Intext Questions
1. For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
Answer
r(av) = −Δ[R]/Δt = −(0.02 − 0.03)/25 = 4 × 10⁻⁴ mol L⁻¹ min⁻¹
In seconds: 4 × 10⁻⁴/60 = 6.67 × 10⁻⁶ mol L⁻¹ s⁻¹
2. In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L⁻¹ to 0.4 mol L⁻¹ in 10 minutes. Calculate the rate during this interval?
Answer
Rate of disappearance of A = −Δ[A]/Δt = (0.5 − 0.4)/10 = 0.01 mol L⁻¹ min⁻¹
Rate of reaction = −½ Δ[A]/Δt = ½ × 0.01 = 0.005 mol L⁻¹ min⁻¹
3. For a reaction, A + B → Product; the rate law is given by, r = k[A]^(1/2)[B]². What is the order of the reaction?
Answer
Order = 1/2 + 2 = 5/2 = 2.5
4. The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?
Answer
Rate = k[X]². New rate = k(3[X])² = 9k[X]². The rate of formation of Y becomes 9 times.
5. A first order reaction has a rate constant 1.15 × 10⁻³ s⁻¹. How long will 5 g of this reactant take to reduce to 3 g?
Answer
t = (2.303/k) log([R]₀/[R]) = (2.303/1.15 × 10⁻³) log(5/3) = 2002.6 × 0.2218 = 444 s
6. Time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
Answer
k = 0.693/t½ = 0.693/60 min = 0.01155 min⁻¹ = 0.693/3600 s = 1.925 × 10⁻⁴ s⁻¹
7. What will be the effect of temperature on rate constant?
Answer
The rate constant increases with temperature; for many reactions it nearly doubles for a 10 K rise. Quantitatively it follows the Arrhenius equation k = A e^(−Ea/RT): a higher T increases the fraction of molecules with energy ≥ Ea, so k increases.
8. The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea.
Answer
log(k₂/k₁) = (Ea/2.303R) × (T₂ − T₁)/(T₁T₂); k₂/k₁ = 2, T₁ = 298 K, T₂ = 308 K
0.3010 = (Ea/(2.303 × 8.314)) × 10/(298 × 308)
Ea = 0.3010 × 19.147 × 298 × 308/10 = 52898 J mol⁻¹ ≈ 52.9 kJ mol⁻¹
9. The activation energy for the reaction 2HI(g) → H₂ + I₂(g) is 209.5 kJ mol⁻¹ at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
Answer
Fraction x = e^(−Ea/RT)
⇒ log x = −Ea/(2.303RT) = −209500/(2.303 × 8.314 × 581) = −18.83
x = 10⁻¹⁸·⁸³ ≈ 1.47 × 10⁻¹⁹
Exercises
1. From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants. (i) 3NO(g) → N₂O(g), Rate = k[NO]² (ii) H₂O₂(aq) + 3I⁻(aq) + 2H⁺ → 2H₂O(l) + I₃⁻, Rate = k[H₂O₂][I⁻] (iii) CH₃CHO(g) → CH₄(g) + CO(g), Rate = k[CH₃CHO]^(3/2) (iv) C₂H₅Cl(g) → C₂H₄(g) + HCl(g), Rate = k[C₂H₅Cl]
Answer
Units of k = (mol L⁻¹)^(1−n) s⁻¹ for order n.
(i) Order 2
⇒ k in L mol⁻¹ s⁻¹ (mol⁻¹ L s⁻¹)
(ii) Order 1 + 1 = 2 → k in L mol⁻¹ s⁻¹
(iii) Order 3/2
⇒ k in mol^(−1/2) L^(1/2) s⁻¹
(iv) Order 1 → k in s⁻¹
2. For the reaction: 2A + B → A₂B the rate = k[A][B]² with k = 2.0 × 10⁻⁶ mol⁻² L² s⁻¹. Calculate the initial rate of the reaction when [A] = 0.1 mol L⁻¹, [B] = 0.2 mol L⁻¹. Calculate the rate of reaction after [A] is reduced to 0.06 mol L⁻¹.
