CBSE Class 12 Chemistry Chemical Kinetics Important Questions Set 08

Very Short Answer Questions (1 Mark)

Question 1. In the reaction aA + bB → products, if concentration of A is doubled (keeping B constant) the initial rate becomes four times and if B is doubled (keeping A constant), the rate becomes double. What is the rate law equation and order of reaction? (Manipur Board 2012)
Answer:
Rate = k [A]² [B], order = 2 + 1 = 3

Teacher's Note:
1. Doubling a concentration and seeing the rate change by 2^x means the order with respect to that reactant is x.

Question 2. Give an example of a zero order reaction and write an expression for its velocity constant. (Jharkhand Board 2011, UP Board 2012)
Answer:
H₂ + Cl₂ \( \xrightarrow{h\nu} \) 2HCl
or 2NH₃ \( \xrightarrow{Pt} \) N₂ + 3H₂.
Rate constant, \( k = \frac{1}{t}\{[A]_0 - [A]\} \).

Teacher's Note:
1. Zero order behaviour of NH₃ on platinum arises because the metal surface is fully covered at high pressure, so adding more gas does not speed up the reaction.

Question 3. Write the expression showing the change of concentration with time in the exponential form for reactions of first order.
Answer:
[A] = [A]₀ e^−kt where [A]₀ is initial concentration, [A] is concentration at time t and k is rate constant.

Teacher's Note:
1. Taking natural logs of this gives ln[A] = ln[A]₀ − kt, the straight-line form used in graphs.

Question 4. If half-life period of a reaction is inversely proportional to initial concentration of the reactant, what is the order of reaction ?
Answer:
Two. This is because for reaction of nth order,
\( t_{1/2} \propto \frac{1}{[A_0]^{n-1}} \). Hence, \( t_{1/2} \propto \frac{1}{[A_0]} \) when n = 2.

Teacher's Note:
1. Put n = 0, 1 and 2 into the general relation to recall all three half-life patterns at once.

Question 5. What is pseudo first order reaction? Give an example of it. (HP Board 2011, 2013)
Answer:
Reactions which are not truly of first order but under certain conditions (one reactant present in large excess) become reactions of first order are called pseudo first order reactions.
Example. CH₃COOC₂H₅ + H₂O \( \xrightarrow{H^+} \) CH₃COOH + C₂H₅OH

Teacher's Note:
1. Inversion of cane sugar in acidic medium is another example you can quote.

Question 6. Radioactive disintegration is a first order reaction. Explain why. (Raj. Board 2011)
Answer:
The rate of disintegration of a radioactive substance is directly proportional to the amount of the substance present. Hence, it is a reaction of first order.

Teacher's Note:
1. This is why every radioactive isotope has a fixed half-life, whatever its starting amount.

Question 7. In the Arrhenius equation k = A exp (−E/RT), A may be termed as rate constant at........ .
Answer:
Infinite temperature.

Teacher's Note:
1. As T → ∞, E/RT → 0 and e⁰ = 1, so k becomes equal to A.

Question 8. Define temperature coefficient of a reaction. (Karnataka Board 2012, Chhattisgarh Board 2012)
Answer:
Temperature coefficient of a reaction is the ratio of the rate constant of the reaction at 308 K to that at 298 K. Its value lies between 2 and 3.

Teacher's Note:
1. In general it is k_(T+10)/k_T, the ratio for a 10 degree rise in temperature.

Question 9. The activation energy of a reaction is zero. Will the rate constant of the reaction depend upon temperature ? Give reason.
Answer:
No. When E_a = 0, k = Ae^−0/RT = A. Thus, k becomes independent of T.

Teacher's Note:
1. Temperature affects k only through the term E_a/RT in the exponent.

Question 10. The rate of a reaction at 30°C is x, then what will be the rate of the same reaction at 60°C ?
Answer:
For every 10° rise of temperature, the rate of reaction is doubled. Hence, 30° \( \xrightarrow{2x} \) 40° \( \xrightarrow{4x} \) 50° \( \xrightarrow{8x} \) 60°C, i.e., rate of reaction will become 8 times.

