Chemical Kinetics MCQs

Rate = change in amount of substance per unit time.

For reactions in solution we express amount as concentration (mol L⁻¹).

Thus rate = Δ[reactant] or Δ[product] divided by Δt.

Hence the correct answer is concentration of reactant or product per unit time.

Molecularity refers to the number of reactant molecules that collide in an elementary step. By definition it must be a whole number (1, 2 or 3) because only whole molecules can participate in a single collision. Hence it cannot be zero (no collision) nor a fractional value. Therefore the correct statement is: It cannot be zero or fractional.

For a first‑order reaction rate law: rate = k[A] ⇒ d[A]/dt = –k[A].

Separate variables and integrate: ∫d[A]/[A] = –k∫dt → ln[A] = –kt + C.

Thus a plot of ln[A] versus t is linear with slope = –k.

Hence the correct answer is –k.

For a zero‑order reaction, rate = k, so concentration falls linearly: [A] = [A]₀ – kt.

Half‑life t½ is the time when [A] = [A]₀/2, giving k t½ = [A]₀/2.

Thus t½ = [A]₀/(2k), which varies directly with the initial concentration.

Hence the half‑life is directly proportional to initial concentration.

Rate constant k ≈ A e^(−Ea/RT); a 10 °C rise changes 1/T slightly, giving Δln k ≈ (Ea/R)(ΔT/T²).

For typical Ea ≈ 50 kJ mol⁻¹ and T≈300 K, Δln k ≈ (5×10⁴/8.314)(10/300²) ≈ 0.7, so k multiplies by e^0.7≈2–3.

Thus a 10 °C increase near room temperature speeds the reaction by roughly a factor of 2 to 3. 2 to 3.

Molecularity refers to how many molecules actually participate in a single elementary step of a reaction.

It is defined as the count of reactant species that collide simultaneously to form products in that step.

Thus the molecularity of an elementary reaction is the number of reacting species colliding in an elementary step.

Rate law for a zero‑order reaction: rate = –d[A]/dt = k (constant).

Integrate –d[A] = k dt → [A] = [A]₀ – kt.

The concentration decreases linearly with time, not exponentially or logarithmically.

Thus the correct expression is [A] = [A]₀ – kt.

For a zero‑order reaction, rate = –d[A]/dt = k, so [A] = [A]0 – kt.

Half‑life t½ is the time when [A] = [A]0/2, thus [A]0/2 = [A]0 – k t½.

Rearranging gives k t½ = [A]0/2, so t½ = [A]0/(2k).

Hence the half‑life is [A0]/2k.

Write Arrhenius equation for each temperature: k1 = A e^(–Ea/RT1) and k2 = A e^(–Ea/RT2).

Take the ratio k2/k1 = e^[–Ea/(R T2) + Ea/(R T1)] = e^[Ea/R (1/T1 – 1/T2)].

Apply common‑log: log(k2/k1) = (Ea/2.303 R)(1/T1 – 1/T2).

Thus the correct option is log(k2/k1) = (Ea/2.303R)(1/T1 - 1/T2).

Collision frequency Z gives the number of collisions per unit time.

Only a fraction of collisions have energy ≥ Ea, given by the Boltzmann factor e^(−Ea/RT).

Thus the effective collisions per unit time = Z × e^(−Ea/RT).

Hence the rate ∝ Z e^(−Ea/RT).

Write the rate expression using stoichiometric coefficients: rate = -(1/νi) d[species]/dt.

For H₂, ν = 3, so rate = -(1/3) d[H₂]/dt.

For NH₃, ν = 2, so rate = (1/2) d[NH₃]/dt.

Equate the two rates: -(1/3) d[H₂]/dt = (1/2) d[NH₃]/dt → -d[H₂]/dt = (3/2) d[NH₃]/dt.

Rate law for a first‑order reaction: rate = k[A] → units of rate = mol L⁻¹ s⁻¹ and concentration = mol L⁻¹.

Thus k = (rate)/(concentration) = (mol L⁻¹ s⁻¹)/(mol L⁻¹) = s⁻¹.

Hence the unit of the rate constant for a first‑order reaction is s⁻¹.

For a first‑order reaction, integrated rate law: ln([A]/[A₀]) = –kt.

Convert natural log to base‑10: log([A]/[A₀]) = –kt/2.303.

Rearrange: log([A₀]/[A]) = (k/2.303) t, which is a straight line.

Thus the slope of the plot log([A₀]/[A]) vs t is k/2.303.

For a first‑order reaction, after each half‑life the amount left is halved: after one half‑life = ½, after two = (½)² = ¼, after three = (½)³.

(½)³ = 1/8, so the fraction remaining after three half‑lives is 1/8.

Hence the correct answer is 1/8.

Activation energy is the barrier that must be overcome for reactants to form the activated complex.

It is the extra kinetic energy that reactant molecules need to acquire before they can collide effectively.

Thus it is the minimum extra energy reactants must gain to react.

Rate = change in concentration per unit time, so it depends on how fast reactants collide.

Increasing temperature raises kinetic energy, raising collision frequency and rate.

Higher concentration gives more collisions, increasing rate.

A catalyst provides an alternative pathway with lower activation energy, also affecting rate; the total time allowed does not change the instantaneous rate. Hence the factor that generally does NOT affect the rate is Total time the reaction is allowed to run.

Rate law for a zero‑order reaction: rate = k (since rate = –d[A]/dt).

Rate has units of concentration per time → mol L⁻¹ s⁻¹.

Thus k must have the same units as the rate, i.e. mol L⁻¹ s⁻¹.

Hence the unit is mol/L/s.

For a zero‑order reaction, rate = k, so d[A]/dt = –k.

Integrating gives [A] = [A]₀ – kt, a linear decrease of concentration with time.

Thus a plot of [A] versus t yields a straight line with slope –k.

The correct answer is [A] vs t.

Half‑life t½ =10 yr means after each 10 yr the amount halves.

After 40 yr there are 40/10 =4 half‑life intervals.

Remaining fraction = (1/2)^4 =1/16.

Thus the fraction remaining is 1/16.

Lower activation energy means a larger fraction of molecules possess energy ≥Ea at a given temperature (Boltzmann distribution).

Since the rate constant k = A e^(–Ea/RT), a smaller Ea makes the exponential term larger.

A larger k gives a higher reaction rate, so the reaction proceeds faster.

Thus the correct answer is faster rate.