FSc Notes Chemistry XI Chapter 11 Reaction Kinetics Exercise Short Questions: 1st Year FSc Chemistry Notes Online Taleem Ilm Hub
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Q1. Multiple Choice Questions:
(i) In zero order reaction, the rate is independent of:
(ii) If the rate equation of a reaction 2A + B -> products is, rate = k[A]2[B], and A is present in large excess, then order of reaction is:
(iii) The rate of reaction:
(iv) With the increase of 10oC temperature the rate of reaction doubles. This increases in rate of reaction is due to:
(v) The unit of the rate constant is the same as that of the rate of reaction in:
Q2. Fill in the blanks:
(i) The rate of an endothermic reaction increases with the increase in temperature.
(ii) All radioactive disintegration nuclear reaction are of First order order.
(iii) For a fast reaction the rate constant is relatively greater and half-life is short.
(iv) The second order reaction becomes First order if one of the reactants is in large excess.
(v) Arrhenius equation can be used to find out activation energy of a reaction.
Q3. Indicate true or false as the case may be:
(i) The half-life of a first order reaction increases with temperature. False
(ii) The reactions having zero activation energies are instantaneous. True
(iii) A catalyst makes a reaction more exothermic. False
(iv) There is difference between rate law and the law of mass action. True
(v) The order of reaction is strictly determined by the stoichiometry of the balanced equation. False
Q4. What is chemical kinetics? How do you compare chemical kinetics with chemical equilibrium and thermodynamics?
Ans. Chemical Kinetics: The study of rates of chemical reactions and factors affecting the rate is called as "chemical kinetics".
Difference between chemical kinetics and chemical equilibrium:
| Chemical Kinetics | Chemical Equilibrium | ||
|---|---|---|---|
| (i) The study of rates and mechanism of reaction is called chemical kinetics. | (i) A state of reversible reaction at which forward and reverse rates becomes equal is called chemical equilibrium | ||
| (ii) Rate study is possible for both reversible and irreversible reactions. | (ii) Chemical equilibrium is the characteristics of reversible reactions only. | ||
| (iii) It is related to rate of reactions and factors affecting it. | (iii) It is related to state of equilibrium and factors affecting it. |
Q5. The rate of chemical reaction with respect to products is written with positive sign, but with respect to reactants is written with a negative sign. Explain it with reference to the following hypothetical reaction.
aA + bB -> cC + dB
Ans. The rate of reaction can be written with respect to reactants or products concentration. The rate of reaction with respect to concentration of reactants is written with negative sign because concentration of reactants decrease with the passage of time.
Rate = -d[A] / dt or -d[B] / dt
The rate of reaction with respect to concentration of products is written with positive sign because concentration of products increases as the time passes.
Rate = +d[C] / dt or +d[D] / dt
Q6. What are instantaneous and average rates? Is it true that the instantaneous rate of 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?
Ans. The rate measured at any moment of reaction during the process is called instantaneous rate, whereas rate of reaction between two time intervals is called average rate.
Instantaneous rate is very fast at the start. It slows down with decreasing concentration of reactants and very slow at the end, but average rate has an intermediate value (neither very slow nor very fast) because it is obtained by dividing the whole change in concentration with time.
It clearly shows that instantaneous rate is faster than average rate at the start but slower than average rate at the end.
Q7. Differentiate between:
(ii) Homogenous and heterogeneous catalysis.
(iii) Fast step and the rate determining step.
(iv) Enthalpy change of reaction and energy if activation of reaction.
Ans. (i) Rate and rate constant of a reaction.