Teacher's Note:
1. A 30 degree rise is three steps of 10 degrees, so the rate goes up by 2³ = 8, not by 2 × 3 = 6.

Question 11. The plot of ln k versus 1/T for a reaction has a slope equal to − 4500 K. What is the activation energy of the reaction ?
Answer:
The plot of ln k vs 1/T has slope = \( -\frac{E_a}{R} \) = − 4500 K
Hence, E_a = 4500 K × 8.314 J K⁻¹ mol⁻¹
= 37413 J mol⁻¹
= 37.4 kJ mol⁻¹

Teacher's Note:
1. Watch the axis: for ln k the slope is −E_a/R, but for log k it is −E_a/2.303R.

Question 12. What are the two necessary conditions for a collision between two molecules to be effective collision?
Answer:
At the time of collision, (i) they should have energy more than threshold value, and (ii) they should have proper orientation.

Teacher's Note:
1. Energy and orientation are both needed; a collision with enough energy but wrong orientation still produces no product.

Short Answer Questions (2 Marks)

Question 13. What are Pseudounimolecular reactions ? Explain with the help of a suitable example. (J & K Board 2011)
Answer:
Reactions which have molecularity two but whose order is one are called pseudo-unimolecular reactions. This happens when one of the reactants is present in large excess.
Example: inversion of cane sugar, C₁₂H₂₂O₁₁ + H₂O \( \xrightarrow{H^+} \) C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (fructose)
Water is in large excess, so its concentration remains constant; Rate = k[C₁₂H₂₂O₁₁], i.e., first order though molecularity is 2.

Teacher's Note:
1. The excess reactant is absorbed into the constant: k′ = k[H₂O].

Question 14. H⁺ catalysed hydrolysis of ester
(CH₃CO₂C₂H₅ + H₂O \( \xrightarrow{H^+} \) CH₃CO₂H + C₂H₅OH)
is a pseudo-unimolecular reaction. Explain. (West Bengal Board 2012)
Answer:
Here two molecules (ester and water) take part in the reaction, so it should be bimolecular, i.e., Rate = k[CH₃CO₂C₂H₅][H₂O].
But water is present in such large excess that its concentration remains practically constant during the reaction. Hence, Rate = k′[CH₃CO₂C₂H₅], where k′ = k[H₂O].
Thus the reaction follows first order kinetics and is called a pseudo-unimolecular reaction.

Teacher's Note:
1. H⁺ is a catalyst here; it is not consumed and does not appear in the overall equation.

Question 15. What do you mean by molecularity and order of a reaction ? Explain why their values are different in the reaction :
CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH.
Answer:
Molecularity is the number of reacting species which collide simultaneously in an elementary reaction. Order is the sum of the powers of the concentration terms in the experimental rate law.
In the given reaction, two molecules react, so molecularity = 2. But water is present in large excess and its concentration remains constant, so Rate = k[CH₃COOC₂H₅], i.e., order = 1.
Hence, molecularity (2) and order (1) are different; it is a pseudo first order reaction.

Teacher's Note:
1. Molecularity counts what collides; order counts what actually changes the rate in the experiment.

Question 16. Define half-life period and average life. Derive relation between them. (J & K Board 2011)
Answer:
Half-life period (t_1/2) is the time in which half of the reactant (or radioactive substance) is used up.
Average life (τ) is the average time for which the nuclei (molecules) of a substance exist before decaying. It is the reciprocal of the rate (decay) constant: τ = \( \frac{1}{k} \).
Since \( t_{1/2} = \frac{0.693}{k} \) and \( \tau = \frac{1}{k} \), we get t_1/2 = 0.693 τ or τ = 1.44 t_1/2.

Teacher's Note:
1. Average life is always longer than half-life (about 1.44 times), a quick check on your answer.

