- iamal
- February 16, 2026
Powerful Rate Constant Unit and Rate Law MCQs Class 12 to Skyrocket Your Chemistry Scores
Understanding chemical kinetics is essential for mastering reaction rates, and practicing rate constant unit and rate law mcqs class 12 is one of the best ways to build conceptual clarity. Aspirants often find the topics of rate constants, rate laws, reaction order, and units of rate constants challenging, but systematic study of rate constant unit and rate law mcqs class 12 can make these topics far simpler and more intuitive.
In chemical kinetics, the rate of a reaction quantifies how fast reactants are converted into products. It depends on factors like concentration, temperature, and catalysts. The mathematical expression that connects these dependences is called the rate law. A key strategy in preparation is to solve many rate constant unit and rate law mcqs class 12 so that aspirants can instantly recognize how rate laws are written and how rate constants vary with reaction order.
Every rate law has a corresponding rate constant, denoted by k. The units of k change depending on the overall order of the reaction. For example, a first-order reaction has a rate constant with units of s⁻¹, while a second-order reaction usually has units of L mol⁻¹ s⁻¹. Knowing these differences is crucial, and this is precisely why repeated practice of rate constant unit and rate law mcqs class 12 helps aspirants internalize these unit patterns.
Typically, a rate law takes the form: rate = k [A]ᵃ[B]ᵇ, where a and b are reaction orders with respect to reactants A and B. The sum (a + b) gives the overall order of the reaction. When aspirants work on rate constant unit and rate law mcqs class 12, they sharpen their ability to derive reaction order from experimental data and identify the correct dependence of rate on concentration.
Understanding the difference between molecularity and reaction order is a frequent area of confusion. Molecularity refers to the number of molecules that collide in an elementary step, and it is always an integer. Reaction order, determined experimentally and often not an integer, is what appears in the rate law. Practicing rate constant unit and rate law mcqs class 12 focuses aspirants’ attention on these subtle but important distinctions.
Rate constant and rate law concepts are not just theoretical. They apply to real-world reactions, such as the decomposition of hydrogen peroxide or enzymatic reactions in biochemistry. For example, in a first-order decay reaction, half-life is constant and independent of initial concentration. In contrast, second-order reactions show half-life dependence on initial concentration. Working through rate constant unit and rate law mcqs class 12 enables aspirants to see how these principles play out in various scenarios, reinforcing both conceptual and numerical understanding.
Temperature effects on rate constants are described by the Arrhenius equation. The activation energy and frequency factor influence k, and temperature changes lead to predictable changes in rate constants. Aspirants often encounter rate constant unit and rate law mcqs class 12 that combine Arrhenius plots with rate law analysis, requiring deeper understanding.
Another important skill is unit analysis. Since unit of rate constant varies with reaction order, errors in units often lead aspirants to incorrect answers in tests. Regular practice with rate constant unit and rate law mcqs class 12 conditions them to check units carefully and avoid common pitfalls.
Graphical methods are also part of kinetics: plotting concentration versus time for zero-order, ln concentration versus time for first-order, and 1/concentration versus time for second-order reactions. Recognizing which plot gives a straight line helps identify reaction order quickly. Rate constant unit and rate law mcqs class 12 frequently test graphical interpretation, making this skill essential.
Reaction mechanisms and rate-determining steps are higher-level topics where rate laws help propose or validate possible pathways for reactions. Advanced rate constant unit and rate law mcqs class 12 require aspirants to link mechanism steps to overall rate expressions.
Rate Constant Unit and Rate Law MCQs Class 12
1.
For the elementary reaction:
3H₂(g) + N₂(g) → 2NH₃(g)
Identify the correct relation among the following relations:
(a) −(3/2) d[H₂]/dt = d[NH₃]/dt
(b) −(2/3) d[H₂]/dt = d[NH₃]/dt
(c) d[NH₃]/dt = −(1/3) d[H₂]/dt
(d) −d[H₂]/dt = d[NH₃]/dt
Answer: (b)
2.
