Chemical Kinetics

• Assertion: all properly oriented collisions give product.
• Reason: all properly oriented collisions lead to bond rearrangements.

The truth of each statement and their relationship.

Collision theory sets two conditions, not one. A collision produces product only if the molecules are correctly oriented and they collide with energy at least equal to the activation energy. Both statements here claim orientation alone is sufficient, and both therefore fail on the same missing condition.

1. Take the assertion. A properly oriented collision between two slow, low-energy molecules simply bounces apart. Orientation alone is not enough. A is false.
2. Take the reason. It makes the same claim in different words — that orientation alone guarantees bond rearrangement. Without the energy to break the old bonds, nothing rearranges. R is false.
3. Both statements are false.

• Three concentration-rate pairs for A → products.
• Rate = k[A]ⁿ, with n to be found.

The order of the reaction.

Pick two rows and see what happens to the rate when the concentration changes by a known factor. If halving the concentration divides the rate by four, the order is 2, because 2² = 4. Choosing the pair with the cleanest ratio — here 3 and 1.5, a factor of exactly 2 — makes this a mental calculation.

1. Take rows 1 and 2. Concentration goes from 3 × 10⁻³ down to 1.5 × 10⁻³ — a factor of ½.
2. Rate goes from 2.5 × 10⁻² down to 6.25 × 10⁻³ — a factor of ¼.
3. So (½)ⁿ = ¼, which gives n = 2.
4. Check against row 3: concentration is 1.5 times row 1, so the rate should be 1.5² = 2.25 times. 2.5 × 10⁻² × 2.25 = 5.625 × 10⁻². ✓ Confirmed.

• Statement I: a first order reaction can be bimolecular.
• Statement II: radioactive decay is first order.

The truth of each statement.

Order and molecularity are independent quantities, and that is the point of statement I. Order is experimental — the sum of the powers in the measured rate law. Molecularity is theoretical — how many species meet in an elementary step. A bimolecular reaction can show first order kinetics whenever one reactant is present in large excess, so its concentration barely changes. Hydrolysis of an ester in excess water is the standard example, called pseudo-first-order.

1. Statement I. Consider ester + water in a large excess of water. Two species react, so it is bimolecular, but [H₂O] is effectively constant, so the measured rate law is rate = k′[ester] — first order. Correct.
2. Statement II. Radioactive decay follows N = N₀e^−λt, which is exactly the first order integrated rate law, and it has a constant half-life. Correct.
3. Both statements are correct.

• First order reaction.
• 75%% complete in 32 minutes.

The time for 50%% completion.

Identical to Q141. For a first order reaction the half-life is constant, so successive halvings take equal times. 75%% complete leaves 25%%, which is two halvings, so 32 minutes covers two half-lives.

1. 75%% complete means 25%% remains.
2. 100% → 50% → 25% is two half-lives.
3. 2 t½ = 32, so t½ = 16 minutes.
4. 50%% completion takes exactly one half-life, so the answer is 16 minutes.

• Four candidate effects of a catalyst.

Which effects are real.

A catalyst provides an alternative path with a lower activation energy. That makes the reaction faster in both directions. But it changes nothing about where the reaction ends up: the reactants and products have the same energies as before, so ΔH, ΔG and the equilibrium position are all untouched. A catalyst changes how fast, never how far.

1. A. Changes Gibbs free energy. ΔG is fixed by the initial and final states, which the catalyst does not touch. False.
2. B. Increases the rate. That is what a catalyst is for. True.
3. C. Lowers the activation energy. This is the mechanism by which B happens. True.
4. D. Changes the enthalpy of reaction. ΔH is the energy difference between reactants and products — also fixed. False.
5. Only B and C stand.

• A concentration-time curve.
• Two times t₁ and t₂ marked, with the corresponding concentration change ΔC.

What the quantity ΔC/Δt represents.

The symbol Δ means a finite change measured across an interval. Dividing a change in concentration by the interval it took gives the average rate over that interval — geometrically, the slope of the straight chord joining the two points. The instantaneous rate is a different object: dC/dt, the slope of the tangent at a single instant.

1. Δ denotes a finite difference, so ΔC is measured between two distinct times t₁ and t₂.
2. ΔC/Δt is therefore the slope of the chord between those two points on the curve.
3. A chord slope averages out everything happening in between, so it is the average rate.
4. The instantaneous rate would require a single point and a tangent, written dC/dt with a d rather than a Δ.

• Statement I concerns whether rate depends on concentration for zero and first order.
• Statement II concerns the units of the rate of reaction.

Which statements are correct.

Two separate traps. Statement I fails because a zero order reaction is defined by its rate being independent of concentration. Statement II fails on a distinction that is easy to slide past: the rate of reaction always has units of mol L⁻¹ s⁻¹, whatever the order, because it is always a concentration divided by a time. It is the rate constant whose units change with order.

1. Statement I. For zero order, rate = k, a constant that does not involve concentration at all. So the claim fails for the zero order half. Incorrect.
2. Statement II. Rate of reaction is always −d[R]/dt, a concentration over a time, so its units are mol L⁻¹ s⁻¹ for every order. Incorrect.
3. For comparison, the rate constant does change: zero order k is in mol L⁻¹ s⁻¹, first order k is in s⁻¹, second order k is in L mol⁻¹ s⁻¹.
4. Both statements are incorrect.

• k = A e^−Eₐ/RT.
• Four statements, one of which is not correct.

Which statement is incorrect.

In the Arrhenius equation Eₐ sits in a negative exponent. So a large Eₐ makes the exponential very small, which makes k small, which makes the reaction slow. The statement claiming a high activation energy implies a fast reaction has the relationship exactly backwards. The other three statements are all standard and correct.

1. A high Eₐ means a big barrier, so only a tiny fraction of molecules can clear it and the reaction is slow. Option A reverses this — this is the incorrect statement.
2. Option B is correct: raising T raises the fraction of molecules above Eₐ, so k rises.
3. Option C is correct: a large Eₐ makes the exponent more sensitive to T, so the temperature dependence is stronger. A reaction with a small barrier is barely affected by warming.
4. Option D is correct: A, the pre-exponential or frequency factor, counts the total collision frequency with the right orientation, before any energy criterion is applied.

• k = A e^−Eₐ/RT.

What the exponential term represents.

The exponential term is the Boltzmann factor: it gives the fraction of molecules whose kinetic energy is at least the activation energy — that is, the fraction able to react. Since A represents the total collision frequency, multiplying it by this fraction gives the rate of the collisions that actually succeed, which is exactly what k should be.

1. A is the total frequency of suitably oriented collisions.
2. k is smaller than A, because only some of those collisions have enough energy.
3. So the exponential must be the fraction that does have enough energy — that is, KE greater than Eₐ.
4. Note that it is always between 0 and 1, which is what a fraction must be, and it grows towards 1 as T rises — more molecules clear the barrier when it is hotter.