Question 3.1
For R → P, [R] decreases from 0.03 mol L⁻¹ to 0.02 mol L⁻¹ in 25 minutes. Find the average rate in minutes and seconds.
Steps only
Rate = (0.03−0.02)/25 = 4.0×10⁻⁴ mol L⁻¹ min⁻¹ = 6.67×10⁻⁶ mol L⁻¹ s⁻¹.
Answer
6.67×10⁻⁶ mol L⁻¹ s⁻¹.
CLASS 12 · CHEMISTRY · CHAPTER 3
Chemical Kinetics
Simple answers to all nine in-text questions and thirty exercises about reaction rates and chemical kinetics.
39 worked answersComplete chapter answers
39 worked answers · short steps first, explanations when you need them
In-text Questions
Textbook pages 66–84
Question 3.2
For 2A → products, [A] falls from 0.5 to 0.4 mol L⁻¹ in 10 minutes. Find the reaction rate.
Steps only
Rate = −½ Δ[A]/Δt = ½(0.1/10) = 0.005 mol L⁻¹ min⁻¹.
Answer
0.005 mol L⁻¹ min⁻¹.
Question 3.3
For rate = k[A]¹ᐟ²[B]², what is the order of the reaction?
Steps only
Order = ½+2 = 2.5.
Answer
2.5.
Question 3.4
Reaction X → Y is second order in X. How does the rate change if [X] is tripled?
Steps only
New rate/old rate = 3² = 9.
Answer
9 times.
Question 3.5
A first-order reaction has k=1.15×10⁻³ s⁻¹. How long does 5 g take to reduce to 3 g?
Steps only
t=(2.303/1.15×10⁻³)log(5/3) ≈ 444 s.
Answer
444 s.
Question 3.6
SO₂Cl₂ has a first-order half-life of 60 minutes. Find its rate constant in s⁻¹.
Steps only
k=0.693/3600=1.925×10⁻⁴ s⁻¹.
Answer
1.925×10⁻⁴ s⁻¹.
Question 3.7
What happens to a reaction's rate constant when temperature increases?
Steps only
Higher temperature raises k; k=Ae^(−Ea/RT).
Answer
k increases.
Question 3.8
The rate doubles when temperature rises from 298 K to 308 K. Find the activation energy.
Steps only
Ea=8.314 ln2 /(1/298−1/308) ≈ 52.9 kJ mol⁻¹.
Answer
52.9 kJ mol⁻¹.
Question 3.9
For 2HI → H₂+I₂, Ea=209.5 kJ mol⁻¹ at 581 K. What fraction of molecules have energy at least Ea?
Steps only
e^(−209500/(8.314×581)) ≈ 1.47×10⁻¹⁹.
Answer
1.47×10⁻¹⁹.
Chapter 3 Exercises
Textbook pages 85–88
Question 3.1
For each given rate law, find its reaction order and the units of k: (i) k[NO]² (ii) k[H₂O₂][I⁻] (iii) k[CH₃CHO]³ᐟ² (iv) k[C₂H₅Cl].
Steps only
Add the powers for order; divide mol L⁻¹ s⁻¹ by concentration^order for k units.
Answer
- (i) 2, L mol⁻¹ s⁻¹;
- (ii) 2, L mol⁻¹ s⁻¹;
- (iii) 1.5, L¹ᐟ² mol⁻¹ᐟ² s⁻¹;
- (iv) 1, s⁻¹.
Question 3.2
For 2A+B→A₂B, rate=k[A][B]² and k=2.0×10⁻⁶ L² mol⁻² s⁻¹. Find the rate at [A]=0.1, [B]=0.2 mol L⁻¹, then when [A]=0.06 mol L⁻¹.
Steps only
r₁=k(0.1)(0.2)²=8.0×10⁻⁹; r₂=k(0.06)(0.2)²=4.8×10⁻⁹ mol L⁻¹ s⁻¹.
Answer
8.0×10⁻⁹ initially; 4.8×10⁻⁹ mol L⁻¹ s⁻¹ later.
Question 3.3
Ammonia decomposes on platinum as a zero-order reaction. Find the production rates of N₂ and H₂ if k=2.5×10⁻⁴ mol L⁻¹ s⁻¹.
Steps only
Rate=k; d[N₂]/dt=k and d[H₂]/dt=3k.
