CH 117 Spring 2015Mock Exam 1

Multiple Choice

  1. What is the average rate of disappearance for 2 O3 (g)  3 O2 (g) if O3 has an initial concentration of 1.25 x 10-3 M which decreases to 9.85 x 10-5 M in 20 seconds?

a). -5.76 x 10-5 M

b). 5.76 x 10-5 M

c). 6.74 x 10-5 M

d). -6.74 x 10-5 M

  1. For the reaction, CuCl2 + H2S  CuS + 2 HCl, the rate can be expressed as Δ[CuS]/ Δt. Which of the following is also correct?

a). rate = -Δ[CuCl2]/Δt

b). rate = -Δ[H2S]/Δt

c). rate = - Δ[HCl]/2Δt this answer choice should be positive!

d). all of these are correct

  1. For the reaction, CH4 + 2 O2 CO2 + 2 H2O, suppose the rate of consumption of CH4 is -3.6 M/s. Which of the following is true?

a).The rate of formation of water is 1.8 M/s.

b). The rate of formation of water is 7.2 M/s.

c). The rate of consumption of O2 is 7.2 M/s.

d). The rate of consumption of O2 is -1.8 M/s.

  1. What is the overall order for a reaction with the rate law of rate = k[A]3[B][C]?

a). 3

b). 4

c). 0

d). 5

  1. What are the units of the rate law constant in the questions above?

a). M-3s-1

b). M-4s-1

c). Ms-1

d). M-5s-1

  1. Experiments were conducted to determine the rate law for the reaction, F2 (g) + 2 ClO2 (g)  2 FClO2 (g). Given the following data, what is the rate law for this reaction?

Trial / [F2] (M) / [ClO2] (M) / Rate (M/s)
1 / 0.10 / 0.010 / 1.2 x 10-3
2 / 0.10 / 0.040 / 4.8 x 10-3
3 / 0.20 / 0.010 / 2.4 x 10-3

a). rate = k[F2][ClO2]2

b). rate = k[F2][ClO2]

c). rate = k[FClO2]2

d). rate = k[F2][ClO2][FClO2]

  1. If the experimental data for the data above were plotted, what would the slope of the line be equal to?

a). 1

b). the negative of the rate constant

c). the rate constant

d). one over the rate constant

  1. What is the rate constant for a first-order chemical reaction, if the reaction has a half-life equal to 85 minutes?

a). 8.15 x 10-3 s-1

b). 1.36 x 10-4 s-1

c). 58.9 s-1

d). 3534.3 s-1

  1. Which of the following is/are true concerning a chemical reaction at equilibrium?
  1. The system will be a mixture of reactant and products.
  2. The rate of the forward and reverse reactions are equal.
  3. The amount of each reactant and product is constant.

a). 1 only

b). 2 only

c). 3 only

d). 1 and 2

e). 1, 2, and 3

  1. The reaction SO2Cl2 SO2 + Cl2 has a rate constant of 0.17 hours-1. If the initial concentration of SO2Cl2 is 1.25 x 10-3 M, how many seconds does it take for the concentration to drop to 3.1 x 10-4 M?

a). 8.20 s

b). 2.9 x 104 s

c). 492 s

d). 7.5 x 103 s

  1. Which of the following is needed for a reaction to occur?

a). Molecules must collide with enough energy.

b). Molecules must be properly oriented and shaped.

c). Effective reactions only take place at relatively high temperatures.

d). A and B only

e). All of the above

  1. Which of the following best describes how physical states of reactants affect reaction rate?

a). Liquids react faster than solids and gases.

b). The smaller the surface area of a reactant, the faster the rate of the reaction.

c). Solids react faster than liquids or gases.

d). Powered reactants react faster than crystalline reactants.

  1. The rate constant for a first-order reaction at 473 K is 7.43 x 10-3 s-1 and is 9.82 x 10-2 s-1 at 873 K. What is the activation energy of this reaction?

a). 857 kJ

b).0.003 kJ

c). 22,200 kJ

d). 22.2 kJ

  1. A mixture of 2.0 mol of CO (g) and 2.0 mol of H2O (g) was allowed to come to equilibrium in a 1 L flask at a high temperature. If Kc = 4.0, what is the molar concentration of H2 (g) in the equilibrium mixture?

