NEET 2027 · Chemistry · Chemical Kinetics · Topic 02 of 15
Arrhenius Equation & Activation Energy
Tier 1 · highest priority. Why heating speeds reactions up, what a barrier really is, and the one formula that carries most of the marks — with four animations and 60 worked questions.
Almost every reaction speeds up when you heat it. That much you already know from
cooking. What the Arrhenius equation adds is the reason, and the reason is surprisingly
sharp: molecules must clear an energy barrier before they can react, and heating does not push them
harder — it simply increases the number that are already fast enough.
Story track
Picture a hill between two valleys. Reactants sit in the left valley. Products sit in the
right valley. To get across, a molecule must climb over the hill. The height of that hill is the
activation energy, Ea.
Now here is the crucial bit. The molecules are not all the same. Some are sluggish, some are
zipping about wildly. At any instant there is a whole spread of energies. Only the molecules that
happen to have more energy than the hill height can make it over. Everyone else bounces back.
So what does heating actually do? It does not give every molecule a shove. It widens the
spread — more molecules end up in the fast tail. The hill is exactly as tall as before. There are
simply more climbers capable of clearing it.
This explains something that would otherwise be baffling: raising the temperature by only ten
degrees roughly doubles the rate. Ten degrees out of three hundred is a change of about
three percent. Why would a three percent change double anything? Because you are not changing the
average molecule at all — you are changing the size of a small, sharply-sensitive tail. Tiny shifts
in a distribution produce huge changes at its edge.
The single sentence to carry into the exam. Temperature does
not lower the barrier; it recruits more molecules capable of crossing it. A catalyst, by
contrast, does lower the barrier. Confusing these two is the most common conceptual error in this
unit.
Building the equation piece by piece
Maths track
k = A · e^(−Ea/RT)
Three parts, each with a job:
e^(−Ea/RT) — the fraction of molecules with energy at least Ea. It is a number
between 0 and 1, and it is usually tiny (10⁻¹⁰ or smaller is normal). This is the factor
temperature acts on.
A — the Arrhenius factor, also called the frequency factor or pre-exponential factor. It
is related to how often molecules collide at all, and it is a constant for a given reaction. It has
the same units as k.
R and T — gas constant 8.314 J K⁻¹ mol⁻¹, and temperature in kelvin, always.
Read the equation as a sentence: rate constant = how often they meet × what fraction of those
meetings are energetic enough.
Turning it into a straight line
Maths track
Exponentials are hard to read off a graph, so take natural logs of both sides:
ln k = ln A − Ea/RT
Rearranged to match y = mx + c:
ln k = (−Ea/R)·(1/T) + ln A
So plotting ln k against 1/T gives a straight, falling line:
slope = −Ea/R, so Ea = −slope × R (positive, since the slope is negative)
intercept on the ln k axis = ln A
In base-10 form, which some papers use: log k = log A − Ea/(2.303RT), so a
log k vs 1/T plot has slope −Ea/2.303R.
The two-temperature formula — your main working tool
Maths track
Write the log form at two temperatures and subtract. The ln A cancels, which is the whole point:
Because A vanishes, you can find Ea from two rate constants alone, without ever knowing
A. That is why nearly every NEET numerical in this unit uses this form.
Sign discipline. The bracket is [1/T₁ − 1/T₂]
with the smaller temperature first. Write it that way every single time and the answer comes
out positive. If your Ea is negative you have flipped the bracket — activation energy is
never negative.
Why 10 °C doubles the rate — worked properly
NCERT states this as a fact. Here is where it comes from. Put k₂/k₁ = 2, T₁ = 298 K, T₂ = 308 K:
log 2 = (Ea/2.303 × 8.314) × (10 / (298 × 308))
Solving gives Ea ≈ 52.9 kJ mol⁻¹. Since a very large number of ordinary reactions have
activation energies near 50 kJ mol⁻¹, the doubling rule works for most of them — but it is a rough
rule, not a law. A reaction with Ea = 100 kJ mol⁻¹ would nearly quadruple over the same
ten degrees.
Reading the energy profile diagram
Story track
The reaction coordinate diagram is a picture of the hill. Four things live on it, and NEET asks
you to tell them apart:
Activation energy (forward), Ea,f — height from the reactant valley to the peak.
Activation energy (backward), Ea,b — height from the product valley to the same peak.
Activated complex — what sits at the peak. Unstable, exists for an instant, cannot
be bottled. It is not an intermediate: an intermediate would sit in a small dip, not on a summit.
ΔH — the difference between the two valleys, and nothing to do with the peak height.
ΔH = Ea,forward − Ea,backward
If the product valley is lower, ΔH is negative — exothermic. If higher, endothermic. The hill can
be enormous in a reaction that is strongly exothermic; height of hill and depth of valley are
independent facts.
Threshold energy is not activation energy. NCERT's footnote is
easy to skim past and is directly examinable:
Threshold energy = activation energy + energy already possessed by the reacting molecules.
Activation energy is the extra the molecules must acquire; threshold is the total they must
end up with.
What a catalyst does to this picture
A catalyst offers a completely different route over the ridge — a lower pass through the hills. It
does not push molecules and does not change how much energy they have.
