Chemical Kinetics Previous Year Questions — NEET Chemistry

3 past-year Chemical Kinetics questions from NEET (Chemistry).

Q56 (2024)

Match List I with List II: <div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;">List-I (Order of reaction)</th><th style="border: 1px solid #888; padding: 8px;">List-II (Unit of rate constant)</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">A. Zero order</td><td style="border: 1px solid #888; padding: 8px;">(I) $\text{mol}^{-1} \text{L s}^{-1}$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">B. First order</td><td style="border: 1px solid #888; padding: 8px;">(II) $\text{mol}^{-2} \text{L}^2 \text{s}^{-1}$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">C. Second order</td><td style="border: 1px solid #888; padding: 8px;">(III) $\text{s}^{-1}$</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">D. Third order</td><td style="border: 1px solid #888; padding: 8px;">(IV) $\text{mol L}^{-1} \text{s}^{-1}$</td></tr></tbody></table></div> Choose the correct answer from the options given below:
  1. (1) A-IV, B-III, C-II, D-I
  2. (2) A-I, B-II, C-III, D-IV
  3. (3) A-IV, B-III, C-I, D-II
  4. (4) A-IV, B-II, C-I, D-III
### Related Formula $$\text{Unit of } k = (\text{mol L}^{-1})^{1-n} \text{ s}^{-1}$$ ### Core Logic $$n = 0 \implies \text{mol L}^{-1} \text{s}^{-1} \quad (IV)$$ $$n = 1 \implies \text{s}^{-1} \quad (III)$$ $$n = 2 \implies \text{mol}^{-1} \text{L s}^{-1} \quad (I)$$ $$n = 3 \implies \text{mol}^{-2} \text{L}^2 \text{s}^{-1} \quad (II)$$ ### Step 1: Conclusion Matches: A-IV, B-III, C-I, D-II. ### Pattern Recognition Substitute $n$ into $(mol/L)^{1-n} s^{-1}$ formula for reaction order. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

Q83 (2024)

For a certain reaction R $\rightarrow$ Product, the plot of concentration [R] vs time has a negative slope as shown: {{IMG1}} The order of reaction is:
  1. (1) 0
  2. (2) 1
  3. (3) 2
  4. (4) 2.5
### Related Formula $$[R] = [R_0] - kt$$ ### Core Logic Integrated rate equation for zero order reaction is $[R] = [R_0] - kt$. A plot of $[R]$ vs $t$ gives a straight line with slope $= -k$ and intercept $[R_0]$. ### Step 1: Conclusion Reaction order is 0. ### Pattern Recognition $[R]$ vs $t$ linear $\rightarrow$ Zero order; $\ln[R]$ vs $t$ linear $\rightarrow$ First order. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

Q88 (2024)

Given below is an expression for the rate constant of a first-order reaction occurring at a certain temperature, T (K): $$\ln k = 14.34 - \frac{1.25 \times 10^4}{T}$$ The energy of activation in $\text{kcal mol}^{-1}$ for the reaction is: (Given: k in $\text{s}^{-1}$, $R = 1.987 \text{ cal mol}^{-1} \text{K}^{-1}$)
  1. (1) 12.42
  2. (2) 14.34
  3. (3) 18.63
  4. (4) 24.84
### Related Formula $$\ln k = \ln A - \frac{E_a}{RT}$$ ### Core Logic Comparing given equation $\ln k = 14.34 - \frac{1.25 \times 10^4}{T}$ with Arrhenius form: $$\frac{E_a}{R} = 1.25 \times 10^4$$ $$E_a = 1.25 \times 10^4 \times 1.987 = 24837.5 \text{ cal/mol} = 24.84 \text{ kcal/mol}$$ ### Step 1: Conclusion Activation energy is $24.84 \text{ kcal mol}^{-1}$. ### Pattern Recognition Slope of $\ln k$ vs $1/T$ equals $-E_a / R$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics
Rankbit System
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%)

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