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Chemical Kinetics formulas

16 formulas and 16 common traps for MHT-CET Chemistry Chemical Kinetics, grouped by subtopic.

Full notes with worked examples

Rate of Reaction, Stoichiometry and Average Rate

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Converting One Species' Rate to Another's: Multiply by the Coefficient Ratio

Rate of reaction

rate=−1ad[A]dt=−1bd[B]dt=1cd[C]dt=1dd[D]dt\text{rate} = -\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}

Writing the Rate Expression, and Reading the Equation Back From It

Signs and reciprocals

reactant: −1coeffd[⋅]dtproduct: +1coeffd[⋅]dt\text{reactant: } -\frac{1}{\text{coeff}}\frac{d[\cdot]}{dt} \qquad \text{product: } +\frac{1}{\text{coeff}}\frac{d[\cdot]}{dt}

Common traps

Multiplying when you should divide

N2_2 at 0.0440.044 means NH3_3 at 0.0880.088; NH3_3 at 0.0880.088 means N2_2 at 0.0440.044. Set up the equal-rates chain and read the direction off the coefficients rather than guessing which way the factor goes.

Coefficient in front instead of as a reciprocal

The term for H2_2 is −13d[H2]dt-\dfrac13\dfrac{d[\text{H}_2]}{dt}, not −3d[H2]dt-3\dfrac{d[\text{H}_2]}{dt}. Option (A) on every N2_2/H2_2/NH3_3 stem is the inverted version.

Rate Law, Order, Molecularity and Rate Expression

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The Rate Law and the Order: Exponents Come From Experiment, Not From the Equation

Rate law

r=k[A]x[B]y,order=x+y (experimental; may be 0, fractional)r = k[A]^x[B]^y,\qquad \text{order} = x + y \ (\text{experimental; may be } 0, \text{ fractional})

The Rate Constant: k = rate/([A]ˣ[B]ʸ), Its Units and Its Properties

Rate constant

k=r[A]x[B]y,[k]=(mol dm−3)1−n s−1k = \frac{r}{[A]^x[B]^y},\qquad [k] = (\text{mol dm}^{-3})^{1-n}\,\text{s}^{-1}

How the Rate Changes When Concentrations Change: Multiply the Factors

Rate factor

r′r=([A]′[A])x([B]′[B])y\frac{r'}{r} = \left(\frac{[A]'}{[A]}\right)^x\left(\frac{[B]'}{[B]}\right)^y

Order Versus Molecularity

The contrast

molecularity∈{1,2,3} (theoretical, per step);order∈R≥0 (experimental)\text{molecularity} \in \{1, 2, 3\} \text{ (theoretical, per step)};\qquad \text{order} \in \mathbb{R}_{\ge 0} \text{ (experimental)}

Common traps

Reading the order off the balanced equation

2N2O5→4NO2+O22\text{N}_2\text{O}_5 \to 4\text{NO}_2 + \text{O}_2 is first order, and H2_2 + Br2_2 is order 32\tfrac32. The coefficients give molecularity for an elementary step; the order is measured.

Squaring the wrong concentration

r=k[A][B]2r = k[A][B]^2 squares [B][B]. With [A]=1[A] = 1 and [B]=0.2[B] = 0.2, squaring the wrong one turns 0.250.25 into 1.251.25 — and both are on the list, because the two rate laws appear on twin stems.

Adding the factors

Doubling A and B in r=k[A][B]2r = k[A][B]^2 gives 2×4=82 \times 4 = 8, not 2+4=62 + 4 = 6. Factors multiply because the rate law is a product.

Calling H₂ + Br₂ 'monomolecular' because the order is fractional

Two molecules collide, so it is bimolecular whatever the order. Options pairing 'monomolecular' with 32\tfrac32 are the planted confusion of the two ideas.

Zero-Order Kinetics

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[A]ₜ = [A]₀ − kt: Constant Rate, k in Concentration per Time

Zero order

[A]t=[A]0−kt,k=[A]0−[A]tt  (mol dm−3 s−1)[A]_t = [A]_0 - kt,\qquad k = \frac{[A]_0 - [A]_t}{t}\ \ (\text{mol dm}^{-3}\,\text{s}^{-1})

Zero-Order Half-Life: t½ = [A]₀/2k, Proportional to the Initial Concentration

Zero-order half-life

t1/2=[A]02kt_{1/2} = \frac{[A]_0}{2k}

Common traps

Forgetting the seconds-to-minutes conversion

0.8240=0.0033\dfrac{0.8}{240} = 0.0033 per second is 0.20.2 per minute; the options are per minute. Convert before matching.

Using 0.693/k

0.693/k0.693/k is the FIRST-order half-life. For zero order the half-life carries [A]0[A]_0; option (B) ak\dfrac{a}{k} forgets the 22.

