Cherryl's Chem Test #3

26 Questions | Total Attempts: 345

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Chemistry Quizzes & Trivia

Practice test #3 for chem II


Questions and Answers
  • 1. 
    For the hypothetical reaction A+3B--2C, the rate should be expressed as:
    • A. 

      Rate=deltal A divided by deta temp

    • B. 

      Rate=7/3 delta B divided by delta temp

    • C. 

      Rate=1/2 delta C divided by delta temp

    • D. 

      -1/2 delta C divided by delta temp

  • 2. 
    The reaction, A+2B--products; has the rate law, rate=k[A][B]3. When the concentration of B is doubled, while that of A is unchanged, by what factor will the rate of reaction increase?
    • A. 

      2

    • B. 

      4

    • C. 

      8

    • D. 

      6

  • 3. 
    A catalyst speeds up a reaction by
    • A. 

      Increasing the number of high-energy molecules

    • B. 

      Increasing the temperature of the molecules in the reaction

    • C. 

      Increasing the number of collisions between molecules

    • D. 

      Decreasing the activation energy for the reaction

  • 4. 
    The graphs below all refer to the same reaction. What order is this reaction?
    • A. 

      Zero-order

    • B. 

      First-order

    • C. 

      Second-order

    • D. 

      Third-order

  • 5. 
    Which one of the following would alter the rate constant (k) for the reaction below? 2A+B--products
    • A. 

      Increasing the concentration of A

    • B. 

      Increasing the concentration of B

    • C. 

      Increasing the temperature

    • D. 

      Measuring k again after the reaction has run a while

  • 6. 
    Which one of the following would alter the rate constant (k) for the reaction below? 2A+B--products
    • A. 

      Increasing the concentration of A

    • B. 

      Increasing the concentration of B

    • C. 

      Increasing the temperature

    • D. 

      Measuring k again after the reaction has run awhile

  • 7. 
    • A. 

      2.51 x 10 -5 kJ/mol

    • B. 

      39.8 kJ/mol

    • C. 

      17.4 kJ/mol

    • D. 

      5.76 x 10 -6 kj/mol

  • 8. 
    I the reaction belwo was shown to be second order with respect to NO2, and zero order with respect to CO, then the rate law would be written as: NO2 (g) + CO (g)---NO(g)+C)23(g)
    • A. 

      Rate=k[NO2]2

    • B. 

      Rate=k[NO2][CO]2

    • C. 

      Rate=k[CO]2

    • D. 

      Rate=k[NO][CO2]

  • 9. 
    A change in temperature from 10 degrees C to 20 degree C is found to double the rate of a given chemical reaction. How did this change affect the reacting molecules?
    • A. 

      It doubled their average velocity

    • B. 

      It doubled their average energy

    • C. 

      It doubled the number of collisions per second

    • D. 

      It double the proportion of molecules possessing at least the minimum energy required for the reaction.

  • 10. 
    Consider a hypothetical reation 2A+B--products. Given the following information concerning the initial rate of the reaction with different initial concentrations: [A] (mol/L) [B] (mol/L) Initial Rate (M/s) Exp. 1 0.020 0.020 4.20 x 10 -3 Exp. 2 0.040 0.020 1.68 x 10 -2 Exp. 3 0.040 0.040 6.72 x 10 -2 What is the rate law that mostly nearly accounts for these data? Rate =
    • A. 

      K[A]2[B]2

    • B. 

      K[A][B]2

    • C. 

      K[A][B]

    • D. 

      K[A]2[B]

  • 11. 
    For the reaction: X2+Y=Z--XY+XZ The rate equation is: rate=k[X2][Y]. Why does the concentration of Z have no effect on the rate?
    • A. 

      The concentration of Z is very small and the others are very large

    • B. 

      Z must react in a step after the rate determining step.

    • C. 

      Z is an intermediate

    • D. 

      The activation energy for Z to react is very high

  • 12. 
    For the chemical reaction system described by the diagram below which statement is true?
    • A. 

      The forward reaction is exothermic

    • B. 

      The reverse reaction is exothermic

    • C. 

      At equilibrium, the activation energy for the forward reaction is equal to the activation energy for the reverse reaction

    • D. 

      The system is at equilibrium

  • 13. 
    The thermal decomposition of acetaldehyde is a second-order reaction. CH2CHO--CH4+CO What is the half-life of acetaldehyde,if the rate constant for the decomposition of acetaldehyde is 6.7 x 10 -6 mmHg*s?
    • A. 

      1.5 x 10 5s

    • B. 

      410s

    • C. 

      5.4 x 10 7s

    • D. 

      520s

  • 14. 
    The equilibrium constant expression for the reaction below is: 2H2S(g)--H2(g)+S2(g)
    • A. 

      Kc=[H2S]2 divided by [H2][S2]

    • B. 