| Rate of Reaction | Rate Constant of Reaction |
|---|---|
| (i) It is the change in concentration of reactant or product per unit time. Rate = dx/dt |
(i) It is the proportionality constant and is the ratio of rate and concentration of reactants. K = Rate / [A]a[B]b |
| (ii) Its units are moles.dm-3s-1. | (ii) Its units are variable depending upon order of reaction. |
| (iii) It is a variable parameter. | (iii) It is always constant under given conditions. |
| Homogenous Catalysis | Heterogenous Catalysis |
|---|---|
| (i) A catalyzed reaction in which catalyst and reaction mixture are in same phase is called homogeneous catalysis. | (i) A catalyzed reactions in which catalyst and reaction mixture are in different phases is called heterogeneous catalysis. |
| (ii) Example: 2SO2(g) + O2 ⇌NO2(g)⇌ 2SO3(g) | (ii) Example: CH2=CH2(g) + H2(g) --Ni(s)-> CH3-CH3(g) |
| (iii) In this case, catalyst is distributed uniformly throughout the system. | (iii) In this case, catalyst is not uniformly distributed throughout the system. |
| Fast Step | Rate Determining Step |
|---|---|
| (i) A step in a multi-step reaction that has high speed and do not affect the overall rate of reaction is called fast step. | (i) A step in a multi-step reaction that is slowest one and controls the overall rate of reaction is called rate determining step. |
| (ii) The reactants in fast step are not involved in rate equation. | (ii) Reactants in rate determining step are involved in rate equation. |
| (iii) Example: NO2 + NO2 --slow--> NO3 + NO NO3 + CO --fast--> NO2 + CO2 Second step is fast step. |
(iii) Example: NO2 + NO2 --slow(RDS)--> NO3 + NO NO3 + CO --fast--> NO2 + CO2 First slow step is rate determining step. |
| Enthalpy Change of Recation | Energy Activation of Reaction |
|---|---|
| (i) Heat change accompanied by chemical reaction is called enthalpy change. | (i) The minimum amount of energy in addition to average K.E., sufficient to convert reactants into products is called activation energy. |
| (ii) It is denoted by 𐤃H. | (ii) It is denoted by Ea. |
| (iii) It is the difference of enthalpies of reactants and products. | (iii) It is the energy barrier between reactants and products. |
| (iv) It may be negative or positive. | (iv) It is always positive. |
Q8. Justify the following statements:
(i) Rate of chemical reaction is an ever changing parameter under the given conditions.
(ii) The reaction rate decreases every moment but rate constant 'k' of the reaction is a constant quantity, under the given conditions.
(iii) 50% of a hypothetical first order reaction completes in one hour. The remaining 50% needs more than one hour to complete.
(iv) The radioactive decay is always a first order reaction.
(v) The unit of rate constant of a second order reaction is dm3 mol-1.s-1, but the unit of rate of reaction is mol.dm-1.s-1.
(vi) The sum of the coefficients of a balanced chemical equation is not necessarily important to give the order of a reaction.
(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. (i) The rate of reaction is directly proportional to concentration of reactants. As the concentration of reactants is maximum at start, the rate is fast. As the time passes, the concentration of reactants decreases and rate of reaction also slows down.
(ii) The rate of reaction decrease every moment with decreasing concentration of reactants. For understanding of rate constant, consider the following hypothetical reaction.
A + B --> C + D
Rate = K [A] [B]
or K + Rate / [A][B]
From above equation, it is clear that rate constant "K" is the ratio of rate of reaction and concentration of reactants. As the concentration of reactants decreases, rate of reaction also decreases but the ratio remains constant.
Hence, rate constant "K" always a constant quantity.
(iii) 50% of hypothetical reaction is completed in one hour and 50% left behind.
From this 50% reaction , 25% will be completed in one hour and 25% will be left behind. Among this 12.5% will still left behind.
Hence this remaining 50% requires more than one hour to complete.
(100% --1hour--> 50% --1hour--> 25% --1hour-->12.5%...)
(iv) During radioactive decay, nuclei of only one type are broken up and no other specie is involved. The rate of this decay depends upon concentration of nuclei of only one type. Hence, it follows first order kinetics.
(v) Equation for rate of reaction is:
Rate = 𐤃C/𐤃t
= Change in conc./Change in time
= moles.dm-3/s1
Rate = moles.dm-3s-1
Second order reaction:
A + B --> C + D
Rate = K [A] [B]
K = Rate / [A] [B]
= moles.dm-3.s-1 / moles.dm-3 x moles.dm-3
K = dm3.mol-1.s-1
(vi) For many reaction, the order is not equal to the sum of co-efficient of balanced because reaction proceeds in more than one steps and order and rate expression is derived from slowest step (rate determining step).
Consider the reaction:
2NO + 2H2 ---> 2H2O + N2
The sum of co-efficients is four and order is three. This order is experimentally determined from mechanism of reaction.
