−1, and the orders of the reaction, with respect to reactants A and B, are 1 and 2, respectively. Determine the rate constant (k) for the reaction using the rate law.
The rate constant(k) = 1.75 M⁻²∙s⁻¹
For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The following data were collected: What is the reaction order with respect to A? What is the reaction order with respect to B? What is the reaction order with respect to C? What is the value of the rate constant k for this reaction? When entering compound units, indicate multiplication of units explicitly using a multiplication dot (multiplication dot in the menu). For example, M−1⋅s−1. Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.55 M of reagent A and 0.90 M of reagents B and C?
The rate of a chemical reaction depends on the concentrations of the reactants. For the general reaction between Aand B, aA+bB⇌cC+dD The dependence of the reaction rate on the concentration of each reactant is given by the equation called the rate law: rate=k[A]m[B]n where k is a proportionality constant called the rate constant. The exponent m determines the reaction order with respect to A, and n determines the reaction order with respect to B. The overall reaction order equals the sum of the exponents (m+n). 2 0 1 k = 1.2×10−3 M−2⋅s−1 3.3×10−4 M/s
Consider the following reaction: 2A+3B→3C+2D The concentrations of product C at three different time intervals are given. Use the following data to determine the rate of reaction in terms of the appearance of product C between time = 0 s and time = 20 s . Time (s) 0 20 40 [C](M) 0.000 0.0240 0.0480
The rate of reaction = 4.00×10−4 M⋅s−1
The rate of the reaction in terms of the "disappearance of reactant" includes the change in the concentration of the reactant, the time interval, and the coefficient of the reactant. Consider the following reaction: 2A+3B→3C+2D The concentrations of reactant A at three different time intervals are given. Use the following data to determine the average rate of reaction in terms of the disappearance of reactant A between time = 0 s and time = 20 s. Time (s) 0 20 40 [A](M) 0.0400 0.0240 0.0180
The rate of reaction = 4.00×10−4 M⋅s−1
± Reaction Rates and Temperature
To use the Arrhenius equation to calculate the activation energy. As temperature rises, the average kinetic energy of molecules increases. In a chemical reaction, this means that a higher percentage of the molecules possess the required activation energy, and the reaction goes faster. This relationship is shown by the Arrhenius equation k=Ae−Ea/RT where k is the rate constant, A is the frequency factor, Ea is the activation energy, R = 8.3145 J/(K⋅mol) is the gas constant, and T is the Kelvin temperature. The following rearranged version of the equation is also useful: ln(k2k1)=(−EaR)(1T2−1T1) where k1 is the rate constant at temperature T1, and k2 is the rate constant at temperature T2.
What is the half-life of the radionuclide that shows the following decay curve?
Total amount remaining after 1 half life = 50% So the time in days taken for this to reach = 3 days The half life of radionulceide = 3 years T1/2 = 3 years
Why don't all collisions between reactant molecules lead to a chemical reaction? Very few collisions involve a collision energy equal to the activation energy. Very few collisions involve a collision energy smaller than the activation energy. Very few collisions involve a collision energy greater than the activation energy. Only a fraction of the collisions have the proper orientation for reaction. All the collisions involve a collision energy smaller than the activation energy. All the collisions have the proper orientation for reaction.
Very few collisions involve a collision energy equal to the activation energy Very few collisions involve a collision energy greater than the activation energy. Only a fraction of the collisions have the proper orientation for reaction.
At 500 ∘C, cyclopropane (C3H6) rearranges to propene (CH3−CH=CH2). The reaction is first order, and the rate constant is 6.7×10−4s−1. If the initial concentration of C3H6 is 5.50×10−2 M . What is the molarity of C3H6 after 35 min ?
[C3H6]t = 1.3×10−2 M
Butadiene (C4H6) reacts with itself to form a dimer with the formula C8H12. The reaction is second order in C4H6. If the rate constant at a particular temperature is 4.0×10−2M−1s−1 and the initial concentration of C4H6 is 1.90×10−2 M . What is its molarity after a reaction time of 1.10 h ?
[C4H6] = 4.7×10⁻³ M At = 0.019 / 1 + (0.019 ∙ 0.04 ∙ (66 min ∙ 60)
In acidic aqueous solution, the purple complex ion Co(NH3)5Br2+ undergoes a slow reaction in which the bromide ion is replaced by a water molecule, yielding the pinkish-orange complex ion : Co(NH3)5(H2O)3+ Co(NH3)5Br2+Purple(aq)+H2O(l)→ Co(NH3)5(H2O)3+Pinkish−orange(aq)+Br−(aq) The reaction is first order in Co(NH3)5Br2+, the rate constant at 25 ∘C is 6.3×10−6 s−1, and the initial concentration of Co(NH3)5Br2+ is 0.100 M. What is its molarity after a reaction time of 18.0 h ?