Answer
Initial rate = 2.0 × 10⁻⁶ × 0.1 × (0.2)² = 8.0 × 10⁻⁹ mol L⁻¹ s⁻¹
[A] used = 0.1 − 0.06 = 0.04 mol L⁻¹; from stoichiometry B used = 0.04/2 = 0.02 mol L⁻¹
⇒ [B] = 0.18 mol L⁻¹
Rate = 2.0 × 10⁻⁶ × 0.06 × (0.18)² = 3.89 × 10⁻⁹ mol L⁻¹ s⁻¹
3. The decomposition of NH₃ on platinum surface is zero order reaction. What are the rates of production of N₂ and H₂ if k = 2.5 × 10⁻⁴ mol⁻¹ L s⁻¹?
Answer
2NH₃
⇒ N₂ + 3H₂. For a zero order reaction, rate = k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹ (the unit of a zero order rate constant is mol L⁻¹ s⁻¹).
Rate = −½ d[NH₃]/dt = d[N₂]/dt = ⅓ d[H₂]/dt
Rate of production of N₂ = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹
Rate of production of H₂ = 3 × 2.5 × 10⁻⁴ = 7.5 × 10⁻⁴ mol L⁻¹ s⁻¹
4. The decomposition of dimethyl ether leads to the formation of CH₄, H₂ and CO and the reaction rate is given by Rate = k[CH₃OCH₃]^(3/2). The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e., Rate = k(p CH₃OCH₃)^(3/2). If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?
Answer
Unit of rate = bar min⁻¹
k = rate/p^(3/2)
⇒ unit = bar min⁻¹/bar^(3/2) = bar^(−1/2) min⁻¹
5. Mention the factors that affect the rate of a chemical reaction.
Answer
(i) Concentration of reactants (pressure for gases), (ii) temperature, (iii) presence of a catalyst, (iv) nature of reactants, (v) surface area of solid reactants, (vi) exposure to radiation/light for photochemical reactions.
6. A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is (i) doubled (ii) reduced to half?
Answer
Rate = k[A]². (i) Rate = k(2[A])² = 4k[A]²
⇒ becomes 4 times. (ii) Rate = k([A]/2)² = ¼ k[A]²
⇒ becomes one-fourth.
7. What is the effect of temperature on the rate constant of a reaction? How can this effect of temperature on rate constant be represented quantitatively?
Answer
The rate constant increases with temperature, roughly doubling for every 10° rise for many reactions, because more molecules acquire energy ≥ Ea (the shaded area under the Maxwell–Boltzmann curve grows).
Quantitatively, by the Arrhenius equation: k = A e^(−Ea/RT), or ln k = ln A − Ea/RT. For two temperatures: log(k₂/k₁) = (Ea/2.303R)[(T₂ − T₁)/(T₁T₂)]. A plot of ln k against 1/T is a straight line of slope −Ea/R.
8. In a pseudo first order reaction in water, the following results were obtained: t/s: 0, 30, 60, 90; [A]/mol L⁻¹: 0.55, 0.31, 0.17, 0.085. Calculate the average rate of reaction between the time interval 30 to 60 seconds.
Answer
r(av) = −Δ[A]/Δt = −(0.17 − 0.31)/(60 − 30) = 0.14/30 = 4.67 × 10⁻³ mol L⁻¹ s⁻¹
9. A reaction is first order in A and second order in B. (i) Write the differential rate equation. (ii) How is the rate affected on increasing the concentration of B three times? (iii) How is the rate affected when the concentrations of both A and B are doubled?
Answer
(i) −d[R]/dt = k[A][B]²
(ii) Rate = k[A](3[B])² = 9k[A][B]²
⇒ 9 times
(iii) Rate = k(2[A])(2[B])² = 8k[A][B]²
⇒ 8 times
10. In a reaction between A and B, the initial rate of reaction (r₀) was measured for different initial concentrations of A and B as given below: A/mol L⁻¹: 0.20, 0.20, 0.40; B/mol L⁻¹: 0.30, 0.10, 0.05; r₀/mol L⁻¹ s⁻¹: 5.07 × 10⁻⁵, 5.07 × 10⁻⁵, 1.43 × 10⁻⁴. What is the order of the reaction with respect to A and B?