Quantity
Catalyst effect
Temperature increase effect
Activation energy Ea
Lowered
Unchanged
Rate constant k
Increased
Increased
Fraction of molecules above Ea
Increased (because the bar drops)
Increased (because the spread widens)
ΔH of the reaction
Unchanged
Essentially unchanged
Equilibrium constant K
Unchanged
Changed
Arrhenius factor A
Unchanged
Unchanged
This table is itself a question. The row that catches people
is the last-but-one: a catalyst leaves K untouched, but heating does shift K. Both speed up
the reaction; only one moves the equilibrium.
Beyond the textbook
Fraction of effective molecules.f = e^(−Ea/RT) gives, directly, the fraction of molecules with energy ≥ Ea.
NCERT's intext 3.9 works this for HI decomposition and gets about 1.47 × 10⁻¹⁹ — an astonishingly
small number, which is exactly why reactions with high barriers appear not to happen at all.
Reading Ea off a given equation. If a question hands
you k = A·e^(−cK/T), compare exponents: Ea/R = c, so
Ea = cR. If it hands you log k = a − b/T, then
Ea = 2.303 R b. Two NCERT exercises (3.26, 3.27) are exactly this,
and they take fifteen seconds once you see the pattern.
Temperature coefficient. Defined as
μ = k(T+10)/k(T), typically between 2 and 3. NCERT states the doubling fact but
never names the quantity.
See it move — four animations
The second animation is the one to spend time on. If the Maxwell–Boltzmann tail makes sense to you, the whole unit follows from it.
ANIM 1
The energy hill — Ea forward, Ea backward and ΔH on one picture
Slide the barrier height and the position of the product valley independently, and watch that they really are independent: you can have a huge hill above a deep valley, or a small hill above a shallow one. Tick the catalyst box and the dashed green route appears — a lower pass over the same ridge. Notice carefully that the two valley levels do not move, which is the picture-proof that a catalyst cannot change ΔH.
ANIM 2
Maxwell–Boltzmann: why ten degrees doubles the rate
The shaded tails are the molecules capable of reacting. Raise the temperature and the peak slides right and flattens — but the total area stays the same, because no molecules are created or destroyed. What changes dramatically is the small shaded region past Ea. Drag Ea to the right and watch the effect become more violent: the higher the barrier, the more temperature-sensitive the reaction. That is the physical meaning of a large activation energy.
ANIM 3
The Arrhenius plot — where slope and intercept come from
Black dots are experimental rate constants at 500, 400, 333, 286 and 250 K. Slide Ea and the line tilts; slide ln A and the line shifts bodily up or down without ever changing its tilt. That is the whole reason this plot is useful — slope carries Ea and only Ea, intercept carries A and only A, so one graph separates two unknowns cleanly.
ANIM 4
Molecules attacking the barrier — a live effectiveness counter
Press Run to begin.
Each dot is a molecule with a fluctuating energy. Red means it currently exceeds Ea, green means it does not. Molecules that arrive at the hill with too little energy bounce straight back and try again later. Turn the temperature up and watch the success percentage climb — this is the Arrhenius exponential happening in front of you, not as a formula but as a counting exercise.
Formula sheet
The starred row is the one you will use most. Learn 2.303 × R = 19.15 as a single number; roughly half the numericals in this unit reduce to a division by 19.15.
Quantity / situation
Formula
When you use it
Arrhenius equation
k = A·e^(−Ea/RT)
The parent equation; use when A and Ea are both known
Natural-log form
ln k = ln A − Ea/RT
Straight-line form; basis of the Arrhenius plot
Base-10 form
log k = log A − Ea/(2.303RT)
When the question is set up in log₁₀
Two-temperature form ★
log(k₂/k₁) = (Ea/2.303R)·[(T₂−T₁)/(T₁T₂)]
The main NEET working tool; A cancels out
Equivalent bracket form
log(k₂/k₁) = (Ea/2.303R)·[1/T₁ − 1/T₂]
Safer for sign discipline — smaller T first
Arrhenius plot
ln k vs 1/T : slope = −Ea/R, intercept = ln A
Graph questions
Base-10 plot
log k vs 1/T : slope = −Ea/2.303R
When the axis is log₁₀ rather than ln
Ea from slope
Ea = −slope × R (= −slope × 2.303R for log₁₀)
Reading a graph
Fraction above Ea
f = e^(−Ea/RT)
'What fraction of molecules can react'
Ea from k = A·e^(−cK/T)
Ea = c × R
Compare exponents directly (NCERT Q 3.26)
Ea from log k = a − b/T
Ea = 2.303 × R × b ; log A = a
Compare coefficients (NCERT Q 3.27)
Activation energies and ΔH
ΔH = Ea(forward) − Ea(backward)
Energy profile questions
Threshold energy
E_threshold = Ea + energy already possessed
Definitional; frequently confused with Ea
Rate doubles per 10 K
corresponds to Ea ≈ 52.9 kJ mol⁻¹ near room temperature
Learn 19.15 — it saves a multiplication every time
60 NEET-type questions with worked solutions
Four graph questions and three assertion–reason questions are included, marked by their coloured left borders. Questions tagged PYQ pattern follow forms that have appeared in NEET/AIPMT papers or come directly from NCERT exercises — exact year attributions are deliberately omitted rather than guessed. In this unit almost every lost mark comes from one of four mechanical slips: using R where 2.303R was needed, forgetting to convert °C to kelvin, flipping the temperature bracket, or leaving Ea in joules when the options are in kilojoules. Check your errors against that list before concluding you did not understand the concept.
