Chemical Kinetics — JEE Mains Chemistry Formulas and Traps

Rate of Reaction and Stoichiometry

Average rate from a change in concentration

Average rate

r_{\text{av}} = -\frac{\Delta[\text{R}]}{\Delta t} = +\frac{\Delta[\text{P}]}{\Delta t}

Rates of different species through the coefficients

Rate of reaction

r = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}

Common traps

Clock reaction: the time and the thiosulphate

The time is recorded the instant the blue colour appears, because that is when the thiosulphate runs out. Thiosulphate must be LESS than KI; with more thiosulphate, iodine never builds up and the blue never comes.

Minutes divided as if they were hours

A change of 0.1 mol L

^{-1}

in 20 minutes is

\dfrac{0.1}{1/3} = 0.3

mol L

^{-1}

h

^{-1}

. Convert the time into the unit the question asks for before dividing.

Two species' rates taken as equal

In

\mathrm{2N_2O_5 \rightarrow 4NO_2 + O_2}

,

\mathrm{NO_2}

forms twice as fast as

\mathrm{N_2O_5}

is used up, not at the same rate. Divide each species' rate by its own coefficient before comparing.

A species' rate reported as the rate of reaction

The rate of reaction is the species' rate divided by its coefficient. If

\mathrm{Br^-}

(coefficient 5) is used up at

5 \times 10^{-4}

mol L

^{-1}

s

^{-1}

, the rate of reaction is

1 \times 10^{-4}

, not

5 \times 10^{-4}

.

Rate Law, Order and Molecularity

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How the rate changes when concentrations change

Ratio of rates

\frac{r_2}{r_1} = \left(\frac{[A]_2}{[A]_1}\right)^{m}\left(\frac{[B]_2}{[B]_1}\right)^{n}

Order from a table of initial rates

Order in A from two runs

m = \frac{\log(r_2/r_1)}{\log([A]_2/[A]_1)} \quad ([B]\ \text{fixed})

Unit of the rate constant for each order, and order versus molecularity

Order	Rate law	Unit of k	Half-life
0	$r = k$	mol L $^{-1}$ s $^{-1}$	$\dfrac{[A]_0}{2k}$ , proportional to $[A]_0$
1	$r = k[A]$	s $^{-1}$	$\dfrac{0.693}{k}$ , independent of $[A]_0$
2	$r = k[A]^2$	L mol $^{-1}$ s $^{-1}$	$\dfrac{1}{k[A]_0}$ , inversely proportional to $[A]_0$
3	$r = k[A]^3$	L $^2$ mol $^{-2}$ s $^{-1}$	Proportional to $\dfrac{1}{[A]_0^2}$
$n$	$r = k[A]^n$	$(\text{mol L}^{-1})^{1-n}$ s $^{-1}$	Proportional to $[A]_0^{\,1-n}$

The unit of k names the order; the half-life column is used again on the zero-order page.

Common traps

Order read from the balanced equation

Coefficients give the exponents only for an elementary (single-step) reaction. Otherwise the order comes from experiment:

\mathrm{2N_2O_5 \rightarrow 4NO_2 + O_2}

is first order, not second.

More solution taken as a faster reaction

The rate depends on concentration. Doubling the volume of the same solution leaves the rate unchanged; adding the same volume of water halves the concentration and lowers the rate.

Two runs where both concentrations changed

If

[A]

and

[B]

both change between two runs, the rate ratio mixes both orders. Choose a pair where only one changes, or divide out the factor from the order you already know.

Rate and k given the same unit

Only for a zero-order reaction. For a first-order reaction the rate is in mol L

^{-1}

s

^{-1}

but

k

is in s

^{-1}

.

A fractional molecularity

Order can be zero or a fraction, because it is measured. Molecularity counts colliding particles in one elementary step, so it is always 1, 2 or 3.

First Order Reactions and Half-Life

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The integrated first-order law and the half-life

Integrated first-order law

k = \frac{2.303}{t}\log\frac{[A]_0}{[A]}\qquad t_{1/2} = \frac{0.693}{k}

Comparing the times to two levels of completion

Ratio of two completion times

\frac{t_1}{t_2} = \frac{\log([A]_0/[A]_1)}{\log([A]_0/[A]_2)}

Exponential form, straight-line plots and rate ratios

Exponential form

[A] = [A]_0\,e^{-kt}\qquad \ln\frac{[A]}{[A]_0} = -kt

Common traps

The percent decomposed used as [A]

70% decomposed leaves 30%. Use

\log\dfrac{100}{30}

, not

\log\dfrac{100}{70}

; the second gives a time far too short.