[Co(NH3)5Br2+] = 0.066 M
The potential energy profile for the one-step reaction AB+CD→AC+BD follows. The energies are in kJ/mol relative to an arbitrary zero of energy. a) Suggest a plausible structure for the transition state. b) Does the reaction rate increase or decrease when the temperature increases? c) Explain why does the reaction rate increase when the temperature increases.
a) see image b) The reaction rate increases. c) The reaction rates of all chemical reactions increase as the temperature increases. This is explained by collision theory.
The exponential term in the Arrhenius equation is equal to the fraction of molecules, f, with kinetic energy greater than or equal to the activation energy: f=e−Ea/(R⋅T) Most scientific calculators have an ex function as the second function of the LN button. A certain reaction with an activation energy of 135 kJ/mol was run at 545 K and again at 565 K . What is the ratio of f at the higher temperature to f at the lower temperature?
f565/f545 = 3
At 25 ∘C, the half-life of a certain first-order reaction is 244 s . What is the value of the rate constant at this temperature?
half -life =244 s rate constant = K relation between half-life and rate constant K = 0.693 / t1/2 = 0.693 / 244 = 0.00284 sec-1 (or) = 2.8 x 10^-3 sec-1 k = 2.84×10−3 /s
Butadiene (C4H6) reacts with itself to form a dimer with the formula (C8H12). The reaction is second order in C4H6. If the rate constant at a particular temperature is 4.0×10−2M−1s−1. How many minutes does it take for the concentration of C4H6 to drop from 1.20×10−2 M to 3.00×10−3 M ?
integrated second order rate law 1/[C₄H₆] = k∙t + 1/[C₄H₆]₀ => t = (1/3.0×10⁻³M - 1/1.2×10⁻²M) / 4.0×10⁻²M⁻¹s⁻¹ = 6250 sec 6250/60= 104 min
What is the value of the rate constant for consumption of HI? (see slide 66)
k = 0.0308 1/(M⋅min)
What is the value of the rate constant?(see slide 6)
k = 1.15×10−2 1M⋅s
Rate constants for the decomposition of gaseous dinitrogen pentoxide are 3.7×10−5 s−1 at 25 ∘C and 1.7×10−3 s−1 at 55 ∘C. 2N2O5(g)→4NO2(g)+O2(g) What is the rate constant at 35 ∘C?
k = 1.4×10−4 s−1
Rate constants for the reaction NO2(g)+CO(g)→NO(g)+CO2(g) are 1.3M−1s−1 at 700 K and 23.0M−1s−1 at 800 K. What is the rate constant at 780 K ?
k = 14 /(M⋅s) 2) What's the rate constant at 730K? Step 1: We need A (preexponential factor) first. Use either Use the relation: k=Ae(-Ea/RT) A=k/(e(-Ea/RT) A=1.3/(-133.768/(8.314X 10^(-3)X700)) A=1.24801 X 10^(10) Step 2: Use the relation: k=Ae(-Ea/RT) (new temperature) k=1.24801 X 10^(10) X e(-133.768/(8.314X 10^(-3)X780)) k=13.7 1/M.s
The half-life of iodine-131, released as a result of the Fukushima Daiichi nuclear disaster, is 8.02 days. What is its decay constant?
k = 8.64×10−2 day−1
The rate constant for a certain reaction is k = 4.20×10−3 s−1 . If the initial reactant concentration was 0.700 M, what will the concentration be after 4.00 minutes?
ln x / 0.700 = -4.2∙10⁻³ ∙ (4∙60)= -1.008 e^x (1.008) = 0.365 = x/0.700 x= 0.365 ∙ 0.700 =0.255 M
At a certain temperature, initial rate data for the decomposition of gaseous N2O5 are as follows: What is the value of the rate constant, including the proper units?
rate of decompo sition = -dr /dt= K[r] K=rate constant 2.4x10^-5=K[0.014] K=2.4x10^-5 /0.014=171.43x10^-5 s^-1
What is the order of the reaction in B?
second-order
What is the overall reaction order?
second-order
What are the units of the rate constant for each of the reactions in Table 12.2 in the textbook? Second reaction Third reaction Fourth reaction
s−1 s−1 M−3⋅s−1 M−1⋅s−1
At what time (in minutes) does the HI concentration reach 0.100 M? (see slide 66)
t = 260 min
In how many minutes does the HI concentration drop from 0.400 M to 0.200 M? (see slide 66)
t = 81.2 min
How many hours are required for 85 % of the Co(NH3)5Br2+ to react?
t = 84 h
The half-life of a reaction, t1/2, is the time required for one-half of a reactant to be consumed. It is the time during which the amount of reactant or its concentration decreases to one-half of its initial value. Determine the half-life for the reaction in Part B using the integrated rate law, given that the initial concentration is 1.95 mol⋅L−1 and the rate constant is 0.0018 mol⋅L−1⋅s−1 .
t1/2 = 1.95 / 1.95 ∙ 0.0018 t1/2 = 555.5 s so the half life for the reaction is t1/2 = 5.4×102 s The faster the reaction, the shorter the half-life for that reaction. The rate of the reaction is proportional to the rate constant; thus, the larger the rate constant, the shorter the half-life.