Q01PYQ pattern
The rate constants of a reaction at 500 K and 700 K are 0.02 s⁻¹ and 0.07 s⁻¹ respectively. The activation energy is nearest to:
(a) 18.2 kJ mol⁻¹
(b) 34.7 kJ mol⁻¹
(c) 8.2 kJ mol⁻¹
(d) 182 kJ mol⁻¹
Given
k₁ = 0.02 s⁻¹ at T₁ = 500 K; k₂ = 0.07 s⁻¹ at T₂ = 700 K
Asked
Ea
Concept
Two rate constants at two temperatures — the A term cancels in the difference.
Formula
log(k₂/k₁) = (Ea/2.303R)·[(T₂−T₁)/(T₁T₂)]
Baby steps
k₂/k₁ = 0.07/0.02 = 3.5, and log 3.5 = 0.544.
Temperature bracket: (700−500)/(700×500) = 200/350000 = 5.714 × 10⁻⁴.
2.303R = 19.15.
0.544 = (Ea/19.15) × 5.714 × 10⁻⁴, so Ea = 0.544 × 19.15 / 5.714 × 10⁻⁴.
Ea = 18230 J mol⁻¹ ≈ 18.2 kJ mol⁻¹.
Answer · (a) about 18.2 kJ mol⁻¹
Shortcut · Compute 2.303R = 19.15 once and reuse it; it turns a three-number multiplication into one.
Q02PYQ pattern
Using the previous data (k = 0.02 s⁻¹ at 500 K, Ea = 18230 J mol⁻¹), the value of the Arrhenius factor A is:
(a) 1.61 s⁻¹
(b) 0.012 s⁻¹
(c) 16.1 s⁻¹
(d) 0.161 s⁻¹
Given
k = 0.02 s⁻¹, T = 500 K, Ea = 18230 J mol⁻¹
Asked
A
Concept
Once Ea is known, substitute back into the parent equation.
Shortcut · A is always larger than k, because the exponential factor is always less than 1. If your A comes out smaller than k, you have flipped the division.
Q03PYQ pattern
The first order rate constant for the decomposition of ethyl iodide at 600 K is 1.60 × 10⁻⁵ s⁻¹. If Ea = 209 kJ mol⁻¹, the rate constant at 700 K is:
(a) 6.36 × 10⁻³ s⁻¹
(b) 1.60 × 10⁻³ s⁻¹
(c) 6.36 × 10⁻⁵ s⁻¹
(d) 3.20 × 10⁻⁵ s⁻¹
Given
k₁ = 1.60 × 10⁻⁵ s⁻¹ at 600 K, Ea = 209 kJ mol⁻¹, T₂ = 700 K
Asked
k₂
Concept
The same two-temperature relation, now solved for the unknown rate constant.
Shortcut · Memorise this result. 'Doubling per 10 K near room temperature' ↔ Ea ≈ 53 kJ mol⁻¹ appears repeatedly.
Q05PYQ pattern
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. The activation energy, assuming it does not change with temperature, is about:
(a) 52.9 kJ mol⁻¹
(b) 26.4 kJ mol⁻¹
(c) 105.7 kJ mol⁻¹
(d) 13.2 kJ mol⁻¹
Given
k₂/k₁ = 4, T₁ = 293 K, T₂ = 313 K
Asked
Ea
Concept
Same relation; note that doubling the temperature gap and squaring the rate ratio roughly cancel.
Shortcut · Quadrupling over 20 K is the same as doubling over each 10 K, so the answer must match the previous question. Spotting that saves the whole calculation.
Q06PYQ pattern
The decomposition of a hydrocarbon follows k = (4.5 × 10¹¹ s⁻¹)·e^(−28000K/T). Its activation energy is:
(a) 232.8 kJ mol⁻¹
(b) 28.0 kJ mol⁻¹
(c) 116.4 kJ mol⁻¹
(d) 536.0 kJ mol⁻¹
Given
k = 4.5 × 10¹¹ e^(−28000K/T)
Asked
Ea
Concept
Compare the given exponent with the standard Arrhenius exponent, term by term.
Formula
k = A e^(−Ea/RT) — so Ea/R = 28000 K
Baby steps
The standard exponent is −Ea/RT; the given exponent is −28000/T.
Matching them: Ea/R = 28000 K.
Ea = 28000 × 8.314 = 232792 J mol⁻¹.
Ea ≈ 232.8 kJ mol⁻¹.
Answer · (a) 232.8 kJ mol⁻¹
Shortcut · Whenever the exponent is written as a plain number over T, multiply that number by R. Ten seconds, one mark.
Q07PYQ pattern
For the first order decomposition of H₂O₂, log k = 14.34 − 1.25 × 10⁴ K/T. The activation energy is:
(a) 239.3 kJ mol⁻¹
(b) 103.9 kJ mol⁻¹
(c) 125.0 kJ mol⁻¹
(d) 14.34 kJ mol⁻¹
Given
log k = 14.34 − 1.25 × 10⁴/T
Asked
Ea
Concept
Compare with the base-10 Arrhenius form; note the extra 2.303 this time.