Expiry time of a drug

A drug that stops working at 50% decomposition expires after ONE half-life. If it falls to one-eighth in 18 months, that is three half-lives, so

t_{1/2} = 6

months and the expiry is 6 months — not 18, and not 9.

67% complete read as two-thirds left

67% complete leaves about one-third, so

t = \dfrac{2.303}{k}\log 3 \approx 1.58\,t_{1/2}

. Using

\log 1.5

(two-thirds left) answers the question for 33% completion.

Times scaled like the percentages

99.9% completion does not take about twice as long as 90%: the logs are 3 and 1, so it takes three times as long.

A first-order reaction that 'completes'

[A] = [A]_0e^{-kt}

is never zero, so a first-order reaction never reaches 100% in a finite time. A statement that it completes in 1000 s is false.

Rate ratio used for any order

A ratio of rates equals a ratio of concentrations only for first order. For second order the rate ratio is the SQUARE of the concentration ratio.

First Order in Gases and Radioactive Decay

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Rate constant of a gas reaction from the total pressure

A(g) → B(g) + C(g)

k = \frac{2.303}{t}\log\frac{p_i}{2p_i - P_t}

Radioactive decay, carbon dating and bacterial growth

Radioactive decay

\frac{N}{N_0} = e^{-\lambda t} = \left(\tfrac12\right)^{t/t_{1/2}},\qquad \lambda = \frac{0.693}{t_{1/2}}

Common traps

The total pressure put into the log

\ln\dfrac{p_i}{P_t}

is wrong: the first-order law needs the pressure of

A

alone. For

A(g) \rightarrow B(g) + C(g)

that is

2p_i - P_t

.

P∞ taken as the initial pressure

At the end every

A

has turned into products. For

A \rightarrow B + C

,

P_\infty = 2p_i

; for

A \rightarrow 2B + C

,

P_\infty = 3p_i

. Divide before using it as

p_i

.

Decay constant rising with temperature

Heating speeds up chemical reactions, not radioactive decay. The decay constant is a property of the nucleus and stays the same at any temperature.

Growth drawn as decay

For bacterial growth

N = N_0e^{kt}

: the plot of

N/N_0

against

t

starts at 1 and curves UPWARD. A falling curve is decay.

Zero Order and Finding the Order

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The zero-order law and its half-life

Zero-order law

[A] = [A]_0 - kt\qquad t_{1/2} = \frac{[A]_0}{2k}

Order from how the half-life depends on the starting concentration

Half-life and order

t_{1/2} \propto [A]_0^{\,1-n}\quad\Rightarrow\quad \frac{t_{1/2}'}{t_{1/2}} = \left(\frac{[A]_0'}{[A]_0}\right)^{1-n}

Identifying zero and first order from the shape of a graph

Plot	Zero order	First order
$[A]$ against $t$	Straight line, slope $-k$	Falling exponential curve that never reaches zero
$\ln[A]$ against $t$	Curve bending downward	Straight line, slope $-k$
$\log\dfrac{[A]}{[A]_0}$ against $t$	Curve bending downward	Straight line through the origin, slope $-\dfrac{k}{2.303}$
Rate against $t$	Horizontal line	Falling exponential curve
Rate against $[A]$	Horizontal line	Straight line through the origin, slope $k$
$t_{1/2}$ against $[A]_0$	Straight line through the origin	Horizontal line

Find the straight plot first; its slope then gives k.

Common traps

One-quarter taken as two half-lives

That holds only for first order. For zero order the second half-life is half the first, so the reactant reaches one-quarter at

1.5\,t_{1/2}

.

Zero-order half-life treated as fixed

t_{1/2} = \dfrac{[A]_0}{2k}

depends on the concentration you start from. Work out

k

from the stated half-life first, then find any later time from

[A] = [A]_0 - kt

.

Seconds compared with minutes

240 s at one pressure and 4.0 min at another are the SAME half-life, which means first order. Convert both to one unit before taking the ratio.

The order changing with concentration

A half-life table is used to find one fixed order. A statement that the order becomes 1 when the concentration is raised is false.

The sign of the slope

\log\dfrac{[A]}{[A]_0}

falls with time, so its slope is

-\dfrac{k}{2.303}

. A statement giving the slope as

+\dfrac{k}{2.303}

for this plot is false.

A rate–time line read as a concentration–time line

A horizontal RATE against time line means zero order. A horizontal CONCENTRATION against time line would mean nothing is reacting. Read the axis label first.

