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 & Explanation
Related Formula
Unit of rate constant (k) = ( molL)¹⁻ⁿ s⁻¹$$\text{Unit of rate constant (}k\text{)} = \left(\frac{\text{mol}}{\text{L}}\right)^{1-n} \text{s}^{-1}$$
where n$n$ is the order of reaction.
Core Logic
For zero order reaction n = 0$\implies n = 0$:
Unit = ( molL)¹ s⁻¹ = mol L⁻¹ s⁻¹$\left(\frac{\text{mol}}{\text{L}}\right)^1 \text{s}^{-1} = \mathrm{mol \, L^{-1} \, s^{-1}}$ (Matches IV)
For first order reaction n = 1$\implies n = 1$:
Unit = ( molL)⁰ s⁻¹ = s⁻¹$\left(\frac{\text{mol}}{\text{L}}\right)^0 \text{s}^{-1} = \mathrm{s^{-1}}$ (Matches III)
For second order reaction n = 2$\implies n = 2$:
Unit = ( molL)⁻¹ s⁻¹ = mol⁻¹ L s⁻¹$\left(\frac{\text{mol}}{\text{L}}\right)^{-1} \text{s}^{-1} = \mathrm{mol^{-1} \, L \, s^{-1}}$ (Matches I)
For third order reaction n = 3$\implies n = 3$:
Unit = ( molL)⁻² s⁻¹ = mol⁻² L² s⁻¹$\left(\frac{\text{mol}}{\text{L}}\right)^{-2} \text{s}^{-1} = \mathrm{mol^{-2} \, L^2 \, s^{-1}}$ (Matches II)
Step 1: Final Match
A arrow$\rightarrow$ IV
B arrow$\rightarrow$ III
C arrow$\rightarrow$ I
D arrow$\rightarrow$ 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$n$ increases by 1, the unit of k$k$ loses one power of concentration (mol/L). The base form is (mol/L)¹⁻ⁿ / s$(mol/L)^{1-n} / s$.
For a certain reaction R arrow$\rightarrow$ Product, the plot of concentration [R] vs time has a negative slope as shown. The order of reaction is :
Displays a straight line graph of [R] versus time with a negative slope, representing the kinetics of the reaction.Displays a straight line graph of [R] versus time with a negative slope, representing the kinetics of the reaction.
The graph given is a plot of [R]$[R]$ (concentration) versus t$t$ (time) yielding a straight line with a negative slope.
Comparing the equation of a straight line y = mx + c$y = mx + c$ with the rate equations:
For a zero-order reaction: [R] = -k t + [R]₀$[R] = -k t + [R]_0$. Here, y = [R]$[R]$, x = t$t$, slope m = -k$m = -k$, and y-intercept c = [R]₀$c = [R]_0$.
For a first-order reaction, the straight line is obtained when plotting ln[R]$\ln[R]$ vs time, not [R]$[R]$ vs time.
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]$[R]$ vs time is a straight line, the reaction order is zero.
Pattern Recognition
[R]$[R]$ vs t$t$ straight line = Zero Order.
ln[R]$\ln[R]$ vs t$t$ straight line = First Order.
1/[R]$1/[R]$ vs t$t$ straight line = Second Order.
Chapter Mix
Class 12 Chemistry: Chemical Kinetics
Q88neet_2026_03_may_morningArrhenius Equation
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$$\ln k = 14.34 - \frac{1.25 \times 10^{4}}{T}$$
The energy of activation in kcal mol⁻¹$mol^{-1}$ for the reaction is :
(Given: k in s⁻¹$s^{-1}$, R = 1.987 cal mol⁻¹$mol^{-1}$K⁻¹$K^{-1}$)
A. 12.42
B. 14.34
C. 18.63
D. 24.84
Solution
Related Formula
k = A e(-Eₐ)/(RT)$$k = A e^{\frac{-E_a}{RT}}$$
Taking natural logarithm on both sides:
ln k = ln A - (Eₐ)/(RT)$$\ln k = \ln A - \frac{E_a}{RT}$$
Core Logic
Compare the given equation with the logarithmic Arrhenius equation:
Given: ln k = 14.34 - (1.25 × 10⁴)/(T)$\ln k = 14.34 - \frac{1.25 \times 10^4}{T}$
Arrhenius: ln k = ln A - (Eₐ)/(R) · (1)/(T)$\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}$
By equating the temperature-dependent term:
The coefficient of 1/T$1/T$ in the ln k$\ln k$ equation is exactly Eₐ/R$E_a/R$. Multiply it by R$R$ (pay attention to units: Joules vs Calories) to get Eₐ$E_a$.
Chapter Mix
Class 12 Chemistry: Chemical Kinetics
More Chemical Kinetics Questions — neet_2026_03_may_morning
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.