Formula
log k = log A − Ea/(2.303RT)
Baby steps
Matching the 1/T coefficients: Ea/(2.303R) = 1.25 × 10⁴ K.
2.303R = 19.15.
Ea = 1.25 × 10⁴ × 19.15 = 239375 J mol⁻¹.
Ea ≈ 239.3 kJ mol⁻¹. (Incidentally log A = 14.34.)
Answer · (a) 239.3 kJ mol⁻¹
Shortcut · log form → multiply by 2.303R = 19.15. ln form → multiply by R = 8.314. Getting these two the wrong way round is the classic trap here.
Q08PYQ pattern
Continuing the previous data (log k = 14.34 − 1.25 × 10⁴ K/T), at what temperature will the half-period be 256 minutes?
(a) 669 K
(b) 512 K
(c) 400 K
(d) 750 K
Given
t½ = 256 min, log k = 14.34 − 1.25 × 10⁴/T, first order
Asked
T
Concept
Convert half-life to k, then invert the given log relation for T.
Formula
k = 0.693/t½ ; T = 1.25 × 10⁴ / (14.34 − log k)
Baby steps
t½ = 256 min = 256 × 60 = 15360 s.
k = 0.693/15360 = 4.512 × 10⁻⁵ s⁻¹.
log k = log(4.512 × 10⁻⁵) = −4.346.
14.34 − (−4.346) = 18.686 = 1.25 × 10⁴/T.
T = 1.25 × 10⁴/18.686 = 669 K.
Answer · (a) about 669 K
Shortcut · Convert minutes to seconds first — the equation's constants assume k in s⁻¹.
Q09PYQ pattern
The decomposition of A has k = 4.5 × 10³ s⁻¹ at 10 °C and Ea = 60 kJ mol⁻¹. At what temperature would k be 1.5 × 10⁴ s⁻¹?
(a) 297 K
(b) 283 K
(c) 320 K
(d) 273 K
Given
k₁ = 4.5 × 10³ s⁻¹ at 283 K, Ea = 60 kJ mol⁻¹, k₂ = 1.5 × 10⁴ s⁻¹
Asked
T₂
Concept
Solve the two-temperature relation for the unknown temperature.
Shortcut · k increased, so T₂ must exceed 283 K. Options (b) and (d) are eliminated before any arithmetic.
Q10PYQ pattern
The activation energy for 2HI(g) → H₂(g) + I₂(g) is 209.5 kJ mol⁻¹ at 581 K. The fraction of molecules with energy equal to or greater than the activation energy is about:
(a) 1.47 × 10⁻¹⁹
(b) 1.47 × 10⁻⁹
(c) 2.09 × 10⁻⁵
(d) 0.209
Given
Ea = 209.5 kJ mol⁻¹, T = 581 K
Asked
fraction with E ≥ Ea
Concept
The exponential factor in the Arrhenius equation IS that fraction.
Shortcut · Typical A values for gas-phase reactions sit around 10¹²–10¹⁴ s⁻¹. An answer far outside that range signals an arithmetic slip.
Q12PYQ pattern
The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. The activation energy is nearest to:
(a) 76.6 kJ mol⁻¹
(b) 52.9 kJ mol⁻¹
(c) 38.3 kJ mol⁻¹
(d) 152 kJ mol⁻¹
Given
t₁₀(298 K) = t₂₅(308 K), first order
Asked
Ea
Concept
Equal times mean the rate constants are in the ratio of the log terms.
Formula
t = (2.303/k)·log(100/(100−x)) ; equal t ⇒ k₂/k₁ = log(100/75)/log(100/90)
Shortcut · Whenever two different completion percentages take the same time, the rate constants are in the inverse ratio of the times — set that up first.
Q13PYQ pattern
Continuing the previous question (Ea ≈ 76.6 kJ mol⁻¹, A = 4 × 10¹⁰ s⁻¹), the rate constant at 318 K is nearest to:
(a) 1.0 × 10⁻² s⁻¹
(b) 4.0 × 10¹⁰ s⁻¹
(c) 1.0 × 10⁻⁵ s⁻¹
(d) 2.5 × 10⁻³ s⁻¹
Given
A = 4 × 10¹⁰ s⁻¹, Ea = 76.6 kJ mol⁻¹, T = 318 K
Asked
k
Concept
Direct substitution into the parent equation, done in logs.
The graph shows ln k plotted against 1/T for a reaction. The activation energy is obtained from the slope as:
(a) Ea = −slope × R
(b) Ea = slope × R
(c) Ea = −slope × 2.303R
(d) Ea = slope/R
Given
Straight line of ln k against 1/T with negative slope
Asked
Expression for Ea
Concept
Match the plotted form to y = mx + c and read the slope.
Formula
ln k = (−Ea/R)(1/T) + ln A
Baby steps
Comparing with y = mx + c, the slope m = −Ea/R.
Rearranging, Ea = −m × R.
Since the slope is negative, the minus sign makes Ea positive, as it must be.
The 2.303 factor appears only if the vertical axis is log₁₀ k rather than ln k — here it is ln k, so option (c) is wrong.
