Arrhenius Theory, Activation Energy, Collision and Related Theories Arrhenius theory, activation energy and its calculation, collision theory of bimolecular gaseous reactions (no derivation).: NEET MCQs with Answers

15 free questions on Arrhenius Theory, Activation Energy, Collision and Related Theories Arrhenius theory, activation energy and its calculation, collision theory of bimolecular gaseous reactions (no derivation). for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

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Question 1

Previous-year question
The addition of a catalyst during a chemical reaction alters which of the following quantities? [NEET 2016]
  1. Entropy
  2. Internal energy
  3. Enthalpy
  4. Activation energy
Show answer
Answer: (D) Activation energy
Explanation: A catalyst changes only the activation energy by opening an alternative path; thermodynamic state functions are unaffected.

Question 2

Previous-year question
The activation energy of a reaction can be determined from the slope of which of the following graphs? [AIPMT 2015]
  1. ln K/T versus T
  2. ln K versus 1/T
  3. T/ln K versus 1/T
  4. ln K versus T
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Answer: (B) ln K versus 1/T
Explanation: Only ln k against 1/T is linear, with slope –Eₐ/R.

Question 3

A reversible reaction is exothermic in the forward direction. When its temperature is raised, what happens to the forward rate constant k_f, the backward rate constant k_b and the equilibrium constant K?
  1. k_f and k_b both rise, k_f by the larger factor, so K increases
  2. k_f rises and k_b falls, so K increases
  3. k_f and k_b both rise, k_b by the larger factor, so K decreases
  4. k_f and k_b both rise by the same factor, so K is unchanged
Show answer
Answer: (C) k_f and k_b both rise, k_b by the larger factor, so K decreases
Explanation: Every rate constant rises with temperature, and the rise is greater for the larger activation energy. Since ΔH = Ea(f) − Ea(b) is negative, the backward reaction has the larger activation energy. Therefore k_b rises by the larger factor, and K = k_f/k_b falls, which agrees with Le Chatelier's principle.

Question 4

Statement I: The conversion of diamond into graphite has a negative ΔG at room conditions, yet it is imperceptibly slow. Statement II: The rate of a process is not determined by its free energy change, because the rate depends on the activation energy. Which is correct?
  1. Both statements are true, and Statement II explains Statement I
  2. Statement I is false, but Statement II is true
  3. Statement I is true, but Statement II is false
  4. Both statements are true, but Statement II does not explain Statement I
Show answer
Answer: (A) Both statements are true, and Statement II explains Statement I
Explanation: A negative ΔG only shows that a change is thermodynamically feasible. It says nothing about how fast the change happens. Diamond persists because the barrier to rearranging its strong covalent network is very high, which is exactly the point made in the second statement.

Question 5

At temperature T, the fraction of collisions having energy equal to or greater than the activation energy of a reaction is 1.0 × 10⁻⁸. If the activation energy does not change, this fraction at temperature 2T (T in kelvin) will be:
  1. 2.0 × 10⁻⁸
  2. 4.0 × 10⁻⁸
  3. 1.0 × 10⁻⁴
  4. 1.0 × 10⁻¹⁶
Show answer
Answer: (C) 1.0 × 10⁻⁴
Explanation: The fraction is e^(−Ea/RT). At 2T it becomes e^(−Ea/2RT), which is the square root of the value at T. The square root of 1.0 × 10⁻⁸ is 1.0 × 10⁻⁴, so doubling T raises the fraction ten thousand times rather than merely doubling it.

Question 6

An exothermic reaction occurs in two elementary steps: P → I (slow) followed by I → Q (fast), where I is a reactive intermediate. Which description fits its energy profile?
  1. Two maxima with the first higher than the second, I at the minimum between them, and Q below P
  2. A single maximum with I at its top, and Q below P
  3. Two maxima with the first higher than the second, I at the minimum between them, and Q above P
  4. Two maxima with the second higher than the first, I at the minimum between them, and Q below P
Show answer
Answer: (A) Two maxima with the first higher than the second, I at the minimum between them, and Q below P
Explanation: Each elementary step has its own transition state, so two steps give two maxima. An intermediate has a real, though short, existence and sits in the valley between them, unlike a transition state, which sits at a peak. The slow step has the larger barrier, so the first maximum is the higher one. An exothermic reaction ends with products below reactants.