Answer
Let rate = k[A]^x[B]^y.
Expts 1 and 2: [A] same, [B] changes from 0.30 to 0.10, rate unchanged
⇒ (3)^y = 1
⇒ y = 0
Expts 2 and 3 (y = 0): 1.43 × 10⁻⁴/5.07 × 10⁻⁵ = (0.40/0.20)^x
⇒ 2.82 = 2^x
⇒ x = log 2.82/log 2 = 1.5
Order with respect to A = 1.5; with respect to B = 0.
11. The following results have been obtained during the kinetic studies of the reaction: 2A + B → C + D. Experiment I: [A] = 0.1, [B] = 0.1, initial rate of formation of D = 6.0 × 10⁻³; II: 0.3, 0.2, 7.2 × 10⁻²; III: 0.3, 0.4, 2.88 × 10⁻¹; IV: 0.4, 0.1, 2.40 × 10⁻² (concentrations in mol L⁻¹, rate in mol L⁻¹ min⁻¹). Determine the rate law and the rate constant for the reaction.
Answer
Let rate = k[A]^x[B]^y.
Expts I and IV ([B] same): 2.40 × 10⁻²/6.0 × 10⁻³ = (0.4/0.1)^x
⇒ 4 = 4^x
⇒ x = 1
Expts II and III ([A] same): 2.88 × 10⁻¹/7.2 × 10⁻² = (0.4/0.2)^y
⇒ 4 = 2^y
⇒ y = 2
Rate law: Rate = k[A][B]²
From expt I: k = 6.0 × 10⁻³/(0.1 × 0.1²) = 6.0 L² mol⁻² min⁻¹ (same value from other experiments).
12. The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: Experiment I: [A] = 0.1, [B] = 0.1, rate = 2.0 × 10⁻²; II: [A] = –, [B] = 0.2, rate = 4.0 × 10⁻²; III: [A] = 0.4, [B] = 0.4, rate = –; IV: [A] = –, [B] = 0.2, rate = 2.0 × 10⁻² (concentrations in mol L⁻¹, rate in mol L⁻¹ min⁻¹).
Answer
Rate = k[A]. From I: k = 2.0 × 10⁻²/0.1 = 0.2 min⁻¹
II: [A] = 4.0 × 10⁻²/0.2 = 0.2 mol L⁻¹
III: Rate = 0.2 × 0.4 = 8.0 × 10⁻² mol L⁻¹ min⁻¹
IV: [A] = 2.0 × 10⁻²/0.2 = 0.1 mol L⁻¹
13. Calculate the half-life of a first order reaction from their rate constants given below: (i) 200 s⁻¹ (ii) 2 min⁻¹ (iii) 4 years⁻¹
Answer
t½ = 0.693/k
(i) 0.693/200 = 3.47 × 10⁻³ s (ii) 0.693/2 = 0.35 min (iii) 0.693/4 = 0.173 years
14. The half-life for radioactive decay of ¹⁴C is 5730 years. An archaeological artifact containing wood had only 80% of the ¹⁴C found in a living tree. Estimate the age of the sample.
Answer
k = 0.693/5730 = 1.209 × 10⁻⁴ year⁻¹
t = (2.303/k) log([R]₀/[R]) = (2.303/1.209 × 10⁻⁴) log(100/80) = 19048 × 0.0969 = 1845 years (about 1845 years old)
15. The experimental data for decomposition of N₂O₅ [2N₂O₅ → 4NO₂ + O₂] in gas phase at 318 K are given below: t/s: 0, 400, 800, 1200, 1600, 2000, 2400, 2800, 3200; 10² × [N₂O₅]/mol L⁻¹: 1.63, 1.36, 1.14, 0.93, 0.78, 0.64, 0.53, 0.43, 0.35. (i) Plot [N₂O₅] against t. (ii) Find the half-life period for the reaction. (iii) Draw a graph between log[N₂O₅] and t. (iv) What is the rate law? (v) Calculate the rate constant. (vi) Calculate the half-life period from k and compare it with (ii).