Answer · (a) Ea = −slope × R
Shortcut · ln on the axis → use R. log₁₀ on the axis → use 2.303R. Check the axis label before anything else.
Q15Graph
Two reactions P and Q are plotted as ln k against 1/T. Line P is steeper than line Q. Which is correct?
(a) P has the higher Ea and is more sensitive to temperature
(b) P has the lower Ea and is more sensitive to temperature
(c) Q has the higher Ea and is more sensitive to temperature
(d) Both have the same Ea but different A
Given
|slope of P| > |slope of Q| on an ln k vs 1/T plot
Asked
Comparison of Ea and temperature sensitivity
Concept
Slope magnitude is Ea/R, so steeper means a larger activation energy.
Formula
slope = −Ea/R
Baby steps
The magnitude of the slope equals Ea/R, so steeper means larger Ea.
P is steeper, so Ea(P) > Ea(Q).
A larger Ea means the exponential e^(−Ea/RT) responds more sharply to a change in T.
So P is also the more temperature-sensitive reaction.
Answer · (a) P has the higher Ea and is more sensitive to temperature
Shortcut · High barrier = steep line = big response to heating. All three go together.
Q16Graph
In the potential-energy versus reaction-coordinate diagram shown, which labelled quantity represents ΔH of the reaction?
(a) Z
(b) X
(c) Y
(d) X + Y
Given
Energy profile with X = reactants to peak, Y = products to peak, Z = between the two levels
Asked
Which represents ΔH
Concept
ΔH is the vertical gap between reactant and product levels only — the peak is irrelevant to it.
Formula
ΔH = Ea(forward) − Ea(backward) = X − Y
Baby steps
X runs from the reactant level up to the peak, so X is the forward activation energy.
Y runs from the product level up to the same peak, so Y is the backward activation energy.
Z spans reactant level to product level, which is exactly the definition of ΔH.
Consistency check: X − Y also equals that gap, and here the products lie above the reactants, so the reaction is endothermic.
Answer · (a) Z
Shortcut · ΔH never touches the peak. If your chosen arrow ends at the summit, it is an activation energy, not ΔH.
Q17Graph
The Maxwell–Boltzmann distribution is drawn at temperatures T and T + 10. Which statement about the curves is correct?
(a) The total area under both curves is the same, but the area beyond Ea is larger at T + 10
(b) The total area is larger at T + 10
(c) The peak moves to lower energy at T + 10
(d) The area beyond Ea is the same for both curves
Given
Distribution curves at two temperatures with Ea marked
Asked
The correct statement
Concept
The area under a probability distribution is fixed at 1; heating only redistributes it.
Formula
Total area = 1 at all temperatures ; fraction beyond Ea = e^(−Ea/RT)
Baby steps
Total probability must equal one at every temperature, so the total area cannot change. That rules out (b).
Raising temperature shifts the peak to higher energy and flattens it, so (c) is wrong.
The fraction beyond Ea is e^(−Ea/RT), which increases as T increases, so (d) is wrong.
Only (a) states both facts correctly, and this pairing is exactly why heating speeds reactions up.
Answer · (a) Total area same; area beyond Ea larger at the higher temperature
Shortcut · 'Total area increases' is always a wrong option on this diagram. Eliminate it on sight.
Q18Assertion–Reason
Assertion (A): A catalyst increases the rate of a reaction but does not change the enthalpy change of the reaction. Reason (R): A catalyst provides an alternative pathway with a lower activation energy, leaving the energies of the reactants and products unaltered.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Given
Statements about catalysts
Asked
Truth values and explanation
Concept
A catalyst acts on the barrier between the valleys, not on the valleys themselves.
Formula
ΔH = Ea(forward) − Ea(backward) ; both change together under a catalyst
Baby steps
Check A: a catalyst lowers Ea, so k and the rate increase. ΔH depends only on reactant and product energies, which the catalyst does not touch. A is true.
Check R: this is precisely NCERT's description of catalytic action. R is true.
Does R explain A? Yes — because only the barrier changes and the two energy levels stay put, ΔH is necessarily unchanged.
Note the catalyst lowers Ea in both directions by the same amount, which is why their difference, ΔH, survives untouched.
Answer · (a) Both A and R are true and R is the correct explanation of A
Q19Assertion–Reason
Assertion (A): Increasing the temperature by 10 K roughly doubles the rate of many reactions, even though the average kinetic energy rises by only a few percent. Reason (R): The rate depends on the number of molecules in the high-energy tail of the distribution, and this tail is extremely sensitive to temperature.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Given
Statements about temperature sensitivity
Asked
Truth values and explanation
Concept
An exponential factor magnifies small changes in the exponent enormously.
Formula
k = A e^(−Ea/RT) — a small change in T produces a large change in the exponential
Baby steps
Check A: raising T from 298 K to 308 K is about a 3% change in absolute temperature, yet the rate roughly doubles. A is true.
Check R: the reacting molecules come from the tail beyond Ea, and that tail grows sharply with temperature. R is true.
Does R explain A? Yes — the disproportion between a small temperature change and a large rate change exists precisely because it is the tail, not the average, that matters.
The same idea shows up on the Maxwell–Boltzmann diagram as the shaded region past Ea roughly doubling.