A 1.0 mg sample of 126Sn decays initially at a rate of 4.6×105 disintegrations/s. What is the half-life of 126Sn in years?
t1/2 = 2.3×10⁵ y
Butadiene (C4H6) reacts with itself to form a dimer with the formula (C8H12). The reaction is second order in C4H6. If the rate constant at a particular temperature is 4.0×10−2M−1s−1. What is the half-life (in minutes) of this reaction when the initial C4H6 concentration is 2.20×10−2 M ?
t½ = 1 / (k∙[C₄H₆]₀) = 1 / ( 4.0×10⁻²M⁻¹s⁻¹ ∙ 2.2×10⁻²M ) = 1136 s = 18.9 min
A reaction of the type A→B+C has a rate constant k=3.6×10−5M/s. What is the order of the reaction? What is the molarity of A after a reaction time of 28.0 min if the initial concentration of A is 9.6×10−2 M ? What is the half-life (in minutes) of the reaction in part (b)?
zeroth order Concentration of A after t time At = 9.6*10^-2 + 3.6*10^-5 * 28 * 60 At = 0.03552 M [A] = 3.6×10−2 M t1/2 = A0/2k t1/2 = 9.6*10^-2/(2*3.6*10^-5) t1/2 = 1333.33 sec t1/2 = 1333.33/60 = 22.22 min t1/2 = 22 min
See above What is the order of the reaction in A?
zeroth-order
If Δ[Br₂]/Δt = 4.6×10−6 M/s , what is the value of Δ[ClO⁻₂]/Δt during the same time interval?
Δ[ClO−2]/Δt = −2.3×10−6 M/s (answer in pic should be negative)
The oxidation of iodide ion by arsenic acid, H3AsO4, is described by the balanced equation 3I−(aq)+H3AsO4(aq)+2H+(aq)→ I3−(aq)+H3AsO3(aq)+H2O(l) If −Δ[I−]/Δt = 5.4×10−4 M/s , what is the value of Δ[I3−]/Δt during the same time interval? What is the average rate of consumption of H+ during that time interval?
Δ[I3−]/Δt = 1.8×10−4 M/s −Δ[H+]/Δt = 3.6×10−4 M/s
Butadiene (C4H6) reacts with itself to form a dimer with the formula C8H12. The reaction is second order in C4H6. If the rate constant at a particular temperature is 4.0×10−2M−1s−1 and the initial concentration of C4H6 is 1.90×10−2 M . What is the time (in hours) when the C4H6 concentration reaches a value of 1.90×10−3 M ?
( 1 / 0.0019 ) - ( 1/0.019) = kt 473.68 = kt t= 473.68 / 0.04 = 11842 sec = 197 min = 3.3 h
The rate constant of a chemical reaction increased from 0.100 s−1 to 2.70 s−1 upon raising the temperature from 25.0 ∘C to 39.0 ∘C . Calculate the value of (1T2−1T1) where T1 is the initial temperature and T2 is the final temperature.
((1/T2)−(1/T1)) = −1.51×10−4 K−1
At 500 ∘C, cyclopropane (C3H6) rearranges to propene (CH3−CH=CH2). The reaction is first order, and the rate constant is 6.7×10−4s−1. If the initial concentration of C3H6 is 5.50×10−2 M . How many minutes does it take for the C3H6 concentration to drop to 1.20×10−2 M ?
(2.303/6.7∙10⁻⁴) ∙log(5.5∙10⁻²/1.2∙10⁻²) = 2272.687857 2272.687857/60= 37.878 t = 38 min
What is the half-life of a first-order reaction with a rate constant of 5.10×10−4 s−1?
0.693/(5.1*10⁻⁴)=1358.8
A zero-order reaction has a constant rate of 3.90×10−4 M/s. If after 80.0 seconds the concentration has dropped to 1.00×10−2 M, what was the initial concentration?
1.00×10−2 M = -(3.90×10−4 M/s ∙ 80.0 s) + [A°] 1.00×10−2 M = -0.0312 + [A°] 1.00×10−2 M - (-0.0312) = [A°] [A°] = 0.0412 M 4.12×10−2 M
The reaction 2X →products is a second-order reaction, and the rate constant is 8.8 × 10−3 1/M·s. If the initial concentration of X is 3.00 M, how many seconds does it take for the concentration of X to drop to 0.700 M?