Question 17. For the reaction C₁₂H₂₂O₁₁ + H₂O \( \xrightarrow{H^+} \) C₆H₁₂O₆ + C₆H₁₂O₆, write
(i) Rate of reaction expression (ii) rate law equation (iii) molecularity (iv) order of reaction
Answer:
(i) Rate = \( -\frac{d[C_{12}H_{22}O_{11}]}{dt} = +\frac{d[C_6H_{12}O_6\,(\text{glucose})]}{dt} = +\frac{d[C_6H_{12}O_6\,(\text{fructose})]}{dt} \)
(ii) Rate = k [C₁₂H₂₂O₁₁]
(iii) Molecularity = 2
(iv) Order = 1

Teacher's Note:
1. Water does not appear in the rate law because its concentration hardly changes – the reason order (1) and molecularity (2) differ.

Question 18. Define the term Activation energy. Why different reactions proceed at different speeds ? (J & K Board 2012)
Answer:
The minimum extra energy which must be absorbed by the reactant molecules so that their energy becomes equal to the threshold energy is called activation energy (E_a = Threshold energy − Average energy of reactants).
Different reactions have different activation energies. A reaction with low activation energy has a larger fraction of molecules with energy equal to or more than the threshold energy, so it is fast; a reaction with high activation energy is slow. Hence, different reactions proceed at different speeds.

Teacher's Note:
1. The fraction of effective molecules is e^−E_a/RT, so even a small difference in E_a changes the rate a lot.

Question 19. The reaction 2H₂(g) + O₂(g) → 2H₂O(g) is feasible. How is that hydrogen and oxygen mixture allowed to stand at room temperature shows no formation of water at all ?
Answer:
Because of high activation energy at room temperature. At room temperature very few molecules of H₂ and O₂ have energy equal to the threshold energy, so the reaction does not take place at a measurable rate. It occurs when energy is supplied (e.g., by a spark).

Teacher's Note:
1. Feasibility (negative ΔG) tells only whether a reaction can occur; activation energy decides how fast it actually occurs.

Question 20. State the role of activated complex in a reaction and state its relation with activation energy.
Answer:
The reactant molecules first form an unstable, high-energy intermediate called the activated complex, which then breaks up to give the products. Only those molecules which can reach the energy of the activated complex form products.
Activation energy = Energy of activated complex — Average energy of reactants.

Teacher's Note:
1. The activated complex sits at the top of the energy profile; it is not an isolable intermediate.

Question 21. What is activation energy ? How is the rate constant of a reaction related to its activation energy ? (CBSE 2007)
Answer:
Activation energy is the minimum extra energy which must be absorbed by the reactant molecules so that their energy becomes equal to the threshold energy.
Rate constant is related to activation energy by the Arrhenius equation, k = Ae^−E_a/RT or \( \log k = \log A - \frac{E_a}{2.303\,RT} \). Thus, lower the activation energy, higher the rate constant.

Teacher's Note:
1. Since E_a sits in a negative exponent, k falls very steeply as E_a rises.

Question 22. Briefly explain the effect of adding catalyst on the rate of a reaction.
Answer:
A catalyst increases the rate of a reaction. It provides an alternative path (mechanism) for the reaction which has lower activation energy. Hence, a larger fraction of molecules has energy equal to or more than the new threshold energy and the rate increases.
A catalyst does not change ΔH or ΔG of the reaction and does not alter the equilibrium constant; it only helps the equilibrium to be attained faster.

Teacher's Note:
1. A catalyst speeds up the forward and backward reactions equally, which is why the position of equilibrium is unchanged.

Question 23. Explain : (i) One gram of pulverised wood burns faster than a one gram piece of wood.
(ii) An increase of 10 K in temperature rarely doubles the kinetic energy of particles but this increase in temperature may be enough to double the rate of reaction.
Answer:
(i) Pulverised wood has a much larger surface area than a single piece of the same mass. Hence, more particles are exposed to oxygen, the number of collisions increases and it burns faster.
(ii) A rise of 10 K increases the average kinetic energy only slightly, but it nearly doubles the fraction of molecules having energy equal to or more than the threshold energy (the shaded area of the energy distribution curve beyond E_a). Hence, the number of effective collisions and so the rate nearly doubles.