Answer · (a) Both A and R are true and R is the correct explanation of A
Q20Assertion–Reason
Assertion (A): The activation energy of a reaction can never be negative. Reason (R): Activation energy is the energy that reactant molecules must acquire in addition to what they already possess, in order to form the activated complex.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Given
Statements about the sign of activation energy
Asked
Truth values and explanation
Concept
The activated complex always sits above the reactants, so the extra energy required is positive.
Formula
E_threshold = Ea + energy already possessed
Baby steps
Check A: the activated complex is an unstable species at the top of the barrier, always above the reactant level, so Ea > 0. A is true.
Check R: this is the textbook definition, including the distinction from threshold energy. R is true.
Does R explain A? Yes — if the extra energy needed to reach the summit were negative, the summit would lie below the starting point, which contradicts the meaning of a barrier.
In practice, a negative Ea from a calculation always signals a flipped temperature bracket, not real chemistry.
Answer · (a) Both A and R are true and R is the correct explanation of A
Q21
Which of the following statements about the Arrhenius factor A is correct?
(a) It has the same units as the rate constant k
(b) It is dimensionless for all reactions
(c) It always has units of s⁻¹
(d) Its units are always mol L⁻¹ s⁻¹
Given
Arrhenius equation k = A e^(−Ea/RT)
Asked
Units of A
Concept
The exponential factor is a pure number, so A must carry all of k's units.
Formula
k = A·e^(−Ea/RT) ; e^(−Ea/RT) is dimensionless
Baby steps
Ea/RT has units of (J mol⁻¹)/(J K⁻¹ mol⁻¹ × K), which all cancel — so the exponent is dimensionless.
Anything raised to a dimensionless power is itself dimensionless, so e^(−Ea/RT) is a pure number.
Therefore A must carry exactly the units of k, whatever they are for that reaction's order.
So A is s⁻¹ for a first order reaction but mol L⁻¹ s⁻¹ for a zero order one — option (c) is too narrow.
Answer · (a) It has the same units as the rate constant k
Q22
For a reaction with activation energy zero, the rate constant:
(a) equals A and is independent of temperature
(b) is zero at all temperatures
(c) increases sharply with temperature
(d) equals A/RT
Given
Ea = 0
Asked
Behaviour of k
Concept
Setting the exponent to zero collapses the Arrhenius equation.
Formula
k = A e^(−Ea/RT)
Baby steps
Put Ea = 0: the exponent becomes zero.
e⁰ = 1, so k = A × 1 = A.
A is a constant for the reaction, so k does not vary with temperature.
Physically: with no barrier, every collision is effective, so heating cannot improve matters.
Answer · (a) equals A and is independent of temperature
Shortcut · Zero barrier means temperature-independent rate. This is the limiting case that checks your understanding of the equation.
Q23PYQ pattern
Which one of the following does a catalyst NOT change?
(a) The equilibrium constant of the reaction
(b) The activation energy
(c) The rate constant
(d) The rate of the backward reaction
Given
Effects of a catalyst
Asked
The unchanged quantity
Concept
A catalyst changes kinetics but never thermodynamics.
Formula
ΔG = −RT ln K — unaffected by a catalyst
Baby steps
A catalyst lowers Ea, so (b) changes.
Lower Ea means higher k, so (c) changes.
It accelerates forward and backward reactions equally, so (d) changes too.
Since ΔG is untouched and K depends only on ΔG, the equilibrium constant is unchanged. The catalyst brings equilibrium sooner, not somewhere different.
Answer · (a) The equilibrium constant of the reaction
Shortcut · Catalyst = kinetics only. Any thermodynamic quantity in the options (ΔG, ΔH, K) is the answer to a 'does not change' question.
Q24
The activation energy of a reaction is 100 kJ mol⁻¹ and A = 6 × 10¹⁴ s⁻¹. At what approximate temperature is the fraction of effective collisions 10⁻¹⁰?
(a) 523 K
(b) 300 K
(c) 1000 K
(d) 425 K
Given
Ea = 100 kJ mol⁻¹, fraction f = 10⁻¹⁰
Asked
T
Concept
Set the exponential factor equal to the given fraction and solve for T.
Formula
f = e^(−Ea/RT), so log f = −Ea/(2.303RT)
Baby steps
log f = log 10⁻¹⁰ = −10.
So −10 = −100000/(19.15 × T).
10 = 100000/(19.15 T), giving 19.15 T = 10000.
T = 10000/19.15 = 522 K ≈ 523 K.
Answer · (a) about 523 K
Q25
For a certain reaction Ea = 0 and A = 3.0 × 10⁵ s⁻¹. The rate constant at 300 K is:
(a) 3.0 × 10⁵ s⁻¹
(b) 1.0 × 10⁵ s⁻¹
(c) 0
(d) 9.0 × 10⁵ s⁻¹
Given
Ea = 0, A = 3.0 × 10⁵ s⁻¹, T = 300 K
Asked
k
Concept
Zero activation energy makes the exponential factor equal to one.
Formula
k = A e^(−Ea/RT) = A when Ea = 0
Baby steps
Exponent = −0/(RT) = 0.
e⁰ = 1.
k = A × 1 = 3.0 × 10⁵ s⁻¹, at 300 K or any other temperature.