Question 7

Reactants A + B (enthalpy level 10 kJ mol⁻¹) can follow two paths. Path 1 passes over a barrier top at 40 kJ mol⁻¹ and gives product C at −30 kJ mol⁻¹. Path 2 passes over a barrier top at 25 kJ mol⁻¹ and gives product D at 0 kJ mol⁻¹. Assuming equal pre-exponential factors, which conclusion is correct?
  1. C is both formed faster and the more stable product
  2. D is formed faster, while C is the more stable product
  3. D is both formed faster and the more stable product
  4. C is formed faster, while D is the more stable product
Show answer
Answer: (B) D is formed faster, while C is the more stable product
Explanation: The activation enthalpy is 30 kJ mol⁻¹ for path 1 and 15 kJ mol⁻¹ for path 2, so D is formed faster. Stability depends on the product's own enthalpy, and C lies lower than D. Speed is set by the barrier height, not by how much energy the reaction releases.
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Question 8

According to collision theory, k = P·Z·e^(−Ea/RT), where P is the probability (steric) factor. Two gas phase reactions at 300 K have the same collision frequency Z. Reaction I has P = 0.50 and Ea = 40.0 kJ mol⁻¹; reaction II has P = 0.050 and Ea = 34.3 kJ mol⁻¹. The ratio k(II)/k(I) is about (take 2.303RT = 5.7 kJ mol⁻¹ at 300 K)
  1. 1
  2. 0.1
  3. 100
  4. 10
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Answer: (A) 1
Explanation: The lower Ea of reaction II is 5.7 kJ mol⁻¹ below that of reaction I, and 5.7 kJ mol⁻¹ equals 2.303RT here, so its exponential term is 10 times larger. Its probability factor is 10 times smaller, since 0.050/0.50 = 0.1. The two effects cancel and k(II)/k(I) is about 1, showing that orientation matters as well as energy.

Question 9

Assertion: For an exothermic reversible reaction, raising the temperature increases both the forward and the backward rate constants, yet the equilibrium constant decreases.
Reason: In an exothermic reaction the backward reaction has the larger activation energy, so its rate constant rises by the larger factor.
  1. The assertion is false but the reason is true.
  2. The assertion is true but the reason is false.
  3. Both assertion and reason are true, but the reason is not the correct explanation of the assertion.
  4. Both assertion and reason are true, and the reason is the correct explanation of the assertion.
Show answer
Answer: (D) Both assertion and reason are true, and the reason is the correct explanation of the assertion.
Explanation: Since ΔH = Ea(forward) − Ea(backward) is negative for an exothermic reaction, Ea(backward) is the larger barrier. By the Arrhenius equation, a rate constant with a larger Ea is more sensitive to temperature. So k(backward) grows faster than k(forward), and K = k(forward)/k(backward) falls. The reason therefore explains the assertion.

Question 10

For a certain reaction, the fraction of molecules having energy equal to or greater than Ea, given by e^(−Ea/RT), is 1 × 10⁻¹⁰ at 300 K. Assuming Ea does not change with temperature, this fraction at 600 K is
  1. 1 × 10⁻²
  2. 1 × 10⁻⁸
  3. 2 × 10⁻¹⁰
  4. 1 × 10⁻⁵
Show answer
Answer: (D) 1 × 10⁻⁵
Explanation: Doubling the absolute temperature halves the exponent Ea/RT. So the new fraction is e^(−Ea/2RT), the square root of the old one. The square root of 1 × 10⁻¹⁰ is 1 × 10⁻⁵, a rise of 100000 times, which shows why rate depends so strongly on temperature.

