Reaction Kinetics Solved Exercise
Solved Exercise from chapter 11 Reaction Kinetics chemistry book 1 for FSC pre engineering and pre medical for Board of Intermediate and Secondary education. Also For Entry Test Preparation for UET, NUST, PIEAS, GIKI, AIR, FAST, WAH University, UHS, other engineering Universities and Medical Colleges.
TEXT BOOK EXERCISE
Q1. Multiple choice questions
(i) In a zero order reaction, the rate is independent of
(a) Temperature of reaction.
(b) Concentration of reactants.
(c) Concentration of products
(d) None of these
(ii) If the rate equation of a reaction 2A + B product is , rate =k[A]2[B]. and A is present in large excess, then order of reaction is
(a) 1 (b) 2
(c) 3 (d) none of these
(iii) The rate of reaction
(a) Increases as the reaction proceeds.
(b) Decreases as the reaction proceeds.
(c) Remains the same as the reaction proceeds.
(d) May decrease or increase as the reaction proceeds.
(iv) With increase of 10oC temperature, the rate of reaction doubles. This increase in rate of reaction is due to:
(a) Decrease in activation energy of reaction.
(b) Decrease in the number of collisions between reactant molecules.
(c) Increase in activation energy of reactants.
(d) Increase in number of effective collisions.
(v) The unit of the rate constant is the same as that of the rate of reaction in
(a) First order reaction
(b) Second order reaction
(c) Zero order reaction
(d) Third order reaction
Ans. (i) b (ii) a (iii) b (iv) d (v) c
Q2. Fill in the blanks with suitable words.
(i) The rate of an endothermic reaction_______wiht the increase in temperature.
(ii) All radioactive disintegration nuclear is of ________order.
(iii) For a fast reaction the rate constant is relatively _______and half-life is ________.
(iv) The second order reaction becomes ________if one of the reactants is in large excess.
(v) Arrhenius equation can be used to find out_________of a reaction.
An. (i) increases (ii) first (iii) large : small
(iv) First order (v) energy of activation
Q3. Indicate TRUE or FALSE as the case may be.
(i) The half-life order reaction increases with temperature.
(ii) The reactions having zero activation energies are instantaneous.
(iii) A catalyst makes a reaction more exothermic.
(iv) There is difference between rate law and the law of mass action.
(v) The order of reaction is strictly determined by the Stoichiometry of the balanced equation.
Ans. (i) False (ii) True (iii) False (iv) True
(v) False
Q4. What is chemical kinetics? How do you compare chemical kinetics with chemical equilibrium and thermodynamics?
Ans. Chemical thermodynamics can be used:
(i) To predict whether or not a reaction will proceed to the right, as written.
(ii) To predict the extent to which a reaction will proceed before reaching a condition of equilibrium.
(iii) To enable the amount of energy theoretically required by or released during reactions to be calculated,
Thermodynamics cab tells us where the position of equilibrium lies between the reactants and the products. It enables us to calculate the equilibrium constants. It has the following limitations.
(i) It can only predict the us where possibility of a reaction but not its success.
(ii) It cannot tell us about the rate at which a reaction will occur. It is because time is not a thermodynamic variable.
(iii) It provides no information about the mechanism of the reaction. This is because the change in any state function is path independent.
Chemical kinetics can be used:
(i) To determinc the rate of reaction, that is the rate at which products are formed or the rate at which reactions are used up in the reaction.
(ii) To understand the factors that affects the rate of a reaction.
(iii) To find the mechanism of a reaction.
Example: When gaseous H2 and O2 are mixed, thermodynamics tell us that H2O should be formed, since H2O is more energetically stable than H2 and O2. It is a well known fact that at room temperature, a mixture of H2 and O2 will not produce H2O. However, if the mixture is sparked, water is produced with explosive violence, thereby proving the prediction of thermodynamics. However, it does not give any indication of how fast the reaction will proceed.
Q5. The fast of a chemical reaction with respect to products is written with positive sing, but with respect to reactants is written with a negative sing. Explain it with reference to the following hypothetical reaction.
aA + bB
cC + dD
cC + dD
Ans. The concentration of reactants decrease with time whereas the concentration of products increases with time.
The rate of disappearance of reactant is always negative. A negative sign in the rate expression indicates that the concentration of the reactant decreases with time.
For the general reaction:
aA + bB
cC + dD
cC + dD
The rate of reaction is given by
Rate =- 
. 
= -
.
= +
.
= +
.
. = -.= +.= +.
Remember that in order to have a unique value of the reaction rate (independent of the concentration term chosen), it is necessary to divide each concentration change by its coefficient in the balanced equation for the reaction.
