Multiple Choice Questions

Consider a second-order reaction in which reactant A decomposes according to the chemical equation 2A→products.The data given below is the time, in minutes, and the corresponding change in the concentration of reactant A for this reaction.t (min) [A]t(M)0.00 0.50020.0 0.38240.0 0.31060.0 0.26080.0 0.224Part CFor a second-order reaction, the rate constant k is the slope of the graph of 1[A] versus t. Based on this information and the data given, calculate the rate constant k for the reaction.Express your answer in M−1⋅min−1 to three significant figures.
  • The rate constant k = 3.10×10^(−2) M−1⋅min−1
  • [HI]t = 1.8×10^(−3) mol/L
  • The rate-determining step is termolecular.
  • Rate of formation of Br2Br2 = 9.96×10^(-5) M/s
An average reaction rate is calculated as the change in the concentration of reactants or products over a period of time in the course of the reaction. An instantaneous reaction rate is the rate at a particular moment in the reaction and is usually determined graphically.The reaction of compound A forming compound B was studied and the following data were collected:Time (s) [A] (M)0.0.0.0.0.0.016What is the average reaction rate betweenands? Express your answer to three significant figures and include the appropriate units.
  • [HI]t = 1.8×10^(−3) mol/L
  • rate = 1.10×10^(−4) M/s
  • k0th = 7.69×10^−5 M/s
  • rate = 1.12×10^(−4) M/s
In a reaction involving reactants in the gas state, how does increasing the partial pressures of the gases affect the reaction rate?
  • The half-life will increase.
  • The rate will increase.
  • k1st = 4.49×10^−2 s−1
  • The lines will have the same negative slope, and the y-intercept of line 2 will be higher than the y-intercept of line 1.
What is the reaction order with respect to C?Express your answer as an integer.
  • bimolecular
  • 3
  • 1
  • 6
Based on the data given and a rate constant of 0.031 M−1⋅min−1, calculate the time at which the concentration of reactant A will be 0.125M .Express your answer in minutes to two significant figures.
  • t = 190 min Using the integrated rate law for a second-order reaction, the time at which the concentration of reactant A will be 0.125M can be calculated ast===(1[A]t−1[A]0)/k(10.125M−10.500 M)/0.031 M−1⋅min−1194 min
  • 0.007 M/s
  • The lines will have the same negative slope, and the y-intercept of line 2 will be higher than the y-intercept of line 1.
  • 0.014 M/s
What was the initial reactant concentration for the reaction described in Part A?Express your answer with the appropriate units.
  • [A]0 = 6.35×10^−2 M
  • k = 1.9×10^(−3) M^(−2)⋅s^(−1)
  • k1st = 4.49×10^−2 s−1
  • rate = 1.10×10^(−4) M/s
Now consider the following reaction and data:H2+2BrCl→2HCl+Br2Time (s) Br2 concentration (M)5 1.515 1.57Part BWhat is the average rate of formation of Br2?Express your answer to three decimal places and include the appropriate units.
  • 0.007 M/s
  • 0.014 M/s
  • 4.0×10^(−4) M/s
  • 50 mph
How does the instantaneous rate of reaction change as the reaction proceeds?
  • rate = 6.8⋅10^(−5) M/s
  • It decreases.
  • M^(−2)⋅s^(−1)
  • bimolecular
Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant figures and include the appropriate units.
  • The half-life will increase.
  • The rate-determining step is termolecular.
  • initial rate = 2.3×10^(−2) M/s
  • rate=k[CH3CH2Br][NaOH]
What are the units of the rate constant k for this reaction?
  • bimolecular
  • It decreases.
  • rate = 6.8⋅10^(−5) M/s
  • M^(−2)⋅s^(−1)
Learning Goal: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 thusrate 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)Consider the reaction2H3PO4→P2O5+3H2OUsing the information in the following table, calculate the average rate of formation of P2O5 between 10.0 and 40.0 s.Time (s) 0 10.0 20.0 30.0 40.0 50.0[P2O5] (M) 0 1.40×10−3 4.40×10−3 6.20×10−3 7.40×10−3 8.00×10−3Express your answer with the appropriate units.