      Kc=[H2S] divided by [H2][S2]

    • C. 

      Kc=[H2][S2] divided by [H2S]2

    • D. 

      Kc=[H2][S2] divided by [H2S]

  • 15. 
    For the following reactions occurring at 500K, arrange them in order of increasing tendency to proceed to completion(least to greatest tendency). 1. 2NOCl--2NO+Cl Kp=1.7 x 10 -2 2. 2SO3--2SO2+O2 Kp=1.3 x 10 -5 3. 2NO2--2NO+O2 Kp=5.9 x 10 -5
    • A. 

      2

    • B. 

      1

    • C. 

      2

    • D. 

      3

  • 16. 
    Consider the two gaseous equilibria: SO2 (g) + 1/20 (g)--SO3(g) K1 2SO3(g)--2SO2(g)+O2 (g) K2 The equilibrium constants K1 and K2 are related by:
    • A. 

      K2=(K1)-2

    • B. 

      K2 2nd power=K1

    • C. 

      K2=(K1)2

    • D. 

      K2=(K1)-1

  • 17. 
    When the following reaction is at equilibrium, which choice is always true? 2NOCl(g)--2NO(g)+Cl2(g)
    • A. 

      [NO][Cl2]=[NOCl]

    • B. 

      [NO]2[Cl2]=[NOCl]2

    • C. 

      [NOCl]=[NO]

    • D. 

      [NO]2[Cl2]=Kc[NOCl]-2

  • 18. 
    At 700K, the reaction: 2SO2(g)+O2--2SO3(g) has an equilibrium constant Kc=4.3 x 10 6th power, and the following concentrations are present: [SO2]=0.010M; [SO2]=10 M; [O2]=0.010M is the mixture at equilibrium, yes or not? if not at equilibrium, in which direction, left to right, or right to lef, will the reaction occur to reach equilibrium/
    • A. 

      Yes

    • B. 

      No, left to right

    • C. 

      No, right to left

    • D. 

      Ther is not enough information to tell

  • 19. 
    For the following reaction at equilibrium, which choice gives a change that will shift the position of equilibrium to favor more products? 2NOBR(g)---2NO(g)+Br2(g) delta Hrxn=+30 kJ
    • A. 

      Increase the pressure

    • B. 

      Remove Br2

    • C. 

      Add more NO

    • D. 

      Lower the temperature

  • 20. 
    Changing which of the following will change the equilibrium constant, K?
    • A. 

      Adding a catalyst

    • B. 

      Doubling the volume

    • C. 

      Decreasing the pressure

    • D. 

      Increasing the temperature

  • 21. 
    For the following reaction at equilibrium in a reaction vessel, which one of the changes below would cause the Br2 concentration to decrease? 2NOBr (g)--2NO(g)+Br2(g)
    • A. 

      Remove some NO

    • B. 

      Remove some NOBr

    • C. 

      Compress the gas mixture into a smaller volume

    • D. 

      Add some Br2

  • 22. 
    Chemical equilibrium has been reached when:
    • A. 

      The forward rate exceeds the reverse rate

    • B. 

      The reverse rate exceeds the forward rate

    • C. 

      The forward rate equals the reverse rate

    • D. 

      Concentration of the products are increasing

  • 23. 
    What is the rate constant for the first order reaction A--B given the following data: Time (s) [A] ---------------- 0 1.76 6 0.88 12 0.44 18 0.22
    • A. 

      K=0.116-1/s=In(0.88-In(1.76)

    • B. 

      K=1.16=In(0.77)-In(2.0)

  • 24. 
    Given that Ea for a certain biological reaction is 48kJ/mol, and that the rate constant is 2.5 x 10 -2 s-1 at 15 degrees C. What is the rate constant at 37 degrees C?
    • A. 

      K2=0.104

    • B. 

      K2=108.5

    • C. 

      K2=220.0

  • 25. 
    The brown gas NO2 and the colorless gas N2O4 exist in equilibrium 2NO2--N2O4 In an experiment, 0.625 mol of N2O4 was introduced into a 5.00 L vessel and was allowed to decompose until it reached equilibrium with No2. The concentration of N2O4 at equilibrium was 0.0750 M. Calculate Kc for the reaction.
    • A. 

      Kc=[N2O4] divided by [NO2]2 =0.075/0.12=7.5

    • B. 

      Kc=[N2O4] divided by [NO2]2=0.065/0.11=6.5

  • 26. 
    Hydrogen iodide decomposes according to the equation: 2HI(g)--H2(g)+I2 Kc=0.0156 at 400 degrees C If 0.550 mol HI was injected into a 2.0 L reaction vessel at 400 degrees C. Calculate the concentrtion of HI at equilibrium.
    • A. 

      HI=0.275-2(0.02748)/square of [HI]=0.22M

    • B. 

      HI=0.175 (0.175)/square of [HI]=0.33