Answer
(i) [N₂O₅] against t gives a curve that falls exponentially (concave upward), starting at 1.63 × 10⁻² mol L⁻¹.
(ii) Half of the initial value = 0.815 × 10⁻² mol L⁻¹. Reading from the curve, this concentration is reached at t ≈ 1440 s, so t½ ≈ 1440 s.
(iii) log[N₂O₅] values: −1.788, −1.866, −1.943, −2.032, −2.108, −2.194, −2.276, −2.367, −2.456 at t = 0 to 3200 s. Plotting these against t gives a straight line with negative slope.
(iv) Since log[N₂O₅] vs t is linear, the reaction is first order: Rate = k[N₂O₅].
(v) Slope = −k/2.303 = (−2.456 − (−1.788))/3200 = −2.09 × 10⁻⁴ s⁻¹
⇒ k = 2.303 × 2.09 × 10⁻⁴ = 4.81 × 10⁻⁴ s⁻¹
(vi) t½ = 0.693/k = 0.693/4.81 × 10⁻⁴ = 1441 s, which agrees closely with the graphical value in (ii).
16. The rate constant for a first order reaction is 60 s⁻¹. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?
Answer
t = (2.303/k) log([R]₀/[R]) = (2.303/60) log 16 = 0.03838 × 1.204 = 4.62 × 10⁻² s
(Alternatively 1/16 = (1/2)⁴, i.e. 4 half-lives = 4 × 0.693/60 = 4.62 × 10⁻² s.)
17. During nuclear explosion, one of the products is ⁹⁰Sr with half-life of 28.1 years. If 1 mg of ⁹⁰Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.
Answer
k = 0.693/28.1 = 0.02466 year⁻¹. For first order decay, [R] = [R]₀ e^(−kt), i.e. log([R]₀/[R]) = kt/2.303.
After 10 years: log(1/[R]) = 0.02466 × 10/2.303 = 0.1071
⇒ [R] = 0.781 mg
After 60 years: log(1/[R]) = 0.02466 × 60/2.303 = 0.6425
⇒ [R] = 0.228 mg
18. For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Answer
t = (2.303/k) log([R]₀/[R])
99% completion: [R] = 0.01[R]₀
⇒ t₉₉ = (2.303/k) log 100 = 2 × 2.303/k
90% completion: [R] = 0.1[R]₀
⇒ t₉₀ = (2.303/k) log 10 = 2.303/k
Hence t₉₉ = 2 × t₉₀.
19. A first order reaction takes 40 min for 30% decomposition. Calculate t½.
Answer
k = (2.303/40) log(100/70) = (2.303/40) × 0.1549 = 8.92 × 10⁻³ min⁻¹
t½ = 0.693/8.92 × 10⁻³ = 77.7 min
20. For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained. t (sec): 0, 360, 720; P (mm of Hg): 35.0, 54.0, 63.0. Calculate the rate constant.
Answer
(CH₃)₂CHN=NCH(CH₃)₂(g)
⇒ N₂(g) + C₆H₁₄(g)
At t: p(reactant) = p₀ − x, total p_t = p₀ + x
⇒ p(reactant) = 2p₀ − p_t
k = (2.303/t) log[p₀/(2p₀ − p_t)]
t = 360 s: k = (2.303/360) log(35/(70 − 54)) = (2.303/360) log 2.1875 = 2.175 × 10⁻³ s⁻¹
t = 720 s: k = (2.303/720) log(35/(70 − 63)) = (2.303/720) log 5 = 2.236 × 10⁻³ s⁻¹
Average k ≈ 2.21 × 10⁻³ s⁻¹
21. The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at a constant volume. SO₂Cl₂(g) → SO₂(g) + Cl₂(g). Experiment 1: Time = 0 s, Total pressure = 0.5 atm; Experiment 2: Time = 100 s, Total pressure = 0.6 atm. Calculate the rate of the reaction when total pressure is 0.65 atm.
