NEET · Chemistry —

Chemical Kinetics appeared 3 times across 1 year — 6.7% of Chemistry. This question is from Arrhenius Equation.

Year 2026 Total
Questions 3 3

Given below is an expression for the rate constant of a first-order reaction occurring at a certain temperature, T (K). ln k = 14.34 - 1.25 × 10⁴T The energy of activation in kcal mol⁻¹ for the reaction is : (Given: k in s⁻¹, R = 1.987 cal mol⁻¹ K⁻¹)

Solution & Explanation

Related Formula
k = A e(-Eₐ)/(RT)

Taking natural logarithm on both sides:

ln k = ln A - (Eₐ)/(RT)
Core Logic

Compare the given equation with the logarithmic Arrhenius equation: Given: ln k = 14.34 - (1.25 × 10⁴)/(T) Arrhenius: ln k = ln A - (Eₐ)/(R) · (1)/(T) By equating the temperature-dependent term:

(Eₐ)/(R) = 1.25 × 10⁴ Eₐ = 1.25 × 10⁴ × R

Given R = 1.987 cal mol⁻¹ K⁻¹.

Step 1: Calculate Activation Energy
Eₐ = 1.25 × 10⁴ × 1.987 Eₐ = 24837.5 cal mol⁻¹ Eₐ = 24.84 kcal mol⁻¹
Pattern Recognition

The coefficient of 1/T in the ln k equation is exactly Eₐ/R. Multiply it by R (pay attention to units: Joules vs Calories) to get Eₐ.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Reference Study Guides

More Chemical Kinetics Previous-Year Questions

Q56 neet_2026_03_may_morning Rate Constant Units
Match List I with List II :
List-I (Order of reaction)List-II (Unit of rate constant)
A. Zero order(I) mol⁻¹ L s⁻¹
B. First order(II) mol⁻² L² s⁻¹
C. Second order(III) s⁻¹
D. Third order(IV) mol L⁻¹ s⁻¹
Choose the correct answer from the options given below :
  • A. A-IV, B-III, C-II, D-I
  • B. A-I, B-II, C-III, D-IV
  • C. A-IV, B-III, C-I, D-II
  • D. A-IV, B-II, C-I, D-III

Solution

Related Formula
Unit of rate constant (k) = ( molL)¹⁻ⁿ s⁻¹

where n is the order of reaction.

Core Logic

For zero order reaction n = 0: Unit = ( molL)¹ s⁻¹ = mol L⁻¹ s⁻¹ (Matches IV)

For first order reaction n = 1: Unit = ( molL)⁰ s⁻¹ = s⁻¹ (Matches III)

For second order reaction n = 2: Unit = ( molL)⁻¹ s⁻¹ = mol⁻¹ L s⁻¹ (Matches I)

For third order reaction n = 3: Unit = ( molL)⁻² s⁻¹ = mol⁻² L² s⁻¹ (Matches II)

Step 1: Final Match

A arrow IV B arrow III C arrow I D arrow II (Note: Option 3 in the source is given as A-IV, B-III, C-I, D-II, although a typo in the raw text output says D-I, the logic leads to D-II).

Pattern Recognition

Every time order n increases by 1, the unit of k loses one power of concentration (mol/L). The base form is (mol/L)¹⁻ⁿ / s.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

Q83 neet_2026_03_may_morning Integrated Rate Equations
For a certain reaction R arrow Product, the plot of concentration [R] vs time has a negative slope as shown. The order of reaction is :
Integrated Rate Equations diagram for Q83 - NEET 2026 Code 12
Displays a straight line graph of [R] versus time with a negative slope, representing the kinetics of the reaction.
Integrated Rate Equations diagram for Q83 - NEET 2026 Code 12
Displays a straight line graph of [R] versus time with a negative slope, representing the kinetics of the reaction.
  • A. 0
  • B. 1
  • C. 2
  • D. 2.5

Solution

Related Formula
[R] = [R]₀ - kt (Zero Order) ln[R] = ln[R]₀ - kt (First Order)
Core Logic

The graph given is a plot of [R] (concentration) versus t (time) yielding a straight line with a negative slope. Comparing the equation of a straight line y = mx + c with the rate equations: For a zero-order reaction: [R] = -k t + [R]₀. Here, y = [R], x = t, slope m = -k, and y-intercept c = [R]₀. For a first-order reaction, the straight line is obtained when plotting ln[R] vs time, not [R] vs time.

Solution for Q83 - Zero Order Plot
Displays a straight line graph of [R] versus time with a negative slope, representing the kinetics of the reaction.

Step 1: Final Conclusion

Since [R] vs time is a straight line, the reaction order is zero.

Pattern Recognition

[R] vs t straight line = Zero Order. ln[R] vs t straight line = First Order. 1/[R] vs t straight line = Second Order.

Chapter Mix

Class 12 Chemistry: Chemical Kinetics

More Chemical Kinetics Questions — neet_2026_03_may_morning

Practice all Chemical Kinetics previous-year questions →

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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)