Mixed, Reasoning & Examiner-Favourite Long Answers

CHEExam Practice & Reasoning

This group blends conceptual 'account for' reasoning, multi-step numericals, and theory comparisons that examiners reuse year after year. The marks lie in stating the precise reason (not just the result) and in showing each calculation step clearly.

First-order rate constant and half-life

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

core relations reused across reasoning problems

Activation energy (two temperatures)

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

T

in kelvin,

R=8.314\,J\,mol^{-1}K^{-1}

Units of k for order n

[k] = (mol\,L^{-1})^{1-n}\,s^{-1}

n

= overall order

Collision-theory rate

\text{rate} = P\,Z\,e^{-E_a/RT}

P

steric (orientation) factor,

Z

collision frequency

First-order $t_{1/2} = \dfrac{0.693}{k}$ is independent of $[A]_0$ because the rate is proportional to $[A]$ , so the fractional decay per unit time is fixed regardless of starting amount.
The unit of $k$ changes with order: for order $n$ , units are $(mol\,L^{-1})^{1-n}\,s^{-1}$ , so zero order is $mol\,L^{-1}s^{-1}$ , first order $s^{-1}$ , second order $L\,mol^{-1}s^{-1}$ .
Order-from-ratio test: for first order $\dfrac{t_{75\%}}{t_{50\%}} = 2$ ; for zero order the ratio is $\dfrac{3}{2}$ — comparing ratios diagnoses the order.
Collision theory: an effective collision needs both energy $\ge E_a$ and correct orientation; rate $= P\,Z\,e^{-E_a/RT}$ , where $P$ is the steric (orientation) factor and $Z$ the collision frequency.
Limitation of collision theory: it treats molecules as hard spheres and ignores their internal structure, so the orientation requirement has to be put in separately through the steric factor $P$ .
For two-temperature $E_a$ problems, apply $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303\,R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$ , then convert temperatures like $27^\circ C$ to $300\,K$ first.
A catalyst speeds both forward and reverse reactions equally and leaves $\Delta H$ (and the equilibrium position) unchanged — it only lowers $E_a$ .
Transition-state (activated-complex) theory improves on collision theory by treating reactants forming an activated complex in quasi-equilibrium, accounting for molecular structure and energy distribution.
Multi-step numericals: first extract $k$ from one piece of data (e.g. a given $t_{1/2}$ or a completion time), then feed that $k$ into the integrated law to find the unknown time or concentration.
When asked to 'account for' a fact, state the underlying reason explicitly — e.g. rate rises with $T$ because the Maxwell-Boltzmann tail beyond $E_a$ grows steeply, not merely because 'molecules move faster'.
Identify reaction order from how $t_{1/2}$ behaves: constant $t_{1/2}$ implies first order, $t_{1/2}\propto[A]_0$ implies zero order, $t_{1/2}\propto 1/[A]_0$ implies second order.
Show working with units at every step; ISC awards method marks even if the final arithmetic slips, so never jump straight to the answer.

Where the marks go

Stating only the result in an 'account for' question without the reason — examiners award marks for the explanation, not the bare fact.
Forgetting to convert $^\circ C$ to $K$ before using the two-temperature Arrhenius equation in multi-step problems.
Confusing collision theory (hard spheres + steric factor) with transition-state theory (activated complex) when listing limitations.
Assuming a catalyst shifts the equilibrium or alters $\Delta H$ — it lowers $E_a$ for both directions equally and leaves equilibrium untouched.
Misapplying $t_{1/2}=0.693/k$ to a zero-order step inside a mixed problem — check the order before choosing the half-life formula.

How the board asks it

Numericalmulti-step extraction of k then integrated first-order law
A first-order reaction is $20\%$ complete in $40$ minutes. Calculate the rate constant $k$ and the time required for the reaction to be $90\%$ complete. $(\log 8 = 0.903,\ \log 10 = 1)$
Give reasons'account for' facts about temperature and catalyst
Account for the following: (i) a small rise of about $10^\circ C$ in temperature can roughly double the rate of a reaction, and (ii) a catalyst increases the rate of a reaction without altering the value of $\Delta H$ for the reaction.
Numericaltwo-temperature arrhenius equation
The rate constant of a reaction at $27^\circ C$ is $1.0 \times 10^{-3}\ s^{-1}$ and at $47^\circ C$ is $2.0 \times 10^{-3}\ s^{-1}$ . Calculate the activation energy $E_a$ for the reaction. $(R = 8.314\ J\,K^{-1}\,mol^{-1},\ \log 2 = 0.301)$
Distinguishcollision theory vs transition-state theory
Distinguish between the collision theory and the transition-state (activated-complex) theory of reaction rates, and state one limitation of the collision theory.
Derive / provehalf-life behaviour and units of k to fix order
For a certain reaction the half-life is found to be independent of the initial concentration $[A]_0$ , and $k$ has the units $s^{-1}$ . Identify the order of the reaction, justifying your answer, and derive the expression $t_{1/2} = \dfrac{0.693}{k}$ .

PreviousCollision Theory, Activation Energy, Energy Profile

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.