Q6. What are instantaneous and average rates? Is it true that the instantaneous rate of a reaction at the beginning of the reaction is greater than average rate and becomes far less than the average rate near the completion of reaction?
Q7. Differentiate between
(i) Rate and rate constant of a reaction
Ans.
Rate of reaction
|
Rate constant of reaction
|
1. “The change in concentration of a reactant or product per unit time is called the rate of reaction.”
2. The reaction takes place in one phase.
3. The mechanism involves the formation of an intermediate substance.
|
1. “The proportionality constant in rate law equation which relates concentration to the rate of reaction is called rate constant of reaction.”
2. It is independent of the concentration of the reactant.
3. Its units depend upon the order of reaction and differs according to order of reaction.
|
(ii) Homogeneous and heterogeneous catalyses
Ans.
Homogeneous catalyses
|
Heterogeneous catalysis
|
1. “The process in which the catalyst and the reactants are in the same phase is called homogeneous catalysis.”
2. The reaction takes place in one phase. The catalyst is uniformly distributed and is itself in the same phase.
3. The mechanism involves the formation of an intermediate substance.
|
1. “The process in which the catalyst and the reaction are in different phases is called heterogeneous catalysis.”
2. The reaction takes place at a phase boundary. The gaseous molecules react at the surface of the solid catalyst.
3. The mechanism involves adsorption. Molecules are bound to ‘active sites’ on a solid surface.
|
(iii) Fast step and the rate determining step
Ans.
Fast step
|
Rate determining step
|
1. “A relatively rapid step of the reaction mechanism is called fast step.”
2. The overall rate of reaction is independent if this step.
3. The number of molecules of each reactant taking part in this step does not appear in the rate equation.
|
1. “The slowest step of the reaction mechanism is called rate determining step.”
2. The rate of this step determines the overall rate of reaction.
3. The number of molecules of each reactant taking part in this step appears in the rate equation.
|
Remember that except the slow step, all other steps of the reaction mechanism are normally faster steps. The terms slow and fast are relative. They don not necessarily imply that the reaction itself is slow or fast.
(iv) Enthalpy change of reaction and energy of activation of reaction
Ans.
Enthalpy change of reaction
|
Energy of activation of reaction
|
1. “The heat change when the reaction is carried out at constant pressure is called enthalpy change of reaction.”
2. It is the amount of heat evolved or consumed in the course of reaction.
3. It is given the symbol,
 H |
1. “ The minimum amount of energy , in addition to the average kinetic energy. Which the reactant molecules must have for effective collisions is called activation energy of reaction.”
2. It is the minimum amount of energy required to initiate a reaction.
3. It is given the symbol, Ea
|
Q8. Justify the following statements
(i) Rate of chemical reaction is an ever changing parameter under the given conditions.
Ans. The rate of chemical reaction change with time. It has a maximum value at the beginning of the reaction, then gradually decreases and finally becomes zero when the reactants are totally converted into products.
(ii) The reaction rate decreases every moment but rate constant ‘k’ of the reaction is constant quantity, under the given conditions.
Ans. The rate of chemical reaction is not uniform. It depends upon the molar concentrations of reactants. The rate of reaction is directly proportional to concentrations of reactants. Since the concentration of reactants decreases every moment, so the rate of reaction, then gradually decreases with time. It has a maximum value at the start of the reaction, then gradually decreases and finally becomes zero. The constant, ‘k’ of the reaction is independent of concentration. So, it remains constant under the given conditions. However, it changes with temperature.
(iii) 50% of a hypothetical first order reaction completes in one hour. The remaining 50% needs more than one hour to complete.
Ans. Calculation of k:
For a first order reaction
T1/2 =
1 hour =
k=0693 hours-1
Calculation of time for remaining 50%
Completion of reaction:
For first order reaction rate equation:
K=
log
log
t= log
log
Where [A]o is initial concentration and [A] is concentration at any time t.
[A]= 
t =
t =
=
=
t =5.6457 hours
Hence remaining 50% needs more than one hour to complete.
(iv) The radioactive decay is always a first order reaction.
Ans. Since the half-life of a first order reaction in constant and is independent of initial concentration of the reaction, so the radio-active decay is always a first order reaction.
(v) The unit of rate constant of a second order reaction is dm3 mol-1s-1 , but the unit of rate of reaction is mol dm-3s-1.
Ans. For second order reaction: A+B Products
Products
Rate =k[A]2
Units of rate are: mol dm-3s-1.