Answer
N₂: 2.5×10⁻⁴; H₂: 7.5×10⁻⁴ mol L⁻¹ s⁻¹.
Question 3.4
For rate=k(pCH₃OCH₃)³ᐟ², pressure is measured in bar and time in minutes. Find the units of rate and k.
Steps only
[k]=(bar min⁻¹)/bar³ᐟ²=bar⁻¹ᐟ² min⁻¹.
Answer
Rate: bar min⁻¹; k: bar⁻¹ᐟ² min⁻¹.
Question 3.5
Name the factors that affect the rate of a chemical reaction.
Steps only
List concentration/pressure, temperature, catalyst, and reactant nature/state/surface area.
Answer
Concentration or pressure, temperature, catalyst, and reactant nature/state/surface area.
Question 3.6
For a reaction second order in one reactant, how does the rate change if its concentration is doubled or halved?
Steps only
2²=4; (1/2)²=1/4.
Answer
4 times; one quarter.
Question 3.7
How does temperature affect the rate constant, and how is the effect shown quantitatively?
Steps only
k increases with T; k=Ae^(−Ea/RT).
Answer
Arrhenius equation: k=Ae^(−Ea/RT).
Question 3.8
For [A]=0.31 mol L⁻¹ at 30 s and 0.17 mol L⁻¹ at 60 s, find the average rate between 30 and 60 s.
Steps only
(0.31−0.17)/30=4.67×10⁻³ mol L⁻¹ s⁻¹.
Answer
4.67×10⁻³ mol L⁻¹ s⁻¹.
Question 3.9
A reaction is first order in A and second order in B. Write the rate equation. How does the rate change if B is tripled, or if both A and B are doubled?
Steps only
Rate=k[A][B]²; ×3²=9; ×2×2²=8.
Answer
9 times; 8 times.
Question 3.10
From the initial-rate data A: 0.20, 0.20, 0.40 mol L⁻¹; B: 0.30, 0.10, 0.05 mol L⁻¹; rates: 5.07×10⁻⁵, 5.07×10⁻⁵, 1.43×10⁻⁴ mol L⁻¹ s⁻¹, find the orders in A and B.
Steps only
Same rate when B changes ⇒ order B=0. Doubling A approximately doubles rate ⇒ order A=1.
Answer
Rate≈k[A]; orders 1 and 0.
Question 3.11
For 2A+B→C+D, use the four initial-rate experiments to find the rate law and k.
Steps only
m=1, n=2; k=0.006/(0.1×0.1²)=6.
Answer
Rate=k[A][B]²; k=6 L² mol⁻² min⁻¹.
Question 3.12
The reaction is first order in A and zero order in B. Complete the missing values in the table.
Steps only
k=0.2 min⁻¹; [A]II=0.20; rateIII=0.08; [A]IV=0.10.
Answer
0.20 mol L⁻¹; 0.08 mol L⁻¹ min⁻¹; 0.10 mol L⁻¹.
Question 3.13
Find first-order half-lives for k values (i) 200 s⁻¹ (ii) 2 min⁻¹ (iii) 4 year⁻¹.
Steps only
0.693/200=0.00347 s; 0.693/2=0.347 min; 0.693/4=0.173 y.
Answer
0.00347 s; 0.347 min; 0.173 year.
Question 3.14
A wooden object has 80% of the carbon-14 found in living wood. If carbon-14 half-life is 5730 years, estimate the object's age.
Steps only
t=(5730/0.693)ln(1/0.8)≈1845 years.
Answer
About 1845 years.
Question 3.15
For N₂O₅ decomposition, use the given concentration-time data to describe the plots, find its half-life and rate law, and calculate k.
Steps only
k=ln(1.63/0.35)/3200≈4.80×10⁻⁴ s⁻¹; t₁/₂=0.693/k≈1444 s.
Answer
First order; k≈4.80×10⁻⁴ s⁻¹; t₁/₂≈1444 s.
Question 3.16
For a first-order reaction with k=60 s⁻¹, how long does it take for the concentration to fall to 1/16 of its initial value?
Steps only
t=ln16/60=0.0462 s.
Answer
0.0462 s.
Question 3.17
A 1 mg amount of strontium-90 has half-life 28.1 years. How much remains after 10 years and 60 years?
Steps only
m=1×2^(−t/28.1): m(10)=0.781 mg; m(60)=0.228 mg.
Answer
0.781 mg; 0.228 mg.