Answer
p(SO₂Cl₂) = 2p₀ − p_t. k = (2.303/t) log[p₀/(2p₀ − p_t)]
k = (2.303/100) log(0.5/(1.0 − 0.6)) = (2.303/100) log 1.25 = 2.23 × 10⁻³ s⁻¹
When p_t = 0.65 atm: p(SO₂Cl₂) = 1.0 − 0.65 = 0.35 atm
Rate = k × p(SO₂Cl₂) = 2.23 × 10⁻³ × 0.35 = 7.8 × 10⁻⁴ atm s⁻¹
22. The rate constant for the decomposition of N₂O₅ at various temperatures is given below: T/°C: 0, 20, 40, 60, 80; 10⁵ × k/s⁻¹: 0.0787, 1.70, 25.7, 178, 2140. Draw a graph between ln k and 1/T and calculate the values of A and Ea. Predict the rate constant at 30° and 50°C.
Answer
T/K: 273, 293, 313, 333, 353
⇒ 10³/T: 3.661, 3.411, 3.193, 3.002, 2.832 K⁻¹
ln k: −14.06, −10.98, −8.27, −6.33, −3.84
The plot of ln k vs 1/T is a straight line. Slope = −Ea/R ≈ −12.1 × 10³ K
⇒ Ea = 12.1 × 10³ × 8.314 ≈ 1.0 × 10⁵ J mol⁻¹ ≈ 100.7 kJ mol⁻¹
Intercept ln A ≈ 30.3
⇒ A ≈ 1.5 × 10¹³ s⁻¹ (graph-read values may differ slightly)
At 30°C (1/T = 3.299 × 10⁻³): ln k = 30.3 − 12115 × 3.299 × 10⁻³ ≈ −9.65
⇒ k ≈ 6.4 × 10⁻⁵ s⁻¹
At 50°C (1/T = 3.095 × 10⁻³): ln k ≈ −7.18
⇒ k ≈ 7.6 × 10⁻⁴ s⁻¹
23. The rate constant for the decomposition of hydrocarbons is 2.418 × 10⁻⁵ s⁻¹ at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.
Answer
log A = log k + Ea/(2.303RT) = log(2.418 × 10⁻⁵) + 179900/(2.303 × 8.314 × 546)
= −4.6166 + 17.2082 = 12.5916
⇒ A = 3.9 × 10¹² s⁻¹
24. Consider a certain reaction A → Products with k = 2.0 × 10⁻² s⁻¹. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L⁻¹.
Answer
Unit of k shows first order. log([A]₀/[A]) = kt/2.303 = (2.0 × 10⁻² × 100)/2.303 = 0.8684
[A]₀/[A] = 7.39
⇒ [A] = 1.0/7.39 = 0.135 mol L⁻¹
25. Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t½ = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?
