Keywords:#Unit of rate constant match list#JEE Main 2026 Morning Q56#Chemical Kinetics Reaction Order#Rate constant units formula
More Chemical Kinetics Previous-Year Questions
Q83neet_2024_05_may_morningZero-Order Kinetics and Graphs
For a certain reaction R arrow$\rightarrow$ Product, the plot of concentration [R] vs time has a negative slope as shown:
Linear plot of reactant concentration vs time with negative slope -k.
The order of reaction is:
Linear plot of reactant concentration vs time with negative slope -k.
A. (1) 0
B. (2) 1
C. (3) 2
D. (4) 2.5
Solution
Related Formula
[R] = [R₀] - kt$$[R] = [R_0] - kt$$
Core Logic
Integrated rate equation for zero order reaction is [R] = [R₀] - kt$[R] = [R_0] - kt$. A plot of [R]$[R]$ vs t$t$ gives a straight line with slope = -k$= -k$ and intercept [R₀]$[R_0]$.
Step 1: Conclusion
Reaction order is 0.
Pattern Recognition
[R]$[R]$ vs t$t$ linear arrow$\rightarrow$ Zero order; ln[R]$\ln[R]$ vs t$t$ linear arrow$\rightarrow$ First order.
Chapter Mix
Class 12 Chemistry: Chemical Kinetics
Q88neet_2024_05_may_morningArrhenius Equation and Activation Energy
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 × 10⁴)/(T)$$\ln k = 14.34 - \frac{1.25 \times 10^4}{T}$$
The energy of activation in kcal mol⁻¹$\text{kcal mol}^{-1}$ for the reaction is:
(Given: k in s⁻¹$\text{s}^{-1}$, R = 1.987 cal mol⁻¹ K⁻¹$R = 1.987 \text{ cal mol}^{-1} \text{K}^{-1}$)
A. (1) 12.42
B. (2) 14.34
C. (3) 18.63
D. (4) 24.84
Solution
Related Formula
ln k = ln A - (Eₐ)/(RT)$$\ln k = \ln A - \frac{E_a}{RT}$$
Core Logic
Comparing given equation ln k = 14.34 - (1.25 × 10⁴)/(T)$\ln k = 14.34 - \frac{1.25 \times 10^4}{T}$ with Arrhenius form:
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