Temperature and the Arrhenius Equation

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Reading Ea and A from the Arrhenius equation or its plot

Arrhenius equation

k = A\,e^{-E_a/RT}\qquad \ln k = \ln A - \frac{E_a}{RT}\qquad \log k = \log A - \frac{E_a}{2.303RT}

Two-temperature form of the Arrhenius equation

Two-temperature form

\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

Ratio of rate constants for two reactions with the same A

Same A, different Ea

\ln\frac{k_2}{k_1} = \frac{E_{a1} - E_{a2}}{RT}

Common traps

The ln form read as the log form

The slope of

\ln k

against

1/T

is

-\dfrac{E_a}{R}

; the slope of

\log k

against

1/T

is

-\dfrac{E_a}{2.303R}

. Using the wrong one puts

E_a

out by a factor of 2.303.

Fraction BELOW the activation energy

e^{-E_a/RT}

is the fraction of molecules with energy equal to or MORE than

E_a

— the ones that can react. A statement calling it the fraction with less than

E_a

is false.

Endothermic reactions slowing on heating

The sign of

\Delta H

does not enter

k = Ae^{-E_a/RT}

. For every reaction

k

rises with temperature, along a curve that bends upward at first.

Celsius in the formula

\dfrac{1}{T}

must be in kelvin. Convert 27 °C to 300 K before subtracting reciprocals, and convert an answer in kelvin back to °C only if the question asks.

The reciprocals subtracted the wrong way round

With

k_2

at the higher temperature

T_2

,

\dfrac{1}{T_1} - \dfrac{1}{T_2}

is positive, and so is

\log\dfrac{k_2}{k_1}

. A negative

E_a

means one of the two was flipped.

ln of the ratio given when log was asked

\log\dfrac{k_2}{k_1} = \dfrac{\Delta E_a}{2.303RT}

and

\ln\dfrac{k_2}{k_1} = \dfrac{\Delta E_a}{RT}

differ by a factor of 2.303. Also keep

\Delta E_a

in joules when

R

is in J K

^{-1}

mol

^{-1}

; kJ puts the answer out by 1000.

Mechanisms, Energy Profiles and Catalysts

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Rate law and activation energy from a mechanism

Composite activation energy

k = \frac{k_1k_2}{k_3}\ \Rightarrow\ E_a = E_{a1} + E_{a2} - E_{a3}

Reading energy profiles and the effect of a catalyst

Barriers and enthalpy

\Delta H = E_{a,f} - E_{a,b}

Common traps

An intermediate left in the rate law

The final rate law contains only species in the overall equation (and any catalyst). Replace an intermediate using the fast equilibrium or the steady state before counting the order.

Rate law written from the overall equation

The exponents come from the SLOW step, not from the overall stoichiometry.

\mathrm{2NO + Br_2 \rightarrow 2NOBr}

is third order because of its mechanism, not because three molecules appear on the left.

A catalyst that changes ΔH

A catalyst lowers both barriers by the same amount, so

\Delta H = E_{a,f} - E_{a,b}

does not change; nor do

\Delta G

and

K

. It cannot make a non-spontaneous reaction occur.

The sign of ΔH flipped

\Delta H = E_{a,f} - E_{a,b}

, forward minus backward. With

E_{a,f} = 120

and

E_{a,b} = 150

kJ mol

^{-1}

,

\Delta H = -30

kJ mol

^{-1}

: exothermic, not

+30

.

Valleys counted as activated complexes

Peaks are activated complexes; the valleys between them are intermediates. The start and end levels are the reactants and products and are neither.

Chemical Kinetics formulas

Rate of Reaction and Stoichiometry

Average rate from a change in concentration

Rates of different species through the coefficients

Common traps

Rate Law, Order and Molecularity

How the rate changes when concentrations change

Order from a table of initial rates

Unit of the rate constant for each order, and order versus molecularity

Common traps

First Order Reactions and Half-Life

The integrated first-order law and the half-life

Comparing the times to two levels of completion

Exponential form, straight-line plots and rate ratios

Common traps

First Order in Gases and Radioactive Decay

Rate constant of a gas reaction from the total pressure

Radioactive decay, carbon dating and bacterial growth

Common traps

Zero Order and Finding the Order

The zero-order law and its half-life

Order from how the half-life depends on the starting concentration

Identifying zero and first order from the shape of a graph

Common traps

Temperature and the Arrhenius Equation

Reading Ea and A from the Arrhenius equation or its plot

Two-temperature form of the Arrhenius equation

Ratio of rate constants for two reactions with the same A

Common traps

Mechanisms, Energy Profiles and Catalysts

Rate law and activation energy from a mechanism

Reading energy profiles and the effect of a catalyst

Common traps

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