Mechanism:
2NO + H2 --slow-> H2O2 + N2 (rate determining step)
H2O2 + H2 --fast-> 2H2
Rate = K[H2]1[NO]2
(vii) Many reactions that consists of more than one steps. There is always one step which is slowest one and controls the overall rate and order of reaction. This slow step is called rate determining step and is experimentally determined. And rate expression is written on the basis of this step.
Experimentally determined mechanism of reaction is:
NO2 + NO2 --slow-> NO3 + NO
NO3 + CO --fast-> NO2 + CO2
and rate expression is:
Rate = K[NO2]2
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.
Ans. Will be updated soon.
Q10. A curve 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.
Ans. Will be updated soon.
Q11. The rate determining step of a reaction is found out from the mechanism of that reaction. Explain with few examples.
Ans. Will be updated soon.
Q12. Discuss the factors which influence the rates of chemical reactions.
Ans. Will be updated soon.
Q13. Explain the following facts about the reaction.
2NO(g) + 2H2(g) ---> 2H2O(g) + N2(g)
(i) The changing concentration 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.
Ans. (i) The effect of changing concentration or rates of this reaction can be studied from following experimental data.
| - | [NO] | [H2] | Rates |
|---|---|---|---|
| 1. | 0.006 | 0.001 | 0.025 |
| 2. | 0.006 | 0.002 | 0.050 |
| 3. | 0.006 | 0.003 | 0.075 |
| 4. | 0.001 | 0.009 | 0.0063 |
| 5. | 0.002 | 0.009 | 0.025 |
| 6. | 0.003 | 0.009 | 0.056 |
The data shows that rate of reaction is changed when concentration of one reactant is changed by keeping the other's constant.
(ii) Above data shows that when rate is studied by changing the concentration of H2 and by keeping the concentration of NO constant. It comes out to be that rate of reaction is directly proportional to the concentration of H2.
Rate = K[H2]1
And individual order with respect to hydrogen is one.
Similarly, rate of reaction is directly proportional to the square of concentration of NO. It is clear from data, when concentration of NO is increased to two times, rate increases four times.
Rate = K[NO2]2
And individual order with respect to NO is 2 and reaction is of 2nd order.
(iii) When concentration of NO is kept constant, the rate of reaction comes out to be directly proportional to concentration of H2 and order is 1.
Rate = K[H2]1
When concentration of H2 is kept constant, the rate of reaction comes out to be directly proportional to square of concentration of NO and order is 2.
Hence, overall rate expression is:
Rate = K[H2]1[NO]2
So, overall order is 3 and reaction is of third order.
Q14. The collision frequency and the orientation of molecules are necessary conditions for determing the proper rate of reaction. Justify the statement.
Ans. Will be updated soon.
Q15. How does Arrhenius equation help us to calculate the energy of activation of reaction?
Ans. Will be updates soon.
Q16. Define the following terms and give examples:
(ii) Heterogeneous catalysis
(iii) Activation of a catalyst
(iv) Auto-catalysis
(v) Catalytic poisoning
(vi) Enzyme catalysis
Ans. Will be updated soon.
Q17. Briefly describe the following with examples:
(ii) A very small amount of a catalyst may prove sufficient to carry out a reaction.
(iii) A finely divided catalyst may prove more effective.
(iv) Equilibrium constant of a reversible reaction is not changed in the present of a catalyst.
(v) A catalyst is specific in its action.
Ans. (i) A catalyst remains chemically unchanged but changes physically at the end of reaction:
Example: (i) Granular catalyst (MnO2) for the decomposition of KClO3 is converted into powder form.
(ii) Shining surfaces of Pt is made dull, when it is used as catalyst for hydrogenation of vegetable oil
(ii) Since, catalyst itself is not used up, it can catalyze a reaction between large amount of reactants.
Example: (i) 1g of Pt can catalyze a reaction between 2.5dm3 of H2 and 1.25dm3 of O2 to form water.
(ii) Thousands of dm3 of H2O2 can decompose in the presence of 1g of colloidal platinum.
(iii) When catalyst is finely divided, it has greater surface area and is more effective.
Example: (i) Colloidal platinum has far more catalystic activity than lump of Pt.
(ii) For hydrogenation of vegetable oils, finely divided Nickel is used.
(iv) When catalyst is added at the equilibrium position of a reversible reaction, it equally speed up the rate to forward and backward reaction. Hence, equilibrium is not disturbed and equilibrium constant remains constant.
