Exercises

EXERCISES

1. Consider the reaction, 2HI → H2 + I2, determine the rate of disappearance of HI when the rate of I2
    formation is 1.8 x 10-6 Ms-1.
  
    Answer : 3.6 × 10-6 Ms-1


2. Because it has a nonpolluting product (water vapor), hydrogen gas is used for fuel aboard the
    space shuttle and may be used by Earth-bound engines in the near future.

   2H2(g) + O2(g) 2H2O(g)
   • Express the rate in terms of changes in [H2],
   [O2] and [H2O] with time.
   • When [O2] is decreasing at 0.23 molL-1 s-1, at
   what rate is [H2O] increasing?
  
   Answer : 0.46molL-1s-1


3. Consider the reaction,

    NO(g) + O2(g) →2NO2(g).
    Suppose that at a particular moment during the
    reaction nitric oxide (NO) is reacting at the rate
    of 0.066 Ms-1

    a) At what rate is NO2 being formed?
    b) At what rate is molecular oxygen reacting?


4. Consider the reaction,

    N2(g) + 3H2(g) → 2NH3(g)
   
    Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the
    rate of 0.074 Ms-1

     a) At what rate is ammonia being formed?
     b) At what rate is molecular nitrogen reacting?

5. ClO2(aq) + 2OH- (aq) → products

    The results of the kinetic studies are given below.

    exp              [ClO2] / M                   [OH-] / M                  Initial rate / Ms-1

     1                     0.0421                         0.0185                        8.21 ×10-3
     2                     0.0522                         0.0185                        1.26 ×10-2
     3                     0.0421                         0.0285                        1.26 ×10-2

    a) Explain what is meant by the order of reaction
    b) Referring to the data determine
         (i) rate law /rate equation
         (ii) rate constant, k
         (iii) the reaction rate if the concentration of both ClO2 and OH- = 0.05 M

   Answer : rate = k [ClO2]2[OH-]
                   k = 250 M-2s-1
                   rate = 3.12 ×10-2 M/s


6. Write rate law for this equation,

     A + B → C

    i) When [A] is double, rate also double. But double the [B] has no effect on rate.

    ii) When [A] is increase 3x, rate increases 3x, and increase of [B] 3x causes the rate to increase 9x.

   iii) Reduce [A] by half has no effect on the rate, but reduce [B] by half causing the rate to be half of the
     initial rate.

   Answer : Rate = k[A]
                 Rate = k[A] [B]2
                 Rate = k[B]

7. C + D → E

    The results of the kinetic studies are given below.

     exp             Initial [ ] (M)         Time interval (Min)          The change in concentration of C (M)

            [C]      [D]

   1                0.10    1.0                        30                                       2.5 x

10-3

      2                0.10    2.0                        30                                       1.0 x 10-2      3                0.05    1.0                       120                                      5.0 x

 

    a) Calculate the rate of reaction for each experiment

    b) Determine the order of reaction with respect to C and D and write the rate law.

    

        of C remains constant.c) State the effect on the reaction rate if the concentration of D is doubled but the concentration   
    Answer : a) 8.33 ×10-5 ,3.33 × 10-4, 4.17×10-5 M min-1 
                   b) rate = k [C] [D]2                   c) rate increase by a factor of 4
 8. The reaction 2A to B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800oC.M to 0.14 M ?

    How long will it take for A to decrease from 0.88

    Answer : t = 66s

9. For the first order decomposition of H2O2(aq) given that k = 3.66 x 10-3 s-1 and [H2O2 ]o = 0.882 M,2 O2] = 0.600 M105.26 s

    determine;

    a) the time at which [H

    b) the [H2O2] after 225 s.

  Answer : a)
                  b) [H2O2] = 0.387 M

10. The decomposition of ethane, C2H6 to methyl radicals is a 1st order reaction with a rate constant-4 s-1 at 700o C.2H6(g) to 2CH3(g)

      of 5.36 x 10

      C

      Calculate the half life of the reaction in minutes.
     
      Answer :  t1/2 = 2 1 . 5 min
11. The decomposition of nitrogen pentoxide is as below;
      
      N
       time, t/min                    0      10      20      30      40      50      60     [N2O5] x 10-4 M      176   124    93      71       53      39     29

    

     The decomposition is first order reaction.

     a)Plot a linear graph to prove it.

     b)From the plot determine rate constant, k

     Answer : k = 0.0296 s-1

12. The decomposition of HI is second order, at 500oC, the halflife of HI is 2.11 min when the initial HI   concentration is 0.10 M. What will be the half-life (in minutes) when the initial HI concentration is 0.010 M? 