Question 11

For a reversible elementary reaction at a fixed temperature, the uncatalysed rate constants are k(forward) = 2.0 × 10⁻³ s⁻¹ and k(backward) = 4.0 × 10⁻⁴ s⁻¹. A catalyst raises k(forward) to 2.0 s⁻¹. The value of k(backward) in the presence of the catalyst is
  1. 4.0 × 10⁻⁷ s⁻¹
  2. 4.0 × 10⁻⁴ s⁻¹
  3. 0.40 s⁻¹
  4. 2.0 s⁻¹
Show answer
Answer: (C) 0.40 s⁻¹
Explanation: The equilibrium constant is K = k(forward)/k(backward) = 5.0, and a catalyst does not change K. A catalyst lowers both barriers by the same amount, so both rate constants rise by the same factor, here 1000. Hence k(backward) = 2.0/5.0 = 0.40 s⁻¹.

Question 12

In a reaction, the potential energy of the reactants is 40 kJ mol⁻¹, that of the products is 15 kJ mol⁻¹ and the threshold energy is 95 kJ mol⁻¹. A catalyst lowers the threshold energy to 70 kJ mol⁻¹. In the presence of the catalyst, the activation energy of the backward reaction and ΔH of the forward reaction are respectively
  1. 55 kJ mol⁻¹ and −25 kJ mol⁻¹
  2. 30 kJ mol⁻¹ and −25 kJ mol⁻¹
  3. 55 kJ mol⁻¹ and −50 kJ mol⁻¹
  4. 80 kJ mol⁻¹ and −25 kJ mol⁻¹
Show answer
Answer: (A) 55 kJ mol⁻¹ and −25 kJ mol⁻¹
Explanation: The backward barrier is measured from the products up to the threshold level, so with the catalyst it is 70 − 15 = 55 kJ mol⁻¹. ΔH is the energy of products minus reactants, 15 − 40 = −25 kJ mol⁻¹. A catalyst lowers the threshold level but leaves the reactant and product levels unchanged, so ΔH stays the same.

Question 13

For a reaction with a constant positive activation energy, let r₁ = k(310 K)/k(300 K) and r₂ = k(610 K)/k(600 K). The correct relation is:
  1. r₂ > r₁ > 1
  2. r₂ > 1 > r₁
  3. r₁ = r₂ > 1
  4. r₁ > r₂ > 1
Show answer
Answer: (D) r₁ > r₂ > 1
Explanation: From the Arrhenius equation, log(k₂/k₁) = (Ea/2.303R)(T₂ − T₁)/(T₁T₂). Both intervals are 10 K, but the product T₁T₂ is much smaller near 300 K. The same temperature rise therefore gives a larger ratio at the lower temperature, while both ratios stay above one.

Question 14

An exothermic overall process occurs in two steps: (i) A + B → X (slow) and (ii) X → AB (fast). Which description of its energy profile is correct?
  1. Step (i) has the larger activation energy, and the energy of AB lies above that of A + B
  2. Step (ii) has the larger activation energy, and the energy of AB lies below that of A + B
  3. Step (i) has the larger activation energy, and the energy of AB lies below that of A + B
  4. Step (ii) has the larger activation energy, and the energy of AB lies above that of A + B
Show answer
Answer: (C) Step (i) has the larger activation energy, and the energy of AB lies below that of A + B
Explanation: A slower step has a higher energy barrier, so the slow first step has the larger activation energy. Because the overall process is exothermic, the product AB must end at a lower energy than A + B. X appears as a minimum between the two maxima.

Question 15

For an exothermic reaction with ΔH = −25 kJ mol⁻¹, the average energy of the reactant molecules is 30 kJ mol⁻¹ and the threshold energy is 95 kJ mol⁻¹. The activation energies of the forward and backward reactions respectively are:
  1. 65 and 90 kJ mol⁻¹
  2. 90 and 65 kJ mol⁻¹
  3. 95 and 70 kJ mol⁻¹
  4. 65 and 40 kJ mol⁻¹
Show answer
Answer: (A) 65 and 90 kJ mol⁻¹
Explanation: The forward activation energy is threshold energy minus reactant energy, 95 − 30 = 65 kJ mol⁻¹. Since ΔH = Ea(f) − Ea(b), the backward value is 65 − (−25) = 90 kJ mol⁻¹. The products lie at 5 kJ mol⁻¹, which also gives 95 − 5 = 90 kJ mol⁻¹.

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