Units of concentration of A: mol dm-3
Therefore, mol dm-3s-1=k(mol dm-3)2
mol dm-3s-1=(mol2 dm-6)
k=
=mol-1dm3s-1
=mol-1dm3s-1
So, the units of rate constant of 2nd order reaction are: dm3 mol-1s-1
(vi) The sum of the coefficients of a balanced chemical equation is not necessarily important to give the order of a reaction.
Ans. The order of a chemical reaction is the sum of the exponents of the concentration terms in the rate equation. It cannot be written by merely looking at the balanced chemical equation. Usually, the powers of concentration terms in the rate equation are different from coefficients of reactants in the balanced equation. The order of a reaction can be determined only by experiment. For example, for N2O5, the rate is proportional to [N2O5] and its power is one although the coefficient of N2O5in the balanced equation is two.
N2O54NO2+O2
4NO2+O2
Rate = k [N2O5]
A reactant whose concentration does not affect the reaction rate is not included in the rate equation. The total number of molecules or atoms taking part in a reaction can never be zero or in fraction while the order of reaction can be zero or in fraction.
(vii) The order of a reaction is obtained from the rate expression of a reaction and the rate expression is obtained from the experiment.
Ans. The rate expression can be only being determined experimentally. It expresses actual dependence of the rate on the concentrations of the reactants. A reactant whose concentration does not affect the reaction rate is not included in the rate equation. The order of reaction is the number of reacting molecules whose concentrations alter as a result of a chemical reaction. Hence the order of a reaction is obtained from the rate equation and the rate expression is obtained from the experiment.
Q9. Explain that half-life method for measurement of the order of a reaction can help us to measure the order of even those reactions which have a fractional order.
Q10. A cure is obtained when a graph is plotted between time on x-axis and concentration on y-axis. The measurement of the slopes of various points give us the instantaneous rates of reaction. Explain with suitable examples.
Q11. The rate determining step of a reaction is found out from the mechanism of that reaction. Explain it with few examples.
Q12. Discuss the factors which influence the rates of chemical reactions.
Q13. Explain the following facts about the reaction.
2NO(g) + 2H2(g) 2H2O(g) +N2(g)
2H2O(g) +N2(g)
(i) The changing concentrations of reactants change the rates of this reaction.
(ii) Individual orders with respect to NO and H2 can be measured.
(iii) The overall order can be evaluated by keeping the concentration of one of the substances constant.
Q14. The collision frequency and the orientation of molecules are necessary conditions for determining the proper rate of reaction. Justify the statement.
Q15. How does Arrhenius equation help us to calculate the energy of activation of a reaction?
Q16. Define the following terms and give examples.
(i) Homogeneous catalysis
(ii) Heterogeneous catalysis
(iii) Activation of a catalyst
(iv) Auto-catalysis
(v) Catalytic poisoning
(vi) Enzyme catalysis.
Q17. Briefly describe the following with examples:
(i) Change of physical state of a catalyst at the end of reaction.
Ans. There may be change in physical state such as the particle size or change in the colour of the catalyst at the end of reaction. For example, granular MnO2 used as a catalyst in the thermal decomposition of KC1O3 is left as fine powder at the end of reaction.
(ii) A very small amount of a catalyst may prove sufficient to carry out a reaction.
Ans. Since a catalyst is not used up in the reaction, a very small amount of catalyst is required. It can catalyses the reaction over and over again. Sometimes a trace of a metal catalyst is required to affect very large amount of reactants. For example, 1 mg of fine platinum powder can convert 2.5 dm3 of H2 and 1.25 dm3 of O2 to water. Dry HC1 and NH3 combine in the presence of trace of moisture to give dense white fume of NH4C1.
(iii) A finely divided catalyst may prove more effective.
Ans. A catalyst is more effective when it is present in a finely divided form than it is used in bulk. With the increase of fine subdivision, the free surface area is increased. As a result, the active sites on the surface are increased. Consequently, the activity of the catalyst is also enhanced. For example, a lump of platinum will have much less catalytic activity than colloidal platinum. Finely divided nickel is a better catalyst than lumps of solid nickel.
(iv) Equilibrium constant of a reversible reaction is not changed in the presence of a catalyst.
Ans. A catalyst for the forward reaction is also a catalyst for the reverse reaction. A catalyst speeds up the rate of both the forward and reverse reactions equally. So the equilibrium constant (k=
) for the reaction remains the same. A catalyst helps the equilibrium to e established earlier. For example, the reaction of N2 and H2 to from NH3 is very slow. In the presence of the catalyst, the equilibrium
) for the reaction remains the same. A catalyst helps the equilibrium to e established earlier. For example, the reaction of N2 and H2 to from NH3 is very slow. In the presence of the catalyst, the equilibrium
N2(g) + 3H2(g) 
2NH2(g)
2NH2(g)
Is reached much sooner but the percentage yield remains uncharged.