CO (g) + H2O (g)  CO2 (g) + H2 (g)

a). 0.67

b). 0.75

c). 1.3

d). 1.0

  1. A 1.20 L flask contains an equilibrium mixture of 0.0168 mol of N2, 0.2064 mol of H2, and 0.0143 mol of NH3. Calculate the equilibrium constant Kc for the reaction.

N2 (g) + 3H2 (g)  2NH3 (g)

a). 1.38

b). 1.99

c). 4.96

d). 4.12

  1. Consider the endothermic reaction:

N2 (g) + O2 (g)  2NO (g) ΔH = 180 kJ

At 2000 K, the equilibrium constant is 5.0 x 10-4. At 2500 K, the value of the equilibrium constant

a). is greater than 5.0 x 10-4

b). is less than 5.0 x 10-4

c). is 5.0 x 10-4

d). depends upon the concentration of oxygen

  1. Consider the equilibrium N2 (g) + O2 (g)2NO (g). At 2300 K the equilibrium constant Kc = 1.7*10-3. Suppose that 0.015 mol NO, 0.25 N2, and 0.25 mol O2 are placed in a 10 L flask, sealed, and heated to 2300 K. Is this reaction at equilibrium?

a). Yes, no shift will occur.

b). No, the reaction must shift towards the products to reach equilibrium.

c). No, the reaction must shift towards the reactants to reach equilibrium.

d). Not enough information to answer this question.

  1. Which of the following best describes the role of a catalyst?

a). A catalyst will affect a reaction by lowering the activation energy and shifting an equilibrium towards the products.

b). A catalyst will affect a reaction by lowering the activation energy and the time it takes for a reaction to reach equilibrium.

c). A catalyst will affect a reaction by lowering the activation energy and have no effect on the equilibrium.

d). A catalyst will not affect a chemical reaction.

Short Answer

  1. Consider the reaction: BaCO3 (s)  BaO (s) + CO2 (g), with an enthalpy of +20.6 kJ. Predict which direction the reaction will shift under each of the following conditions.
  1. Add BaCO3– no change since BaCO3 is a solid
  2. Add CO2 – shift left
  3. Remove BaO– no change
  4. Raise the temperature– shift right
  5. Increase the volume of the reaction flask– shift right
  1. The following mechanism has been proposed for the reaction: 2 SO2 + O2 2 SO3.

Step 1: 2 NO2 + 2 SO2 2 NO + 2 SO3fast, in equilibrium

Step 2: 2 NO + O2 2 NO2slow

Answer the following pertaining to this mechanism.

  1. List any intermediates.NO, NO2
  2. Determine the overall rate law for this reaction.

There is no correct answer to this question, since intermediates appear in the overall rate law regardless of how you manipulate it, but I will take you through the reasoning anyways.

First right the rate law for each of the steps, remember that the slow step determines rate.

For step 1, since it is in equilibrium we can write both a forward rate and a reverse rate and those will be equal to each other  rate = kf[NO2]2[SO2]2 = kr[NO]2[SO3]2.

For step 2: rate = k[NO]2[O2].

You want to base your overall rate law on the slow step, but the slow step has an intermediate present in it, so we will have to use the fast step to get rid of the intermediate in the rate law. (Replace [NO] with an equivalent expression that does not involve intermediates.) To do this, we should solve for [NO] in the fast step rate laws and use that equivalence to replace it in the slow step.

[NO]2 = kf[NO2]2[SO2]2/kr[SO3]2 normally we would replace the [NO]2 term in the slow step rate law with this expression, however this new expression also has an intermediate present (NO2), so there is no way to write this rate law without having an intermediate in it.

If this question were asked on the test, there would be a rate law that you could determine, so follow the steps above and make sure your overall rate law has no intermediates in it.