Question 3.18
Show that the time for 99% completion of a first-order reaction is twice the time for 90% completion.
Steps only
t₉₀=2.303/k; t₉₉=4.606/k.
Answer
t₉₉=2t₉₀.
Question 3.19
A first-order reaction decomposes by 30% in 40 minutes. Find its half-life.
Steps only
t₁/₂=0.693/[(1/40)ln(1/0.7)]≈77.7 min.
Answer
77.7 min.
Question 3.20
Azoisopropane decomposes to two gas molecules from one gas molecule. Total pressure is 35 mm Hg initially and 54 mm Hg after 360 s. Find k.
Steps only
pA=70−54=16; k=ln(35/16)/360≈2.17×10⁻³ s⁻¹.
Answer
2.17×10⁻³ s⁻¹.
Question 3.21
SO₂Cl₂ decomposes at constant volume. Total pressure changes from 0.5 atm initially to 0.6 atm after 100 s. Find the reaction rate when total pressure is 0.65 atm.
Steps only
k=ln(0.5/0.4)/100=2.23×10⁻³; pSO₂Cl₂=0.35 atm; rate=kp=7.81×10⁻⁴ atm s⁻¹.
Answer
7.81×10⁻⁴ atm s⁻¹.
Question 3.22
For N₂O₅, rate constants are supplied at 0, 20, 40, 60 and 80°C. Find A and Ea from a graph of ln k against 1/T, then predict k at 30 and 50°C.
Steps only
Linear Arrhenius fit: Ea≈101 kJ mol⁻¹, A≈1.46×10¹³ s⁻¹; k303≈6.43×10⁻⁵, k323≈7.63×10⁻⁴ s⁻¹.
Answer
Ea≈101 kJ mol⁻¹; A≈1.46×10¹³ s⁻¹; k(30°C)=6.43×10⁻⁵, k(50°C)=7.63×10⁻⁴ s⁻¹.
Question 3.23
At 546 K, k=2.418×10⁻⁵ s⁻¹ and Ea=179.9 kJ mol⁻¹. Find the pre-exponential factor A.
Steps only
A=2.418×10⁻⁵ exp[179900/(8.314×546)]≈3.93×10¹² s⁻¹.
Answer
3.93×10¹² s⁻¹.
Question 3.24
For A→products, k=2.0×10⁻² s⁻¹ and [A]₀=1.0 mol L⁻¹. Find [A] after 100 s.
Steps only
[A]=1.0e⁻²=0.135 mol L⁻¹.
Answer
0.135 mol L⁻¹.
Question 3.25
Sucrose has first-order half-life 3.00 hours. What fraction remains after 8 hours?
Steps only
(1/2)^(8/3)=0.1575.
Answer
0.1575.
Question 3.26
For k=(4.5×10¹¹ s⁻¹)e^(−28000 K/T), find Ea.
Steps only
Ea=28000×8.314=232792 J mol⁻¹≈232.8 kJ mol⁻¹.
Answer
232.8 kJ mol⁻¹.
Question 3.27
Given log k=14.34−1.25×10⁴ K/T, find Ea and the temperature for a half-life of 256 minutes.
Steps only
Ea=2.303R×1.25×10⁴≈239 kJ mol⁻¹; T≈669 K.
Answer
239 kJ mol⁻¹; 669 K.
Question 3.28
k=4.5×10³ s⁻¹ at 10°C and Ea=60 kJ mol⁻¹. At what temperature is k=1.5×10⁴ s⁻¹?
Steps only
T₂=[1/283−(8.314/60000)ln(1.5×10⁴/4.5×10³)]⁻¹≈297 K.
Answer
297 K (24°C).
Question 3.29
The time for 10% completion at 298 K equals the time for 25% completion at 308 K. If A=4×10¹⁰ s⁻¹, find Ea and k at 318 K.
Steps only
Ea≈76.7 kJ mol⁻¹; k₃₁₈=4×10¹⁰e^(−Ea/(8.314×318))≈1.03×10⁻² s⁻¹.
Answer
76.7 kJ mol⁻¹; 1.03×10⁻² s⁻¹.
Question 3.30
A reaction rate becomes four times larger when temperature rises from 293 K to 313 K. Find Ea, assuming it is constant.
Steps only
Ea=8.314 ln4/(1/293−1/313)≈52.9 kJ mol⁻¹.
Answer
52.9 kJ mol⁻¹.