(iv) A catalyst is specific in its action.
Ans. A catalyst cans catalyst only a specific reaction. When a particular catalyst works for one reaction it may not necessarily work for any other reaction. It may increase the rate of one reaction but not increase the rate of another reaction. For example, MnO2 can catalyse the decomposition of KC1O3 but not that of NH3. The decomposition of NH3 is catalysed by a hot tungsten wire.
2KC1O3(g) 
2KC1(s) + 3O2(g)
2KC1(s) + 3O2(g)
2NH3(g) 
N2(g) + 3H2(g)
N2(g) + 3H2(g)
Remember that transition metals catalyse reaction of different types.
For example, the decomposition of aqueous solutions of hydrogen peroxide (H2O2) is catalysed by MnO2 or colloidal platinum.
2H2O2(l) 2H2O(l) + O2(g)
2H2O(l) + O2(g)
Q18. What are enzymes? Give examples in which they act as catalyst. Mention the characteristics of enzyme catalysis.
Q19. In the reaction of NO and H2, it was observed that equimolecular mixture of gases at 340.5 mm pressure was half changed in 102 seconds. In another experiment with an initial pressure of 288 mm of Hg, the reaction was half completed in 140 seconds. Calculate the order of reaction.
Ans. Initial pressure of reactants = Initial concentrations of reactants.
Given: a1=340.5 mm : t1=102 seconds
A2=288 mm : t2=140 seconds
Order of reaction, n=?
Formula used:
n =1 +
n =1 +
=1 + 1.89=2.89
=1 + 1.89=2.89
n =3
Hence the reaction of third order.
Q20. A study of chemical kinetiecs of a reaction
A + B products
products
Gave the following data at 25oC. Calculate the rate law
Exp. [A] [B] Rate
|
1 1.00 0.15 4.2 x 10-6
2 2.00 0.15 8.4 x 10 -6
3 1.00 0.2 5.6 x 10-6
|
Solution: Examination of the data shows that when the concentration of B is kept constant 0.15. The concentration of A is doubled form 1.00 to 2.00; the rate doubles from 4.2 x 10-6 tp 8.4 x 10-6. Mathematically, the rate of reaction with respect to A is directly proportional to the first power of the concentration of A. this is,
Rate
[A]
[A]
Examination of the experiment 1 and 3 shows that when the concentration of A is kept constant 1.00, the concentration of B is increased from 0.15 to 0.2, the rate of the reaction with respect to B is also increased form 4.1 x10-6 to 5.6 x10-6. Mathematically, the rate of reaction with respect to B is directly proportional to the first power of the concentration of B. that is,
Rate [B]
[B]
Since the rate of a reaction is proportional to the product of concentration of the reactants, so,
Rate [A][B]
[A][B]
Rate=k[A][B]
This equation is known as the rate law for the reaction.
Q21. Some reaction taking places around room temperature have activation energies around 50kJ mol-1.
(i) What is the value of the factor 
at 25oC.
at 25oC.
Ans. Ea=50kJ mol-1 =50000 Jmol-1
R=8.3143 J K-1 mol-1
At 25oC, the factor =
=
=1.72 x10-9
===1.72 x10-9
(ii) Calculate this factor at35oC, and 45oC note the increase in this factor for every 10oC rise in temperature.
Ans. At 25oC, the factor ==
=3.314 x10-9
===3.314 x10-9
At 25oC, the factor ==
=6.12 x 10-9
===6.12 x 10-9
(iii) prove that for every 10oCrise in temperature, the factor doubles and so rate constant also doubles.
Ans. At 25oC, the factor =1.72 x 10-9
=1.72 x 10-9
At 35oC, the factor=3.314 x 10-9
=3.314 x 10-9
At 45oC, the factor=6.12 x 10-9
=6.12 x 10-9
Since for every 10oC rise in temperature, the factor becomes doubles and so rate constant also doubles.
Q22. H2 and 12 react to produce HI. Following data for rate constant at various temperatures (K) have been collected.
Temperature(K) Rate constant( cm3 mol-1s-1)(K)
|
500 6.814 x 10-4
550 2.64 x 10-2
600 0.56 x 10o
650 7.31 x 10o
700 66.67 x 10o
|
(i) Pot a graph between 
on x-axis and log k on the y-axis.
on x-axis and log k on the y-axis.
(ii) Measure the slope of this straight line and calculate the energy for actives this reaction.
Ans. 8326.32: 160.6 Jmol-1.
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