For the chemical reaction:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
The correct option is:
(a) 3 d[H₂]/dt = 2 d[NH₃]/dt
(b) −(1/3) d[H₂]/dt = −(1/2) d[NH₃]/dt
(c) −d[N₂]/dt = 2 d[NH₃]/dt
(d) −d[N₂]/dt = (1/2) d[NH₃]/dt
Answer: (d)
3.
For a reaction scheme:
A →(k₁) B →(k₂) C
If the rate of formation of B is set to be zero, then the concentration of B is given by:
(a) k₁k₂[A]
(b) (k₁/k₂)[A]
(c) (k₁ − k₂)[A]
(d) (k₂ + k₁)[A]
Answer: (b)
4.
For an elementary chemical reaction:
A₂ ⇌ 2A
The expression for d[A]/dt is:
(a) 2k₁[A₂] − k₋₁[A]²
(b) k₁[A₂] − k₋₁[A]²
(c) 2k₁[A₂] − 2k₋₁[A]²
(d) k₁[A₂] + k₋₁[A]²
Answer: (c)
5.
2N₂O₅(g) → 4NO₂(g) + O₂(g)
Initial [N₂O₅] = 3.00 mol L⁻¹, after 30 min it becomes 2.75 mol L⁻¹.
The rate of formation of NO₂ is:
(a) 4.67 × 10⁻² mol L⁻¹ min⁻¹
(b) 1.667 × 10⁻² mol L⁻¹ min⁻¹
(c) 8.33 × 10⁻² mol L⁻¹ min⁻¹
(d) 2.083 × 10⁻² mol L⁻¹ min⁻¹
Answer: (b)
6.
Burning coal:
C(s) + O₂(g) → CO₂(g)
The rate increases by:
(a) Decreasing oxygen concentration
(b) Powdering coal
(c) Decreasing temperature
(d) Providing inert atmosphere
Answer: (b)
7.
The rate depends upon two concentration variables. What is the order?
(a) 1
(b) 3
(c) 2
(d) 0
Answer: (c)
8.
For A(g) → 2B(g) + C(g), first order reaction:
The integrated rate equation is:
(a) (2.303/t) log (P₀/(P₀ − Pₜ))
(b) (2.303/t) log (2P₀/(3P₀ − Pₜ))
(c) (2.303/t) log (P₀/(2P₀ − Pₜ))
(d) (2.303/t) log (2P₀/(2P₀ − Pₜ))
Answer: (b)
9.
Assertion: Rate is never negative.
Reason: Minus sign shows product concentration decreasing.
(a) Both correct, Reason correct explanation
(b) Both correct, Reason not explanation
(c) Assertion correct, Reason incorrect
(d) Both incorrect
Answer: (c)
10.
For A + 2B → C + D, rate expression:
(a) −d[A]/dt = −(1/2) d[B]/dt
(b) d[A]/dt = −(1/2) d[B]/dt
(c) −d[A]/dt = (1/2) d[B]/dt
(d) d[A]/dt = (1/2) d[B]/dt
Answer: (a)
11.
At what temperature rate doubles from 300 K?
(a) 329
(b) 307.7
(c) 292.03
(d) 323.5
Answer: (c)
12.
2A + B + C → A₂B + C
Rate after reaction of A:
(a) 1.28 × 10⁻⁶
(b) 1.28 × 10⁻⁸
(c) 1.28 × 10⁻¹⁰
(d) 1.28 × 10⁶
Answer: (c)
13.
Instantaneous rate for 3A + 2B → 5C:
(a) +1/3 d[A]/dt = −1/2 d[B]/dt = +1/5 d[C]/dt
(b) −1/3 d[A]/dt = −1/2 d[B]/dt = +1/5 d[C]/dt
(c) −1/3 d[A]/dt = +1/2 d[B]/dt = −1/5 d[C]/dt
(d) +1/3 d[A]/dt = −1/2 d[B]/dt = −1/5 d[C]/dt
Answer: (b)
14.