     Answer : 21 minutes

13. The rate constant for the first-order decomposition of  N2O5(g) at 100oC is 1.46 x10-1s-1.

      a) If the initial concentration of N2O5 in a reaction vessel is 4.5 ×10-3 mol/L, what will the concentration 

          be 20.0 s after the decomposition begins?

      b) What is the half-life (in s) of N2O5 at 100oC?

     

      Answer : a) 0.00024 M

                    b) 4.75 s

14. For the reaction A + B produce C + D, the enthalpy change of the forward reaction is + 21 kJ/mol. The activation energy of the forward reaction is  84 kJ/mol.

a) What is the activation energy of the reverse reaction ?

b) Sketch the reaction profile of this reaction.

15. In the presence of platinum as a catalyst, hydrogen iodide decomposes to form hydrogen and  iodine. The activation energy for this reaction is 58 kJ mol-1. Calculate the ratio of the rate constant at 30oC and 20oC.

Answer : 0.46

16. The results of the decomposition of N2O5 at two different temperature were recorded as;10-5       308                            6.61 x

a) Base on the unit of the rate constant, k, determine the order of the reaction.

b) Find the value of E

Answer : a) first
              b) 102 kj per mola and A for the reaction. 10-5

Temperature(K)           rate constant, k (s-1)

       298                            1.74 x 

2O5(g) to 2 NO2(g) + ½ O2(g)10-3

ha! jawab, jawab....(4.4)

1.	Phosgene, COCl2, one of the poison gases used during World War I, is formed from chlorine and carbon monoxide. The mechanism is thought to proceed by: step 1: Cl + CO → COCl step 2: COCl + Cl2 → COCl2 + Cl a. Write the overall reaction equation. b. Identify any reaction intermediates. c. Identify any catalysts.
	a. The overall reaction: step 1: Cl + CO → COCl step 2: COCl + Cl2 → COCl2 + Cl overall: CO + Cl2 → COCl2 b. The reaction intermediate is COCl - it is produced during the first step but immediately used up in the second step. c. The catalyst is Cl - a catalyst can be identified by the fact that it is added as a reactant but emerges (as a product) unchanged; thus it will always first appear as a reactant but will later (not necessarily the next step) appear as a product in the reaction. Since it appears on both sides of the equation it will be cancelled out of the net or overall equation, just as are reaction intermediates.
2.	We have typically been simplifying our potential energy curves somewhat; for multistep reactions, potential energy curves are more accurately shown with multiple peaks. Each peak represents the activated complex for an individual step. Consider the PE curve for a two-step reaction:
	Answers a) What is ΔH for the overall reaction? - 20 kJ b) What is ΔH for the first step of the reaction mechanism? + 20 kJ c) What is ΔH for the second step of the reaction mechanism? -40 kJ d) What is ΔH for the overall reverse reaction? +20 kJ e) What is Ea for the first step? +80 kJ f) What is Ea for the second step? +40 kJ g) Which is the rate-determining step - step 1 or step 2? How do you know? step 1 because it has the largest activation energy, Ea h) What is Ea for the reverse of step 1? +60 kJ i) Is the overall reaction endothermic or exothermic? exothermic

ha! jawab, jawab....(4.2&4.3)

1.	Consider the following reaction that occurs between hydrochloric acid, HCl, and zinc metal: HCl(aq) + Zn(s) → H2 (g) + ZnCl2 (aq) Will this reaction occur fastest using a 6 M solution of HCl or a 0.5 M solution of HCl? Explain.
	Solution: The reaction will occur fastest with 6 M HCl, because it is more concentrated than the 0.5 M solution. In the more concentrated solution there are more moles of HCl present - with a higher concentration of reacting particles, collisions will occur more frequently, leading to a faster rate of reaction.
2.	Again consider the reaction between hydrochloric acid and zinc. How will increasing the temperature affect the rate of the reaction? Explain.
	Solution: Increasing the temperature will most likely increase the rate of the reaction, for two reasons: Particles will move around faster at the higher temperature and thus will collide more frequently, resulting in a faster rate of reaction. Particles will collide with more force. Thus, more particles will likely have sufficient energy (Ea) to reach the activated complex and thus have a successful collision.
3.	Based on the following kinetic energy curves, which reaction will have a faster rate - A or B? Explain. Also, which reaction, A or B, would benefit most in terms of increased rate if the temperature of the system were increased?
	A. B.
	Solution: Reaction B would be faster than Reaction A because it has a lower threshold energy (activation energy). Thus, more particles have at least the minimum amount of energy required for a successful reaction. Reaction A would benefit most by an increase in temperature. Reaction B already have the majority of particles above the threshold energy; having more particles above the threshold would not make a significant difference in the rate.

ha! jawab, jawab....(4.1)

1.