Teacher's Note:
1. The rate depends on the high-energy tail of the Maxwell–Boltzmann curve, which grows much faster than the average energy.

Question 24. Explain the effect of (a) Concentration of the reactants (b) Presence of a catalyst on the rate of a chemical reaction. (Hr. Board 2011)
Answer:
(a) Greater the concentration of reactants, greater is the number of collisions per unit time and hence greater is the rate of reaction (Rate = k[A]^x[B]^y). The rate decreases as the reaction proceeds because the concentration of reactants decreases.
(b) A catalyst increases the rate of reaction by providing an alternative path of lower activation energy, so that more molecules can cross the energy barrier.

Teacher's Note:
1. Concentration changes the rate but not the rate constant; a catalyst changes the rate constant itself.

Question 25. Define the following terms : (i) Collision frequency (ii) Half-life period of a reaction. (Hr. Board 2011)
Answer:
(i) Collision frequency is the number of collisions taking place per second per unit volume of the reaction mixture.
(ii) Half-life period is the time in which the concentration of a reactant is reduced to half of its initial value.

Teacher's Note:
1. Collision frequency alone cannot give the rate, since only a tiny fraction of collisions are effective.

Question 26. Define the following terms : (i) Order of reaction (ii) Threshold energy. (Bihar Board 2011)
Answer:
(i) Order of reaction is the sum of the powers of the concentration terms of the reactants in the experimentally determined rate law.
(ii) Threshold energy is the minimum energy which the colliding molecules must have in order that the collision may result in a chemical reaction.

Teacher's Note:
1. Threshold energy = average energy of reactants + activation energy; keep the two terms distinct.

Question 27. Define the following terms : (i) Activation energy (ii) Temperature coefficient (iii) Half-life period of a reaction. (Assam Board 2012)
Answer:
(i) Activation energy is the minimum extra energy which must be absorbed by the reactant molecules so that their energy becomes equal to the threshold energy.
(ii) Temperature coefficient is the ratio of the rate constants of a reaction at two temperatures differing by 10°, usually 308 K and 298 K; its value lies between 2 and 3.
(iii) Half-life period is the time in which half of the reactant is converted into products.

Teacher's Note:
1. A temperature coefficient of about 2 is the basis of the rule that the rate doubles for every 10 degree rise.

Short Answer Questions (3 Marks)

Question 28. Derive the relationship between rate constant and half-life period of a first order reaction. (Karnataka Board 2012)
Answer:
For a first order reaction, \( k = \frac{2.303}{t}\log\frac{[R]_0}{[R]} \)
At t = t_1/2, [R] = \( \frac{[R]_0}{2} \)
\( \therefore\; k = \frac{2.303}{t_{1/2}}\log\frac{[R]_0}{[R]_0/2} = \frac{2.303}{t_{1/2}}\log 2 \)
\( t_{1/2} = \frac{2.303 \times 0.3010}{k} \) or \( t_{1/2} = \frac{0.693}{k} \)

Teacher's Note:
1. Substituting [R] = [R]₀/2 is the key step; the [R]₀ then cancels, showing why t_1/2 is independent of initial concentration.

Question 29. Derive the equation for the rate constant of a first order reaction and show that the time required for the completion of half of the first order reaction is independent of initial concentration. (Bihar Board 2012, UP Board 2012, Assam Board 2012)
Answer:
For the first order reaction R → P, Rate = \( -\frac{d[R]}{dt} = k[R] \) or \( \frac{d[R]}{[R]} = -k\,dt \)
Integrating, ln [R] = −kt + I
At t = 0, [R] = [R]₀, so I = ln [R]₀. Hence, ln [R] = −kt + ln [R]₀
or \( k = \frac{1}{t}\ln\frac{[R]_0}{[R]} = \frac{2.303}{t}\log\frac{[R]_0}{[R]} \)
Half-life: at t = t_1/2, [R] = [R]₀/2
\( t_{1/2} = \frac{2.303}{k}\log 2 = \frac{0.693}{k} \)
Since t_1/2 does not contain [R]₀, the time for half completion is independent of initial concentration.