In Haber process, effect of Mo and CO:
(a) Increases and decreases
(b) Decreases and decreases
(c) Decreases and increases
(d) Both increase
Answer: (a)
15.
Data-based rate expression question:
(a) k[D₂]¹[A]²
(b) k[D₂]²[A]¹
(c) k[D₂]¹[A]¹
(d) k[D₂]²[A]²
Answer: (a)
16.
Unit of rate constant for rate = k[A]¹ᐟ²[B]³ᐟ²:
(a) mol⁻¹ L s⁻¹
(b) s⁻¹
(c) mol⁻² L s⁻¹
(d) mol² L⁻¹ s⁻¹
Answer: (a)
17.
2P + Q → products; doubling P:
(a) Doubled
(b) Halved
(c) Quadrupled
(d) Same
Answer: (c)
18.
Faster reaction, smaller is:
(a) Rate constant
(b) Concentration
(c) Half-life
(d) Energy
Answer: (c)
19.
2SO₂ + O₂ → 2SO₃
Rate of O₂ disappearance = 2 × 10⁻⁴
Rate of SO₃ formation:
(a) 2 × 10⁻⁴
(b) 4 × 10⁻⁴
(c) 6 × 10⁻⁴
(d) 8 × 10⁻⁴
Answer: (b)
20.
2O₃ → 3O₂ mechanism
Rate law:
(a) k[O₃]²
(b) k[O₃]²[O₂]⁻¹
(c) k[O₃][O₂]
(d) k[O₂][O]
Answer: (b)
21.
Incorrect statement:
(a) Rate law cannot be determined experimentally
(b) Complex reactions have fractional order
(c) Biomolecular reaction involves two species
(d) Molecularity applies to elementary reactions
Answer: (a)
22.
(a) Mechanism may change
(b) Enthalpy unchanged
(c) Cannot change order
(d) Not consumed
Answer: (c)
23.
Average rate for 2SO₂ + O₂ → 2SO₃:
(a) Δ[SO₂]/Δt
(b) −Δ[O₂]/Δt
(c) (1/2)Δ[SO₂]/Δt
(d) Δ[SO₃]/Δt
Answer: (b)
24.
BrO₃⁻ reaction rate conversion:
(a) 0.01
(b) 0.3
(c) 0.03
(d) 0.005
Answer: (c)
25.
First-order decomposition (40.5 min, 24%):
(a) 3.9 × 10⁻²
(b) 7.0 × 10⁻³
(c) 5.2 × 10⁻³
(d) 10.5 × 10⁻²
Answer: (b)
26.
r = k[A]²; doubling A:
(a) Doubled
(b) Quadrupled
(c) 8 times
(d) Unchanged
Answer: (b)
27.
Does not influence rate:
(a) Molecularity
(b) Temperature
(c) Concentration
(d) Nature
Answer: (a)
28.
Catalyst in reversible reaction affects:
(a) Forward
(b) Forward and reverse
(c) Reverse
(d) None
Answer: (b)
29.
For the reaction:
5Br⁻(aq) + 6H⁺(aq) + BrO₃⁻(aq) → 3Br₂(aq) + 3H₂O(l)
If −Δ[BrO₃⁻]/Δt = 0.01 mol L⁻¹ min⁻¹, then Δ[Br₂]/Δt in mol L⁻¹ min⁻¹ is:
(a) 0.01
(b) 0.3
(c) 0.03
(d) 0.005
Answer: (c)
30.
In a first-order reaction, it takes 40.5 minutes for the reactant to be 24% decomposed. Find the rate constant of the reaction.
(a) 3.9 × 10⁻² min⁻¹
(b) 7.0 × 10⁻³ min⁻¹
(c) 5.2 × 10⁻³ min⁻¹
(d) 10.5 × 10⁻² min⁻¹
Answer: (b)