Which one of the following reactions would you expect to be fastest at room temperature and why?

	Solution:
Pb2+(aq) + 2 Cl-(aq) →PbCl2 (s)	fastest - ions in aqueous solution react very quickly; all are in the same phase
Pb(s) + Cl2 (g) → PbCl2 (s)	slower - one of the reactants is a solid

 

---

2.

Consider the following reactions. Which do you predict will occur most rapidly at room conditions? Slowest?

	Solution:
C2H6 (g) + O2 (g) → 2 CO2 (g) + 3 H2O(g)	slow due to covalent bonding (unless the reaction is highly exothermic)
Fe(s) + O2 (g) → Fe2O3 (s)	slowest - solid reactant (Fe); this reaction describes the rusting of iron
H2O(l) + CO2 (g) → H2CO3 (g)	slow due to covalent bonding
2 Fe3+(aq) + Sn2+(aq) → 2 Fe2+(aq) + Sn4+(aq)	fastest - ions in solution react very quickly

ha! jawab, jawab....(3)

1.	Answer the following questions based on the potential energy diagram shown here: Does the graph represent an endothermic or exothermic reaction? Label the postion of the reactants, products, and activated complex. Determine the heat of reaction, ΔH, (enthalpy change) for this reaction. Determine the activation energy, Eafor this reaction. How much energy is released or absorbed during the reaction? How much energy is required for this reaction to occur?
	Solution The graph represents anendothermic reaction the reactants, products, and activated complex are shown
	c. ΔH = +50 kJ. Since this in an endothermic reaction, ΔH will have a positive value. ΔH is the difference in energy between the energy levels of the initial reactants (50 kJ) and the final products (100 kJ), and does not depend on the actual pathway (remember Hess's Law?) d. Ea = +200 kJ. Activation energy is the amount of energy required to go from the energy level of the reactants (50 kJ) to the highest energy point on the graph, the activated complex (250 kJ). e. 50 kJ of energy are absorbed during this endothermic reaction (this is the value of ΔH) f. 200 kJ of energy are required for this reaction to occur ( Ea). Even though ΔH is only 50 kJ, enough energy must be supplied to reach the activated complex, or the reaction will not occur.
2.	Sketch a potential energy curve that is represented by the following values of ΔH and Ea. You may make up appropriate values for the y-axis (potential energy). ΔH = -100 kJ and Ea = 20 kJ Is this an endothermic or exothermic reaction?
	Solution: Since ΔH is a negative number, we know that the reaction is exothermic. Therefore, begin by sketching an exothermic potential energy graph:
	Next, assign values to the y-axis that will satisfy our requirements, namely ΔH = -20 kJ and Ea = +60kJ Remember: ΔH is the difference between energy levels of the reactants and products - this difference will equal -20 kJ. Ea is the difference between energy levels of the initial reactants and the peak of the curve, the activated complex.
3.	In the next unit we will be discussing reactions that are reversible, and can go in either the forward or reverse directions. For example, hydrogen gas and oxygen gas react to form water, but water can also be broken down into hydrogen and oxygen gas. We typically write a reaction that can be reversed this way, using the double arrow symbol ( or ↔): 2 H2 + O2 ↔ 2 H2O This reaction is exothermic in the forward direction: 2 H2 + O2 ↔ 2 H2O + 285 kJ but endothermic in the reverse direction: 2 H2O + 285 kJ ↔ 2 H2 + O2 Consider a general reversible reaction such as: A + B ↔ C + D
	Given the following potential energy diagram for this reaction, determine ΔH and Ea for both the forward and reverse directions. Is the forward reaction endothermic or exothermic?
	Solution: You should immediately see that the forward reaction is exothermic - the products (C + D) are at a lower energy level than the reactants.	ΔHforward = -20 kJ the difference in energy level between the reactants and products. Ea forward = +60 kJ amount of energy required to go from the initial reactants A + B to the activated complex, the peak of the graph. ΔHreverse = +20 kJ difference in energy from C + D and A + B. Notice the value of ΔH does not change, only the sign. Ea reverse = +80 kJ amount of energy required to go from C + D to the activated complex.
4.	Sketch a potential energy diagram for a general reaction A + B ↔ C + D Given that ΔHreverse = -10 kJ and Ea forward = +40 kJ
	Solution: Begin by determining whether the forward reaction (A + B → C + D) is endothermic or exothermic: Since ΔHreverse = -10 kJ you can determine that ΔHforward = +10 kJ (same value, just change the sign). Since ΔHforward is positive you know that the forward direction of the reaction is endothermic. Begin by sketching a general endothermic potential energy curve that is endothermic. Don't worry about drawing to scale at this time:
	At what energy level should you put the activated complex, the top of the peak? Since Ea forward = +40 kJ, you know that from the initial reactants (A + B) you need to go up 40 energy units. Pick any value for a starting energy level (let's use a starting point of 10 here), then add 40 to find the peak of your graph. Return to ΔH to determine the energy level of the products. Since ΔHforward = +10 kJ, add 10 to our starting value to determine the energy level of the products:

ha! jawab, jawab....(2)

1.	Nitrogen monoxide reacts with hydrogen gas to produce nitrogen gas and water vapour. The mechanism is believed to be: Step 1: 2 NO → N2O2 Step 2: N2O2 + H2 → N2O + H2O Step 3: N2O + H2 → N2 + H2O For this reaction find the following: the overall balanced equation any reaction intermediates
	Solution To find the overall balanced equation, cross out substances that appear in equal numbers on both sides of the reaction and add together like items on the same side of the equation: Step 1: 2 NO → N2O2 Step 2: N2O2 + H2 → N2O + H2O Step 3: N2O + H2 → N2 + H2O   Net Reaction: 2 NO + 2 H2 → N2 + 2 H2O To identify the reaction intermediates, look for substances that first appear on the product side of the equation, but then appear in the next step as a reactant. In this example there are two reaction intermediates - N2O2 and N2O.
2.	Give two reasons why most molecular collisions do not lead to a reaction.
	Solution: The collision may not have the correct orientation or the necessary energy.
3.	An important function for managers is to determine the rate-determining steps in their business processes. In a certain fast-food restaurant, it takes 3 minutes to cook the food, 1.5 minutes to wrap the food, and 5 minutes to take the order and make change. How would a good manager assign the work to four employees?
	Solution: Assign two workers to take the orders since that is the rate determining step.

ha! jawab, jawab....(1)

1.	In the following decomposition reaction,  2 N2O5 → 4 NO2 + O2 oxygen gas is produced at the average rate of 9.1 × 10-4 mol · L-1 · s-1. Over the same period, what is the average rate of the following: the production of nitrogen dioxide the loss of nitrogen pentoxide
	Solution: From the equation we see that for every 1 mole of oxygen formed, four moles of nitrogen dioxide are produced. Thus, the rate of production of nitrogen dioxide is four times that of oxygen: rate NO2 production = 4 × (9.1 × 10-4 mol · L-1· s-1)   = 3.6 × 10-3 mol · L-1· s-1 Nitrogen pentoxide is consumed at twice the rate that oxygen is produced: rate loss of N2O5 = 2× (9.1 × 10-4 mol · L-1· s-1)   = 1.8 × 10-3 mol · L-1· s-1
2.	Consider the following reaction: N2(g) + 3 H2(g) → 2 NH3(g) If the rate of loss of hydrogen gas is 0.03 mol · L-1· s-1, what is the rate of production of ammonia?
	Solution: From the balanced equation we see that there are 2 moles NH3 produced for every 3 moles H2 used. Thus: rate NH3 production = 2 3 × (0.03 mol · L-1· s-1)   = 0.02 mol · L-1· s-1

11.3 Reaction Kinetics

11.3 Factor Affecting Reaction Rate

Objectives

1.    Explain factors affecting reaction rate;

              -Concentration / Pressure   -Temperature

              -Catalyst                              -Particle size

    2. Explain the effect of temperature on reaction

        rate using Maxwell-Boltzmann distribution

        curve.

    3. Explain the effect of catalyst on activation

       energy based on energy profile diagram for

       exothermic and endothermic reactions.

   4. Relate rate constant to temperature and activation energy using the Arrhenius   

       equation

      

       k = Ae-Ea/RT or ln k = ln A –Ea/RT

   5. Determine k, Ea, T and A using Arrhenius

       equation by calculation and graphical method.

Ln (k1 / k2) = Ea/R (1/T2 – 1/T1)

Factors Affecting Reaction rate

a) Effect of Concentration

o When concentration of reactant increases, frequency

of collision also increases.

o More particles present in the same volume, they are more likely to collide

o The probability of effective collisions increase

frequency of effective collision = collision frequency x fraction of molecules with

                                                                                           sufficient energy

o Rate of reaction increases.