Example: 2SO2 + O2 ⇌V2O5⇌ 2SO3
(v) When a particular catalyst works for one reaction, it may not necessarily work for any other reaction. If different catalysts are used for same reactants then the product may change.
Example: (i) HCOOH --Al2O3-> H2O+ CO
(ii) HCOOH --Cu-> H2 + CO2
Q18. What are enzymes? Give examples in which they act as catalyst. Mention the characteristics of enzyme catalyis.
Ans. Will be updated soon.
Q19. In the reaction of NO and H2, it was observed that equimolecular mixture of gases at 340.5 mm of Hg, was half changed in 102 seconds. In another experiment with an initial pressure of 288 mm Hg, the reaction was half completed in 140 seconds. Calculate the order of reaction.
Ans. Initial pressure of NO and H2 are the initial concentration of reactants and denoted by "a" and a2". The half life period t1 and t2. a1 = 340.5 mm Hg
Half-life period t1 = 102 seconds
a2 = 288 mm Hg
Half-life period t2 = 140 seconds
n = 1 + [log(t1/t2) / log(a2/a1)]
Putting the values:
n = 1 + [log(102/140) / log(288/340.5)]
n = 1 + [log(0.728) / log(0.845)]
n = 1 + [-0.1378 / -0.731]
n = 1 + [1.88] = 2.88
The values of n is very close to three, so, the reaction is third order.
Q20. A study of chemical kinetics of a reaction A + B --> Products gave the following data at 25oC. Calculate the rate law.
| [A] | [B] | Rate |
|---|---|---|
| 1.00 | 2.00 | 1.00 |
| 0.15 | 0.15 | 0.2 |
| 4.2x10-6 | 8.4x10-6 | 5.6x10-6 |
Ans. The rate of this chemical reaction is directly proportional to A and B concentrations with certain exponents. We have to determine those exponents.
In first two experiments its is clear that when the concentration of A is doubled, the rate is doubled. Hence, rate of reaction with respect to A is with exponent as unity.
So, Rate ∝ [A]1
Reaction is first order with respect to A.
Compare experiment 1 and 3. When the concentration of B is increased from 0.15 to 0.2, the rate increases from 4.2x10-6 to 5.6x10-6. The increase is proportional to the first power of B;
Rate ∝ [B]1
In other words, the reaction is first order with respect to B.
The overall rate law is:
Rate ∝ [A][B]
Rate = K[A][B]
So, the reaction is of second order.
Q21. Some reactions taking place around room temperature have activation energies around 50KJmol-1:
(i) What is the value of the factor e-Ea/KT at 25oC?
(ii) Calculate this factor at 35oC and 45oC and note the increase in this factor for every 10oC rise in temperature.
(iii) Prove that for every 10oC rise in temperature, the factor doubles and so rate constant also doubles.
Ans. (i) To calculate the exponential factor ar 25oC, we put Ea and T in Arrhenius factor.
(ii) Exponential factor at 35oC:
R = 8.3143JK-1mol-1
T = 35oC = 308K
Putting values:
=e-50000/(8.3143x308)
=e-19.52
=3.31x10-9
(ii) Exponential factor at 45oC:
R = 8.3143JK-1mol-1
T = 45oC = 318K
Putting values:
=e-50000/(8.3143x318)
=e-18.91
=6.12x10-9
The results indicate that:
Exponential factor at 35oC: = 3.31x10-9
Exponential factor at 45oC: = 6.12x10-9
(iii) Rate = Ae-Ea/RT
Since the factor A is almost constant for a particular reaction, hence, when the exponential factor doubles, then rate is doubled. For every 10oC increase of temperature, the rate has almost doubled.
Q22. H2 and I2 react to produce HI. Following data for rate constant at various temperatures (K) have been collected:
(i) Plot a graph between 1/T on x-axis and log k on the y-axis.
(ii) Measure the slope of this straight line and calculate the energy for activation of this reaction.
| Temp. (K) | Rate Constant (cm3mol-1s-1) (k) |
|---|---|
| 500 | 6.814x10-1 |
| 550 | 2.64x10-2 |
| 600 | 0.56x100 |
| 650 | 7.31x100 |
| 700 | 66.67x100 |
Ans. Will be updated Soon.
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