Answer
k = 0.693/3 = 0.231 h⁻¹. log([R]₀/[R]) = kt/2.303 = 0.231 × 8/2.303 = 0.8024
[R]₀/[R] = 6.345
⇒ fraction remaining [R]/[R]₀ = 0.158
26. The decomposition of hydrocarbon follows the equation k = (4.5 × 10¹¹ s⁻¹) e^(−28000 K/T). Calculate Ea.
Answer
Comparing with k = A e^(−Ea/RT): Ea/R = 28000 K
Ea = 28000 × 8.314 = 232792 J mol⁻¹ = 232.8 kJ mol⁻¹
27. The rate constant for the first order decomposition of H₂O₂ is given by the following equation: log k = 14.34 − 1.25 × 10⁴ K/T. Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
Answer
Compare with log k = log A − Ea/(2.303RT): Ea/(2.303R) = 1.25 × 10⁴ K
Ea = 1.25 × 10⁴ × 2.303 × 8.314 = 2.393 × 10⁵ J mol⁻¹ = 239.3 kJ mol⁻¹
t½ = 256 min = 15360 s
⇒ k = 0.693/15360 = 4.51 × 10⁻⁵ s⁻¹
⇒ log k = −4.346
−4.346 = 14.34 − 1.25 × 10⁴/T
⇒ T = 1.25 × 10⁴/18.686 = 669 K
28. The decomposition of A into product has value of k as 4.5 × 10³ s⁻¹ at 10°C and energy of activation 60 kJ mol⁻¹. At what temperature would k be 1.5 × 10⁴ s⁻¹?
Answer
log(k₂/k₁) = (Ea/2.303R)(1/T₁ − 1/T₂); T₁ = 283 K
log(1.5 × 10⁴/4.5 × 10³) = log 3.333 = 0.5229 = (60000/19.147)(1/283 − 1/T₂)
1/283 − 1/T₂ = 1.669 × 10⁻⁴
⇒ 1/T₂ = 3.534 × 10⁻³ − 0.167 × 10⁻³ = 3.367 × 10⁻³
⇒ T₂ = 297 K (about 24°C)
29. The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 10¹⁰ s⁻¹. Calculate k at 318 K and Ea.
Answer
t = (2.303/k₂₉₈) log(100/90) = (2.303/k₃₀₈) log(100/75)
k₃₀₈/k₂₉₈ = log(4/3)/log(10/9) = 0.1249/0.0458 = 2.73
log 2.73 = (Ea/2.303R) × (308 − 298)/(298 × 308)
⇒ 0.4362 = (Ea/19.147) × 1.0896 × 10⁻⁴
Ea = 7.67 × 10⁴ J mol⁻¹ = 76.7 kJ mol⁻¹
At 318 K: log k = log A − Ea/(2.303RT) = 10.602 − 76660/(19.147 × 318) = 10.602 − 12.591 = −1.989
⇒ k = 1.0 × 10⁻² s⁻¹
30. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Answer
log(k₂/k₁) = (Ea/2.303R)(T₂ − T₁)/(T₁T₂)
⇒ log 4 = (Ea/19.147) × 20/(293 × 313)
Ea = 0.6021 × 19.147 × 293 × 313/20 = 52860 J mol⁻¹ = 52.86 kJ mol⁻¹
Important Extra Questions
Distinguish between order and molecularity of a reaction.
Answer
Order is the sum of powers of concentration terms in the experimentally found rate law; it can be zero, fractional or whole and applies to elementary and complex reactions. Molecularity is the number of reacting species colliding simultaneously in an elementary step; it is always a whole number (1, 2 or 3) and has no meaning for a complex reaction as a whole.
Show that for a zero order reaction t½ = [R]₀/2k.
Answer
For zero order, [R] = [R]₀ − kt. At t½, [R] = [R]₀/2
⇒ [R]₀/2 = [R]₀ − kt½
⇒ t½ = [R]₀/2k. So t½ is directly proportional to the initial concentration.
What is a pseudo first order reaction? Give an example.
Answer
A reaction whose true order is higher but which behaves as first order because one reactant is in large excess, so its concentration hardly changes. Example: acid hydrolysis of ethyl acetate, CH₃COOC₂H₅ + H₂O
⇒ CH₃COOH + C₂H₅OH, where rate = k′[CH₃COOC₂H₅]; inversion of cane sugar is another example.
How does a catalyst increase the rate of a reaction? Does it change ΔG?
Answer
A catalyst provides an alternative pathway with lower activation energy, so more molecules can cross the barrier and the rate increases. It does not change ΔG or the equilibrium constant; it only helps equilibrium to be reached faster by speeding up forward and backward reactions equally.
According to collision theory, why do all collisions not lead to products?
Answer
Only effective collisions form products. For a collision to be effective, the colliding molecules must have energy at least equal to the threshold energy and must be properly oriented. This is expressed as Rate = P Z(AB) e^(−Ea/RT), where P is the probability (steric) factor.