Answer · (a) 3.0 × 10⁵ s⁻¹
Q26
If the activation energy of a reaction is doubled while the temperature is held constant, the rate constant:
(a) is reduced to k²/A
(b) is doubled
(c) is halved
(d) is unchanged
Given
Ea → 2Ea at constant T
Asked
New rate constant
Concept
Doubling the exponent squares the exponential factor.
Formula
k = A e^(−Ea/RT) ; k' = A e^(−2Ea/RT) = A(e^(−Ea/RT))²
Baby steps
Original: k = A·x where x = e^(−Ea/RT), so x = k/A.
New: k' = A·x² = A × (k/A)².
k' = k²/A.
Since x is much smaller than 1, squaring it makes the rate constant dramatically smaller — doubling the barrier is catastrophic for the rate, not merely halving.
Answer · (a) is reduced to k²/A
Shortcut · The exponential means changes in Ea act multiplicatively, never additively. 'Doubled Ea → halved k' is always wrong.
Q27
Two reactions have the same A but activation energies of 40 and 80 kJ mol⁻¹. At the same temperature:
(a) The one with 40 kJ mol⁻¹ has the larger rate constant
(b) The one with 80 kJ mol⁻¹ has the larger rate constant
(c) Both have equal rate constants
(d) The comparison depends on the order
Given
Same A, Ea = 40 and 80 kJ mol⁻¹, same T
Asked
Which has the larger k
Concept
With A fixed, k depends only on the exponential, which shrinks as Ea grows.
Formula
k = A e^(−Ea/RT)
Baby steps
A larger Ea makes the exponent −Ea/RT more negative.
A more negative exponent means a smaller exponential factor.
With A the same for both, the smaller barrier gives the larger k.
So the 40 kJ mol⁻¹ reaction is faster.
Answer · (a) The one with 40 kJ mol⁻¹ has the larger rate constant
Q28
The activation energy of the forward reaction is 60 kJ mol⁻¹ and ΔH is −20 kJ mol⁻¹. The activation energy of the backward reaction is:
(a) 80 kJ mol⁻¹
(b) 40 kJ mol⁻¹
(c) 20 kJ mol⁻¹
(d) 60 kJ mol⁻¹
Given
Ea(forward) = 60 kJ mol⁻¹, ΔH = −20 kJ mol⁻¹
Asked
Ea(backward)
Concept
ΔH is the difference between the two activation energies.
Formula
ΔH = Ea(f) − Ea(b), so Ea(b) = Ea(f) − ΔH
Baby steps
Ea(b) = 60 − (−20).
= 60 + 20 = 80 kJ mol⁻¹.
Sanity check: the reaction is exothermic, so the products lie lower, meaning the climb back up must be steeper. 80 > 60 is consistent.
Answer · (a) 80 kJ mol⁻¹
Shortcut · Exothermic → backward barrier is the larger. Endothermic → forward barrier is the larger. Use this to check your sign before choosing.
Q29
For an endothermic reaction with ΔH = +30 kJ mol⁻¹, the minimum possible activation energy of the forward reaction is:
(a) 30 kJ mol⁻¹
(b) 0
(c) 15 kJ mol⁻¹
(d) 60 kJ mol⁻¹
Given
ΔH = +30 kJ mol⁻¹, endothermic
Asked
Minimum Ea(forward)
Concept
The peak must lie at or above the higher of the two valleys.
Formula
Ea(f) = ΔH + Ea(b), with Ea(b) ≥ 0
Baby steps
Ea(f) = ΔH + Ea(b) = 30 + Ea(b).
The smallest possible backward barrier is zero.
So the smallest possible forward barrier is 30 kJ mol⁻¹.
Physically: you must at minimum supply the energy difference between the valleys; you can never climb less than that.
Answer · (a) 30 kJ mol⁻¹
Shortcut · For any endothermic reaction, Ea(forward) ≥ ΔH always. This bound alone answers several questions.
Q30PYQ pattern
Threshold energy of a reaction is:
(a) activation energy plus the energy already possessed by the reacting molecules
(b) the same as the activation energy
(c) activation energy minus the energy of the reactants
(d) always equal to ΔH
Given
Definition question
Asked
Meaning of threshold energy
Concept
NCERT's footnote definition — activation energy is the extra needed, threshold is the total required.
Formula
Threshold energy = Ea + energy possessed by the reacting species
Baby steps
Reactant molecules already carry some energy at the given temperature.
The activated complex requires a certain total energy to form — this total is the threshold energy.
Activation energy is the shortfall, the extra amount that must still be acquired.
So threshold = activation energy + energy already possessed.
Answer · (a) activation energy plus the energy already possessed by the reacting molecules
Q31
A reaction has k₁ at T₁ and k₂ at T₂ with T₂ > T₁. Which is always true?
(a) k₂ > k₁
(b) k₂ < k₁
(c) k₂ = k₁
(d) The comparison depends on the order of the reaction
Given
T₂ > T₁, positive Ea
Asked
Comparison of rate constants
Concept
For any positive activation energy, k rises monotonically with temperature.
Formula
k = A e^(−Ea/RT)
Baby steps
As T increases, Ea/RT decreases in magnitude.
So −Ea/RT becomes less negative, and the exponential grows.