  • The rate constant k = 3.10×10^(−2) M−1⋅min−1
  • Rate of formation of P2O5 = 2.00×10^(−4) M/s
  • Rate of formation of Br2Br2 = 9.96×10^(-5) M/s
  • Rate of consumption of H+ = −1.99×10^(−4) Ms
What is the average reaction rate betweenand 1200.s ?Express your answer to three significant figures and include the appropriate units.
  • [A]0 = 6.35×10^−2 M
  • rate = 1.10×10^(−4) M/s
  • rate = 6.8⋅10^(−5) M/s
  • Rate of consumption of H+ = −1.99×10^(−4) Ms
How do the two graphs compare if the activation energy of the second reaction is higher than the activation energy of the first reaction but the two reactions have the same frequency factor?
  • The rate-determining step is termolecular.
  • The lines will have the same negative slope, and the y-intercept of line 2 will be higher than the y-intercept of line 1.
  • The lines will have the same y-intercept, negative slope direction, and the slope of line 2 will be steeper than the slope of line 1.
  • t = 190 min Using the integrated rate law for a second-order reaction, the time at which the concentration of reactant A will be 0.125M can be calculated ast===(1[A]t−1[A]0)/k(10.125M−10.500 M)/0.031 M−1⋅min−1194 min
The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they resemble the equation for a straight line,y=mx+b.Order Integrated Rate Law Graph Slope0 [A]=−kt+[A]0 [A] vs. t −k1 ln[A]=−kt+ln[A]0 ln[A] vs. t −k2 1[A]= kt+1[A]0 1[A] vs. t kThe reactant concentration in a zero-order reaction was 5.00×10−2M after 175s and 4.00×10−2M after 305s . What is the rate constant for this reaction?Express your answer with the appropriate units.
  • rate = 1.10×10^(−4) M/s
  • [HI]t = 1.8×10^(−3) mol/L
  • k1st = 4.49×10^−2 s−1
  • k0th = 7.69×10^−5 M/s
What would happen to the rate if [O2] were doubled?
  • double
  • It decreases.
  • quadruple
  • M^(−2)⋅s^(−1)
Suppose that a certain biologically important reaction is quite slow at physiological temperature (37 ∘C) in the absence of a catalyst.Assuming that the collision factor remains the same, by how much must an enzyme lower the activation energy of the reaction in order to achieve a 3×105-fold increase in the reaction rate?Express your answer using one significant figure.
  • [HI]t = 1.8×10^(−3) mol/L
  • Eauncatalized−Eacatalized = 30 kJ/mol
  • Rate of formation of Br2Br2 = 9.96×10^(-5) M/s
  • The rate constant k = 3.10×10^(−2) M−1⋅min−1
What can you conclude from the fact that the plot of ln p versus t is linear?
  • the rate constant increases as E a decreases.
  • The reaction is first order in CH3NC
  • The rate will increase.
  • The rate-determining step is termolecular.
Based on your answer to Part B, what is the average rate of formation of HCl?Express your answer to three decimal places and include the appropriate units.
  • 0.014 M/s
  • 0.007 M/s
  • 4.0×10^(−4) M/s
  • t = 190 min Using the integrated rate law for a second-order reaction, the time at which the concentration of reactant A will be 0.125M can be calculated ast===(1[A]t−1[A]0)/k(10.125M−10.500 M)/0.031 M−1⋅min−1194 min
How does the half-life of a second-order reaction change as the reaction proceeds?
  • The half-life will increase.
  • It decreases.
  • The rate will increase.
  • The rate-determining step is termolecular.
0 h : 0 m : 1 s

Multiple Choice Questions
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