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Mastering chemical kinetics graph

Mastering Chemical Kinetics Graphs | ChemCa.in
Physical Chemistry / Chemical Kinetics

Mastering Kinetics Graphs

A comprehensive visual guide to Zero, First, Second, and n-th Order reaction graphs.

In chemical kinetics, graphical analysis is the most reliable way to determine the order of a reaction. By plotting different functions of concentration against time, we look for the plot that yields a straight line ($y = mx + c$).

1 Zero Order Reactions

Rate Law: Rate = $k[A]^0 = k$ (Constant rate)

Integrated Rate Equation: $[A] = [A]_0 - kt$

Half-Life ($t_{1/2}$): $t_{1/2} = \frac{[A]_0}{2k}$

Concentration vs. Time

$y = mx + c \implies [A] = -kt + [A]_0$

Zero Order: Concentration vs Time Graph A Cartesian graph plotting Concentration [A] on the y-axis against Time t on the x-axis. The plot is a straight line sloping downwards from a y-intercept, indicating a constant rate of decrease. [A] Time (t) [A]₀ Slope = -k

Rate vs. Concentration

Rate is independent of concentration

Zero Order: Rate vs Concentration A Cartesian graph plotting Rate on the y-axis against Concentration on the x-axis. The plot is a perfectly horizontal straight line, showing that rate does not change with concentration. Rate [A] Rate = k (Constant)

2 First Order Reactions

Rate Law: Rate = $k[A]^1$

Integrated Rate Equation: $\ln[A] = \ln[A]_0 - kt$   (or $\log[A] = \log[A]_0 - \frac{kt}{2.303}$)

Half-Life ($t_{1/2}$): $t_{1/2} = \frac{\ln 2}{k} \approx \frac{0.693}{k}$ (Independent of initial concentration)

$\ln[A]$ vs. Time

$y = mx + c \implies \ln[A] = -kt + \ln[A]_0$

First Order: Natural Log of Concentration vs Time A graph plotting ln[A] against time. It shows a straight line sloping downwards, demonstrating that the natural logarithm of concentration decreases linearly with time. $\ln[A]$ Time (t) $\ln[A]_0$ Slope = -k

Half-life vs. Initial Conc.

Half-life is a constant

First Order: Half-life vs Initial Concentration A graph plotting half-life (t1/2) on the y-axis against initial concentration on the x-axis. It is a horizontal line, proving half-life is independent of the starting amount. $t_{1/2}$ $[A]_0$ Constant ($0.693/k$)

Note on Exponential Decay: If you plot plain concentration $[A]$ vs. time $t$ for a first-order reaction, you get an exponential decay curve (not a straight line) represented by $[A] = [A]_0 e^{-kt}$. Radioactive decay strictly follows first-order kinetics.

3 Second Order Reactions

Rate Law: Rate = $k[A]^2$

Integrated Rate Equation: $\frac{1}{[A]} = \frac{1}{[A]_0} + kt$

Half-Life ($t_{1/2}$): $t_{1/2} = \frac{1}{k[A]_0}$

$1/[A]$ vs. Time

$y = mx + c \implies \frac{1}{[A]} = kt + \frac{1}{[A]_0}$

Second Order: Inverse Concentration vs Time A graph plotting 1/[A] against time. It shows a straight line sloping upwards from a positive y-intercept, unique to second-order reactions. $1/[A]$ Time (t) $1/[A]_0$ Slope = +k

Rate vs. $[A]^2$

Direct proportionality to the square

Second Order: Rate vs Concentration Squared A graph plotting Rate against [A] squared. It is a straight line passing directly through the origin. Rate $[A]^2$ Slope = k

4 General Formula (n-th Order)

To generalize for any reaction order $n$ (where $n \neq 1$), the half-life dependence on the initial concentration is given by a master equation:

Master Half-Life Relation

$$ t_{1/2} \propto \frac{1}{[A]_0^{n-1}} $$

Order (n) Straight Line Plot Slope Half-Life Dependency
0 $[A]$ vs $t$ $-k$ $t_{1/2} \propto [A]_0$
1 $\ln[A]$ vs $t$ $-k$ $t_{1/2} \propto [A]_0^0$ (Constant)
2 $1/[A]$ vs $t$ $+k$ $t_{1/2} \propto 1/[A]_0$

Knowledge Check

10 Practice MCQs on Chemical Kinetics Graphs

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