With A fixed, k must therefore increase.
Order of the reaction plays no part in this — it only affects the units of k, not its temperature dependence.
Answer · (a) k₂ > k₁
Q32
The plot of log k against 1/T for a reaction has a slope of −5000 K. The activation energy is:
(a) 95.8 kJ mol⁻¹
(b) 41.6 kJ mol⁻¹
(c) 5.0 kJ mol⁻¹
(d) 239 kJ mol⁻¹
Given
Slope of log k vs 1/T = −5000 K
Asked
Ea
Concept
Note the axis is log₁₀, so the 2.303 must be included.
Formula
slope = −Ea/(2.303R), so Ea = −slope × 2.303R
Baby steps
Ea = 5000 × 2.303 × 8.314.
2.303 × 8.314 = 19.15.
Ea = 5000 × 19.15 = 95750 J mol⁻¹.
Ea ≈ 95.8 kJ mol⁻¹.
Answer · (a) about 95.8 kJ mol⁻¹
Shortcut · Option (b) is what you get by using R instead of 2.303R — the trap for anyone who skipped the axis label.
Q33
The rate constant of a reaction increases by a factor of 3 when the temperature rises from 300 K to 310 K. The activation energy is nearest to:
Shortcut · Learn log 2 = 0.301, log 3 = 0.477, log 4 = 0.602, log 5 = 0.699. These four cover most exam ratios.
Q42
Which statement about the activated complex is INCORRECT?
(a) It can be isolated and stored under suitable conditions
(b) It exists at the maximum of the potential energy curve
(c) It has a very short lifetime
(d) The energy required to form it from reactants is the activation energy
Given
Statements about the activated complex
Asked
The incorrect statement
Concept
An activated complex sits at an energy maximum, which makes it inherently unstable.
Formula
—
Baby steps
Statements (b), (c) and (d) are all standard textbook descriptions and are correct.
A species at an energy maximum has no stability — any displacement sends it downhill, either forward to products or back to reactants.
So it cannot be isolated or stored. Statement (a) is incorrect.
Contrast this with an intermediate, which sits in a local energy minimum and can sometimes be detected or even isolated.
Answer · (a) It can be isolated and stored under suitable conditions
Shortcut · Activated complex = peak = never isolable. Intermediate = dip = sometimes isolable.
Q43
If the temperature of a reaction is raised from 27 °C to 37 °C and the rate doubles, then at 47 °C the rate relative to that at 27 °C will be approximately:
(a) 4 times
(b) 3 times
(c) 2 times
(d) 8 times
Given
Rate doubles per 10 °C rise, from 300 K upwards
Asked
Relative rate after 20 °C rise
Concept
Each successive 10-degree rise multiplies the rate again — the effect compounds.
Formula
k(T + 20)/k(T) ≈ 2 × 2
Baby steps
27 °C → 37 °C is one 10-degree step, giving a factor of 2.
37 °C → 47 °C is a second 10-degree step, giving another factor of 2.
Total factor = 2 × 2 = 4.
Note this is approximate, since the doubling rule itself is only a rule of thumb.
Answer · (a) about 4 times
Shortcut · Temperature effects multiply, they do not add. Two steps give 2², three steps give 2³.
Q44
For a reaction, ln A = 20 and Ea = 100 kJ mol⁻¹. The value of ln k at 400 K is:
(a) −10.1
(b) 20.0
(c) 10.1
(d) −30.1
Given
ln A = 20, Ea = 100 kJ mol⁻¹, T = 400 K
Asked
ln k
Concept
Direct substitution into the natural-log form; note R is used, not 2.303R.
Formula
ln k = ln A − Ea/RT
Baby steps
RT = 8.314 × 400 = 3325.6 J mol⁻¹.
Ea/RT = 100000/3325.6 = 30.07.
ln k = 20 − 30.07 = −10.07 ≈ −10.1.
Answer · (a) about −10.1
Shortcut · Negative ln k simply means k is less than 1 — perfectly normal, not an error.
Q45
The rate constant doubles when the temperature is raised from T to T + 10. For the same reaction, raising the temperature from T + 10 to T + 20 will:
(a) roughly double the rate constant again
(b) quadruple the rate constant
(c) leave it unchanged
(d) halve it
Given
Rate constant doubles over the first 10 K step
Asked
Effect of the next 10 K step
Concept
Each further 10 K step gives roughly another doubling, though the factor drifts slightly downward at higher temperature.
Formula
log(k₂/k₁) = (Ea/2.303R)·[(T₂−T₁)/(T₁T₂)]
Baby steps
The bracket depends on T₁T₂, which grows slowly as temperatures rise.
So the second step gives a slightly smaller factor than the first, but only slightly.
To the accuracy of the rule of thumb, the rate constant doubles again.
This is why the effect compounds as roughly 2ⁿ over n successive steps.
Answer · (a) roughly double the rate constant again
Q46
A reaction is found to have an activation energy of 0 kJ mol⁻¹ in the forward direction. Its ΔH must be:
(a) negative or zero
(b) positive
(c) exactly zero
(d) impossible to determine
Given
Ea(forward) = 0
Asked
Sign of ΔH
Concept
With no forward barrier, the products cannot lie above the reactants.