Teacher's Note:
1. Write the constant of integration step clearly (I = ln [R]₀); examiners award a mark for it.

Question 30. Derive integrated rate equation for rate constant for a first order reactants. What would be the units of the first order rate constant, if the concentration is expressed in moles per litre and time in seconds ? (Pb. Board 2011, Hr. Board 2011)
Answer:
For R → P, \( -\frac{d[R]}{dt} = k[R] \) or \( \frac{d[R]}{[R]} = -k\,dt \)
Integrating, ln [R] = −kt + I. At t = 0, [R] = [R]₀, so I = ln [R]₀.
∴ ln [R] = −kt + ln [R]₀ or \( k = \frac{2.303}{t}\log\frac{[R]_0}{[R]} \)
Units of k = \( \frac{\text{mol L}^{-1}\text{ s}^{-1}}{\text{mol L}^{-1}} \) = s⁻¹

Teacher's Note:
1. The integrated form can also be written as [R] = [R]₀e^−kt; know both forms.

Question 31. Define energy of activation. Draw diagram of energy profile to show the influence of a positive catalyst on the energy of activation of a reaction. (Karnataka Board 2012)

Answer:
Energy of activation is the minimum extra energy which must be absorbed by the reactant molecules so that their energy becomes equal to the threshold energy.
A positive catalyst provides an alternative path (mechanism) of lower activation energy (E_a′ < E_a), as shown in the energy profile diagram. Hence, more molecules can cross the energy barrier and the rate increases.
The energies of reactants and products (and so ΔH) are not changed by the catalyst.

Teacher's Note:
1. In the diagram keep the reactant and product levels the same for both paths; only the height of the peak changes.

Question 32. What is the effect of temperature on the rate of reaction ? Explain giving reasons.
Answer:
The rate of a reaction increases with increase in temperature. For most reactions, the rate becomes nearly double for every 10° rise in temperature.
Reason: On increasing the temperature, the kinetic energy of the molecules increases. As a result, the fraction of molecules having energy equal to or more than the threshold energy (activation energy) increases sharply. Hence, the number of effective collisions increases and so the rate increases.
Quantitatively, the effect is given by the Arrhenius equation, k = Ae^−E_a/RT.

Teacher's Note:
1. The main reason is the rise in the fraction of high-energy molecules, not just the increase in the number of collisions.

Question 33. Write Arrhenius equation showing the effect of temperature on the reaction rate. What do different symbols signify? How does it help in the calculation of activation energy of a reaction ?
Answer:
Arrhenius equation: k = Ae^−E_a/RT
where k = rate constant, A = frequency factor (pre-exponential factor), E_a = activation energy, R = gas constant and T = absolute temperature.
Taking logs, \( \log k = \log A - \frac{E_a}{2.303\,RT} \). A plot of log k vs 1/T is a straight line with slope = \( -\frac{E_a}{2.303\,R} \), from which E_a can be calculated.
Alternatively, if k₁ and k₂ are rate constants at T₁ and T₂, \( \log\frac{k_2}{k_1} = \frac{E_a}{2.303\,R}\left(\frac{T_2 - T_1}{T_1 T_2}\right) \), which gives E_a.

Teacher's Note:
1. Use R = 8.314 J K⁻¹ mol⁻¹ and temperatures in kelvin; E_a then comes out in J mol⁻¹.