124 s
What is the half-life of neon-19 if a sample with an initial decay rate of 1.36 × 105 disintegrations/s has a decay rate of 4.50 × 102 disintegrations/s after 144 s?
17.5 s
How is the rate of formation of NH3 related to the rate of consumption of N2?
2
Ammonia is manufactured in large amounts by the reaction N2(g)+3H2(g)→2NH3(g)
3 (answer in pic is backward. Should read 3/1, not ⅓)
At 500 ∘C, cyclopropane (C3H6) rearranges to propene (CH3−CH=CH2). The reaction is first order, and the rate constant is 6.7×10−4s−1. If the initial concentration of C3H6 is 5.50×10−2 M . How many minutes does it take for 30% of the C3H6 to react?
30% reacted means 70% remained. t = (2.303/k)(log(100/70)) t = (2.303/6.7∙10⁻⁴)(log(100/70)) t = 532.4465878 seconds t = 532.4465878 / 60 t = 8.87 min
A certain first-order reaction has a rate constant of 6.30×10−3 s−1. How long will it take for the reactant concentration to drop to 18 of its initial value?
330 s
Consider the following elementary steps that make up the mechanism of a certain reaction: 3X→E+F E+M→F+N What is the overall reaction? Which species is a reaction intermediate? What is the rate law for step 1 of this reaction? What is the rate law for step 2 of this reaction?
3X+M→2F+N E Rate = k*[X]³ Rate = k*[E][M]
At 500 ∘C, cyclopropane (C3H6) rearranges to propene (CH3−CH=CH2). The reaction is first order, and the rate constant is 6.7×10−4s−1. If the initial concentration of C3H6 is 5.50×10−2 M . How many minutes will it take for the concentration of cyclopropane to drop to 6.25% of its initial value?
A = Aoe- kt A = 0.0625 x 0.0550 = 3.44x10-3 M Ao = 0.0550M k = 6.7×10−4s−1 t =? ln A = ln Ao - kt ln A/Ao = -kt ln 3.44x10-3 /0.0550 = -kt ln 0.0625= -kt -2.77 = -6.7×10−4 ∙ t -2.77/(-6.7∙10⁻⁴) = t t = 4138.19 sec t = 4138.19/60 = 68.99 min
Which of the following are correct for zero-order reactions? The concentration of the reactants changes nonlinearly. A zero-order reaction slows down as the reaction proceeds. A higher concentration of reactants will not speed up the reaction. The rate of reaction does not equal the rate constant. The units for the rate constant and the rate of reaction are the same.
A higher concentration of reactants will not speed up the reaction. The units for the rate constant and the rate of reaction are the same. Also... For zero-order reactions the rate of reaction does not change over time. Zero-order reactions are typically found when a material that is required for the reaction to proceed is saturated by the reactants. A zero-order reaction is independent of the concentration of the reactants. Zero-order reactions have the following features: 1. The concentration versus time graph is a straight line with a negative slope; this negative slope represents the rate constant of a reaction. 2. The rate of the reaction is equal to the rate constant. 3. The units of k and the rate of reaction are mol⋅L−1⋅s−1. Photochemical reactions and surface reactions are examples of zero-order reactions.
To understand zero-order reactions. Chemical kinetics is the study of the speed with which a chemical reaction occurs and the factors that affect this speed. The speed of a reaction is the rate at which the concentrations of reactants and products change. This relationship is expressed through rate laws. For a general reaction aA+bB⟶gG+hH, the rate law is given as rate of reaction= k[A]m[B]n where the coefficients m and n are experimentally determined while a, b, g, and h are stoichiometric coefficients unrelated to m and n.
A zero-order reaction has a rate law in which the sum of the exponents m+n+... is equal to zero. Thus, the rate of reaction is independent of the concentration of the reactants. For a zero-order reaction, the graph of reactant concentration versus time is a straight line. Therefore, the rate-law equation for a zero-order reaction can be compared to that of a straight line, y=mx+b.
Consider the two reactions in the table. What is the order of the first reaction in BrO3−? What is the order of the first reaction in Br−? What is the order of the first reaction in H+? What is the order of the second reaction in H2? What is the order of the second reaction in I2? What is the overall reaction order for the first reaction? What is the overall reaction order for the second reaction?
A) 1 B) 1 C) 2 D) 1 E) 1 F) 4 G) 2
What is the rate constant of a first-order reaction that takes 354 seconds for the reactant concentration to drop to half of its initial value?
A:- Equation for first order reaction half life, t1/2 = 0.693 / k 354 sec. = 0.693 / k k = 0.693 / 354 sec. k = 0.0019576 sec-1 k=1.96∙10⁻³ s⁻¹
Calculate the value of ln(k2k1) where k1 and k2 correspond to the rate constants at the initial and the final temperatures as defined in part A.
Answer is not negative. ln(k2/k1) = 3.30