First-Order Kinetics, Rate Constant and Half-Life

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k = 0.693/t½ and k = rate/[A]: the Half-Life Is Fixed, the Unit Is time⁻¹

First-order constant

t1/2=0.693k,k=r[A]  (s−1),slope of log⁡[A]t vs t=−k2.303t_{1/2} = \frac{0.693}{k},\qquad k = \frac{r}{[A]}\ \ (\text{s}^{-1}),\qquad \text{slope of } \log[A]_t \text{ vs } t = -\frac{k}{2.303}

k = (2.303/t) log([A]₀/[A]ₜ): Percent Decomposed, and the Time to 90%, 99%, 99.9%

Integrated first-order law

k=2.303tlog⁡10[A]0[A]tt90%=2.303k, t99%=4.606k, t99.9%=6.909kk = \frac{2.303}{t}\log_{10}\frac{[A]_0}{[A]_t} \qquad t_{90\%} = \frac{2.303}{k},\ t_{99\%} = \frac{4.606}{k},\ t_{99.9\%} = \frac{6.909}{k}

Counting Half-Lives: Fraction Left After n Half-Lives Is (1/2)^n

Half-life counting

[A]t[A]0=(12)n,t=n t1/2\frac{[A]_t}{[A]_0} = \left(\frac12\right)^{n},\qquad t = n\,t_{1/2}

Common traps

Hours left as hours

A half-life of 2.52.5 h with the answer wanted in s−1^{-1} needs 90009000 s in the denominator. 0.6932.5=0.277\dfrac{0.693}{2.5} = 0.277 h−1^{-1} is right in the wrong unit and matches nothing on the list.

Using the percent decomposed as [A]ₜ

60% decomposed means log⁡10040\log\dfrac{100}{40}, not log⁡10060\log\dfrac{100}{60}. The second gives 0.01020.0102, which is option (A) — exactly half the answer.

Scaling the time with the concentration

0.1→0.0250.1 \to 0.025 takes the same 3030 min as 0.8→0.20.8 \to 0.2 would; a first-order half-life does not shrink for smaller amounts. 7.57.5 min is the option for the student who scaled.

Reaction Mechanism, Intermediates and Rate-Determining Step

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The Rate-Determining Step Writes the Rate Law

Rate from the slow step

slow step aA+bB→… ⇒ r=k[A]a[B]b\text{slow step } aA + bB \to \dots \ \Rightarrow\ r = k[A]^a[B]^b

Intermediates: Made in One Step, Used in the Next; Catalysts: Used, Then Regenerated

Cancelling species

intermediate: produced then consumed;catalyst: consumed then regenerated\text{intermediate: produced then consumed;}\qquad \text{catalyst: consumed then regenerated}

Common traps

Writing the rate law from the overall equation

2NO2Cl→2NO2+Cl22\text{NO}_2\text{Cl} \to 2\text{NO}_2 + \text{Cl}_2 suggests k[NO2Cl]2k[\text{NO}_2\text{Cl}]^2, option (D); the slow step is unimolecular and the law is first order.

Picking the product that appears in both steps

Cl−\text{Cl}^- is formed in BOTH ClO−^- steps and never consumed — it is a product, not an intermediate. The intermediate is the one that disappears again.

Temperature Dependence, Arrhenius and Collision Theory

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The Arrhenius Equation and Its Plot: Slope −Ea/2.303R, Intercept log A

Arrhenius

k=A e−Ea/RT,log⁡10k=log⁡10A−Ea2.303R⋅1Tk = A\,e^{-E_a/RT},\qquad \log_{10}k = \log_{10}A - \frac{E_a}{2.303R}\cdot\frac{1}{T}

Two Temperatures: log(k₂/k₁) = (Ea/2.303R)·(T₂ − T₁)/(T₁T₂)

Two-temperature Arrhenius

log⁡10k2k1=Ea2.303R(T2−T1T1T2)\log_{10}\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{T_2 - T_1}{T_1 T_2}\right)

Collision Theory: Only Effective Collisions Count

Effective collisions

rate=Z⋅e−Ea/RT⋅P(Z collision frequency, P orientation factor)\text{rate} = Z \cdot e^{-E_a/RT} \cdot P \qquad (Z \text{ collision frequency, } P \text{ orientation factor})

Common traps

Dropping the 2.303 with a base-10 plot

The slope is −EaR-\dfrac{E_a}{R} only for ln⁡k\ln k. With log⁡10k\log_{10}k it is −Ea2.303R-\dfrac{E_a}{2.303R}; option (B) inverts it and drops the sign.

Dividing the half-life by the ratio at the LOWER temperature

Cooling from 400400 K to 300300 K slows the reaction, so the half-life gets LONGER: 900×12.25900 \times 12.25, not 900/12.25900 / 12.25. Decide the direction before touching the ratio.

'Collisions are fewer than the observed rate'

It is the other way round: collisions vastly outnumber reactions, which is the whole point of the effective-collision idea. Option (C) states the inversion.

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