   [reactants] increases,the frequency of collision increases

o This observation correlates with the RATE LAW that has been previously  discussed…

    Reaction rate = k [ A ]x [ B ]y …

                             (A & B = reactants)

                            (x & y = rate order)

o Based upon this equation,

                       Reaction rate directly proportional to the concentration of reactants

o REMINDER!

    Only in zero order reactions, the rate of reaction is not

    dependant upon the concentration of the reactants.

    (depending on its rate order)

    Reaction rate = k [ A ]0 = k (constant)

b) Effect of Temperature

o At a higher temperature, molecules have higher kinetic energy and move at higher speed

o more collisions will occur in a given time

o Furthermore, the higher the KE, the higher the energy of the collisions

o So more molecules will have energy greater than Ea

o effective collision also will be increased

o thus the reaction rate increases.

Distribution of Kinetic of Molecules(Maxwell-Boltzmann Distributions)

The collision frequency is higher temperature because the fraction of reactant molecules with activation energy is higher.

Maxwell-Boltzmann Distributions

• The figure shows the distribution KE of gaseous molecule at temperature T1 and T2.

• At higher temperature, the fraction of molecules with energy greater than Ea increases.axwel-Boltzmann Distributions

• Thus, rate of reaction increase with increase of

temperature.

Take note:

• Area under the curve ∝ to the total number of molecules

area under the curve is the same, bcoz no. of molecules is the same.

Maxwel-Boltzmann Distributions

• There are a wide range of molecular energy

A few molecules have low and high KE, most have value in the middle.

• The shape of the graph changes with temperature

At higher temperature the peak of the graph moves towards the right to a higher KE value as the curve broadens out.

• At higher temperatures, the fraction of molecules with higher energy increase

Average energy increase at higher temperature, but fraction of molecules with low energy decreases. At higher temperature, fraction of molecules with energy

greater than Ea increase.

Effect Of Temperature

ARRHENIUS EQUATION

• the effect of temperature on the rate constant, k:

                                k = A e -Ea⁄RT

Where…

k = rate constant

A = frequency factor

(is a measure of the probability of a favorable collision)

e = natural log exponent

Ea = activation energy for the reaction (kJ/mol)

R = universal gas constant (8.314 J mol-1 K-1)

T = absolute temperature (T in Kelvin)

Arrhenius Equation - Derivation

  k = A e^ -Ea⁄RT

ln k = ln (A.e^ –Ea/RT)

ln k = ln A – Ea/RT

ln k = ln A - Ea / R (1/T)

   y = C + mx

Example

The table below gives the rate constants, k for the reaction between potassium hydroxide and bromoethane at different temperatures.

K(M^-1s^-1)	T(k)
0.63	322
2.50	331
10.0	347
22.6	353

a) Using a graphical method, calculate the activation energy (kJmol-1) for this reaction.

b) What is the overall order of reaction? Explain

c) Calculate the initial rate of reaction at 330 K when the concentrations for both KOH and   CH3CH2Br are 0.1M.

Solution……

1/T	0.0031	0.0030	0.0029	0.0028
ln k	-0.46	0.92	2.30	3.12

a) Slope = Ea/R

              = 12492

Ea = 12492 × 8.314

     = 1.04 ×105 Jmol-1

     = 104 kJmol-1

b) Second order, unit of k = M-1s-1

 c) ln k = -Ea/R(1/T) + ln A

     ln k = -12492 (1/330) + 38.45

         k = 1.81 M^-1s^-1

Rate = k [KOH][CH3CH2Br]

        = 1.81 x 0.1 x 0.1

        = 1.81 x 10-2 Ms-1

Arrhenius Equation – Further Derivation

Rate constant, k varies with T.

The ratio of rate constant at two different T can be calculated.

ln k1 = ln A – Ea / RT1……………………….(1)

ln k2 = ln A – Ea / RT2……………………….(2)

Equation (1) minus (2) gives :

ln ( k1/k2) = Ea / R (1/T2-1/T1)

Effect Of Catalyst

• A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed.

• It increases the reaction rate by providing an alternative reaction pathway of which having lower activation energy.

• A catalyst provides a different reaction mechanism.

A catalyst provides an alternative pathway for the reaction to occur (----curve) which has a lower activation energy.

When Ea decreases, k increases, reaction rate increases

Effect Of Surface Area

• For reactions that occur on a surface that is for solid, the rate increases as the surface

area is increased.

• A larger surface area increases the contact area between the reactants thus increases

the frequency of collision and the probability of effective collision between the molecules

of the reactants.