Question 34. How is rate constant of a reaction affected by temperature ? Write the name of this relation. How will you calculate activation energy of a reaction with the help of this relation ?
Answer:
The rate constant of a reaction increases with increase in temperature; it becomes nearly double for every 10° rise in temperature.
The relation is the Arrhenius equation: k = Ae^−E_a/RT or \( \log k = \log A - \frac{E_a}{2.303\,RT} \)
Calculation of E_a: (i) Plot log k against 1/T; the slope of the straight line = \( -\frac{E_a}{2.303\,R} \), so E_a = −2.303 R × slope.
(ii) Or measure k₁ at T₁ and k₂ at T₂ and use \( \log\frac{k_2}{k_1} = \frac{E_a}{2.303\,R}\left(\frac{T_2 - T_1}{T_1 T_2}\right) \).

Teacher's Note:
1. In the two-temperature form, keep the larger k with the higher T so that both sides stay positive.

Question 35. Write a short note on ‘collision theory of reaction rates’.
Answer:
According to collision theory, the reactant molecules are assumed to be hard spheres and a reaction takes place only when they collide with each other.
The number of collisions per second per unit volume of the reaction mixture is called the collision frequency (Z).
Every collision is not effective. A collision leads to a reaction only if (i) the colliding molecules have energy equal to or more than the threshold energy and (ii) they have proper orientation.
Thus, Rate = P Z_AB e^−E_a/RT, where Z_AB is the collision frequency of reactants A and B, e^−E_a/RT is the fraction of molecules with energy ≥ E_a and P is the probability (steric) factor which takes care of proper orientation.

Teacher's Note:
1. Two conditions, energy and orientation: the exponential term handles energy and P handles orientation.

Question 36. Write short notes on : (i) Activation energy (ii) Arrhenius equation. (MP Board 2013)
Answer:
(i) Activation energy: Molecules react only if they collide with energy equal to or more than the threshold energy. The minimum extra energy which the reactant molecules must absorb to reach the threshold energy is called activation energy (E_a = Threshold energy − Average energy of reactants). Lower the activation energy, faster is the reaction. A catalyst lowers the activation energy.
(ii) Arrhenius equation: It gives the effect of temperature on the rate constant: k = Ae^−E_a/RT, where A is the frequency factor and e^−E_a/RT is the fraction of molecules having energy ≥ E_a. In log form, \( \log k = \log A - \frac{E_a}{2.303\,RT} \); a plot of log k vs 1/T is a straight line of slope −E_a/2.303R, which is used to find E_a.

Teacher's Note:
1. Learn the two-temperature form too: log(k₂/k₁) = (E_a/2.303R)[(T₂ − T₁)/T₁T₂].

Long Answer Questions (5 Marks)

Question 37. (a) Illustrate graphically the effect of catalyst on activation energy.
(b) Catalysts have no effect on the equilibrium constant. Why ?
(c) The decomposition of A into product has value of k as 4.5 × 10³ s⁻¹ at 10°C and activation energy is 60 kJ mol⁻¹. Calculate the temperature at which the value of k will be 1.5 × 10⁴ s⁻¹.

Answer:
(a) A catalyst provides an alternative path of lower activation energy (E_a′ < E_a), as shown in the energy profile diagram. The energies of reactants and products remain the same.
(b) A catalyst lowers the activation energy of the forward and the backward reactions to the same extent. Hence it increases the rates of both reactions equally, and the equilibrium constant (K = k_f/k_b) remains unchanged; the catalyst only helps the equilibrium to be attained faster.
(c) k₁ = 4.5 × 10³ s⁻¹, T₁ = 283 K, k₂ = 1.5 × 10⁴ s⁻¹, E_a = 60000 J mol⁻¹
\( \log\frac{k_2}{k_1} = \frac{E_a}{2.303\,R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) \)
\( \log\frac{1.5\times10^{4}}{4.5\times10^{3}} = \frac{60000}{2.303 \times 8.314}\left(\frac{1}{283} - \frac{1}{T_2}\right) \)
0.5229 = 3133.6 \( \left(\frac{1}{283} - \frac{1}{T_2}\right) \) ⇒ \( \frac{1}{T_2} \) = 3.534 × 10⁻³ − 1.669 × 10⁻⁴ = 3.367 × 10⁻³
T₂ = 297 K = 24°C

Teacher's Note:
1. Convert 10°C to 283 K and kJ to J before substituting; most errors in Arrhenius sums come from these two steps.

Question 38. (a) A reaction is second order in A and first order in B
(i) Write the differential rate equation
(ii) How is the rate affected on increasing the concentration of A three times?
(iii) How is the rate affected when the concentration of both A and B are doubled?
(b) A first order reaction takes 40 minutes for 30% decomposition. Calculate t_1/2 for this reaction. (Given log 1.428 = 0.1548) (CBSE 2013)
Answer:
(a) (i) \( -\frac{d[A]}{dt} \) = k [A]² [B]
(ii) Rate = k(3[A])²[B] = 9k[A]²[B], i.e., rate increases 9 times.
(iii) Rate = k(2[A])²(2[B]) = 8k[A]²[B], i.e., rate increases 8 times.
(b) \( k = \frac{2.303}{t}\log\frac{100}{100-30} = \frac{2.303}{40}\log 1.428 = \frac{2.303 \times 0.1548}{40} \) = 8.9 × 10⁻³ min⁻¹
\( t_{1/2} = \frac{0.693}{k} = \frac{0.693}{8.9\times10^{-3}} \) = 77.87 min

Teacher's Note:
1. 30% decomposition means 70% remains, so the ratio is 100/70 = 1.428, not 100/30.

Question 39. (a) For a first order reaction show that time required for 99% completion is twice the time required for the completion of 90% of the reaction.
(b) Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation :
\( \log k = \log A - \frac{E_a}{2.303\,R}\left(\frac{1}{T}\right) \)
where E_a is the activation energy. When a graph is plotted for log k vs 1/T, a straight line with a slope of − 4250 K is obtained. Calculate E_a for the reaction (R = 8.314 J K⁻¹ mol⁻¹) (CBSE 2013)
Answer:
(a) For 99% completion, \( t_{99\%} = \frac{2.303}{k}\log\frac{100}{100-99} = \frac{2.303}{k}\log 100 = \frac{2.303 \times 2}{k} \)
For 90% completion, \( t_{90\%} = \frac{2.303}{k}\log\frac{100}{100-90} = \frac{2.303}{k}\log 10 = \frac{2.303}{k} \)
\( \therefore\; \frac{t_{99\%}}{t_{90\%}} = 2 \), i.e., t_99% = 2 × t_90%.
(b) Slope = \( -\frac{E_a}{2.303\,R} \) = − 4250 K
E_a = 2.303 × 8.314 J K⁻¹ mol⁻¹ × 4250 K = 81375 J mol⁻¹ = 81.38 kJ mol⁻¹

Teacher's Note:
1. In (b) the plot is of log k, so the factor 2.303 must be included; with ln k it would be dropped.

Teacher's Note:
1. In long answers, link the two parts: temperature changes the rate mainly by changing the fraction of molecules that can cross the activation energy barrier.

Question 41. (a) Draw the plot of ln k vs 1/T for a chemical reaction. What does the intercept represent ? What is the relation between slope and E_a ?
(b) A first order reaction takes 30 minutes for 20% decomposition. Calculate t_1/2 (log 2 = 0.3010). (CBSE 2019)

Answer:
(a) \( \ln k = \ln A - \frac{E_a}{RT} \). The plot of ln k vs 1/T is linear (see figure). Intercept = ln A, Slope = −E_a/R.
(b) \( k = \frac{2.303}{30}\log\frac{100}{100-20} = \frac{2.303}{30}\log 1.25 \) = 7.4 × 10⁻³ min⁻¹
\( t_{1/2} = \frac{0.693}{k} = \frac{0.693}{7.4\times10^{-3}} \) = 93.65 min

Teacher's Note:
1. log 1.25 = log 10 − 3 log 2 = 1 − 0.903 = 0.097; this trick lets you use the given log 2 value.