A first order reaction: k = 0.2 s⁻¹. Time for 99% completion is closest to
A 5 s
B 10 s
C 23 s
D 46 s
99% completion → 1% remains. t=(2.303/k)log(100)= (2.303/0.2)*2 ≈ 23.0 s.
For zero order reaction, rate constant can be found from
A k = ([A]₀−[A])/t
B k = (2.303/t)log([A]₀/[A])
C k = (1/t)(1/[A] − 1/[A]₀)
D k = 0.693/t½
From [A]=[A]₀−kt.
For first-order, if [A] falls from 0.8 M to 0.2 M in time t, then [A] falls from 0.4 M to 0.1 M in
A t/2
B t
C 2t
D 4t
Same ratio (0.8→0.2 is 4 times drop; 0.4→0.1 also 4 times). For first order, time depends on ratio, not absolute.
For first order reaction, if t90 is time for 90% completion, then t99 is
A equal to t90
B about 2 times t90
C about 3 times t90
D about 10 times t90
t ∝ log([A]₀/[A]). For 90% completion ratio=10; for 99% ratio=100 → log 100 is double log 10.
In first order kinetics, t½ is related to t3/4 (time for 75% completion) as
A t3/4 = t½
B t3/4 = 2t½
C t3/4 = 3t½
D t3/4 = 4t½
75% completion means 25% remains = (1/2)².
The integrated rate equation is used to find
A equilibrium constant
B activation energy only
C concentration-time relationship
D entropy change
Integrated laws connect concentration with time for each order.
If a reaction follows second-order kinetics, a plot of [A] vs t will be
A straight line
B exponential
C curve (nonlinear)
D horizontal
Second order gives linearity in 1/[A] vs t, not [A] vs t.
If k has units L mol⁻¹ s⁻¹, the reaction is likely
A zero order
B first order
C second order
D third order
This unit corresponds to second order.
For first order reaction, if k increases, then t½
A increases
B decreases
C unchanged
D becomes zero always
t½ = 0.693/k (inverse relation).
For second order, if k increases, then t½
A increases
B decreases
C unchanged
D independent of k
t½ = 1/(k[A]₀).
A reaction completes in finite time only if it is
A first order
B zero order
C second order
D any order
Zero order reaches [A]=0 at t=[A]₀/k; higher orders approach zero asymptotically.
A first order reaction never truly reaches [A]=0 because
A k becomes zero
B it stops at equilibrium always
C exponential decay approaches zero asymptotically
D rate becomes negative
Exponential curves approach zero but don’t hit exactly zero in finite time.
If concentration falls to one-eighth, the number of half-lives passed is
A 1
B 2
C 3
D 8
(1/2)³ = 1/8.
For a first order reaction, time for 50% completion is
A proportional to initial concentration
B independent of initial concentration
C depends on pressure only
D depends on volume only
t½ depends only on k.
A linear plot is obtained for first order by plotting
A [A] vs t
B ln[A] vs t
C [A]² vs t
D rate vs t
ln[A] = ln[A]₀ − kt.
In a first order reaction, when 75% reactant is consumed, fraction remaining is
A 1/2
B 1/4
C 3/4
D 1/8
75% consumed means 25% remains.
The expression 2.303 appears in first-order kinetics due to conversion from
A mol to L
B ln to log10
C s to min
D K to °C
ln x = 2.303 log10 x.
For second-order reaction, if [A] becomes half, t equals
A 0.693/k
B 1/(k[A]₀)
C [A]₀/(2k)
D 2.303/k
That is the second-order half-life expression.
If a reaction shows straight-line plot of 1/[A] vs t, the slope increases when
A k decreases
B k increases
C temperature decreases always
D [A]₀ increases
Slope = k.
For first-order, if k = 0.1 min⁻¹, time for 50% completion is
A 1 min
B 6.93 min
C 10 min
D 13.86 min
t½ = 0.693/0.1 = 6.93 min.
For first order, time for 25% completion (i.e., 75% remaining) is
A 0.5 t½
B 1 t½
C 2 t½
D 3 t½
To go from 100% to 75%, ratio = 100/75 = 4/3; time is less than half-life; approximately 0.415 t½, closest is 0.5 t½ (exam approximation).
For first order, time for 12.5% remaining is
A 1t½
B 2t½
C 3t½
D 4t½
12.5% = 1/8 = (1/2)³.
A reaction follows zero order with k = 2×10⁻³ mol L⁻¹ s⁻¹ and [A]₀ = 0.1 M. Time to complete is
A 25 s
B 50 s
C 500 s
D 5000 s
t = [A]₀/k = [A]₀/k = 0.1/(2×10⁻³)=50; wait carefully: 0.1 / 0.002 = 50 s. So correct is B actually.
✅ Fix: Correct Answer: B. 50 s
Explanation: Zero order completion time t = [A]₀/k = 50 s.
The equation 1/[A] = 1/[A]₀ + kt represents
A first order
B second order
C zero order
D pseudo-zero order
This is the integrated form for second-order (single reactant).
If k is very small, reaction will be
A very fast
B very slow
C independent of time
D explosive
Smaller k means slower reaction under same conditions.
For a first-order reaction, if k = 0.02 s⁻¹, then t½ is
A 3.465 s
B 34.65 s
C 346.5 s
D 0.3465 s
t½ = 0.693/0.02 = 34.65 s.
The integrated rate equation helps to calculate
A rate constant from concentration-time data
B equilibrium constant from pressure data
C ΔH from calorimetry only
D pH from buffer equation only
You plug concentration at time t to compute k.
If a first-order reaction’s concentration drops from 1.0 M to 0.5 M in 10 min, then k is
A 0.0693 min⁻¹
B 0.693 min⁻¹
C 6.93 min⁻¹
D 0.00693 min⁻¹
k = 0.693/t½ = 0.693/10 = 0.0693 min⁻¹.
If a second-order reaction has t½ = 20 s when [A]₀ = 0.5 M, then k equals
A 0.1 L mol⁻¹ s⁻¹
B 1 L mol⁻¹ s⁻¹
C 0.01 L mol⁻¹ s⁻¹
D 10 L mol⁻¹ s⁻¹
t½ = 1/(k[A]₀) ⇒ k = 1/(20×0.5)=1/10=0.1.
For first order, if [A] becomes 10% of initial, time taken equals
A 1 t½
B 2 t½
C 3.32 t½
D 10 t½
10% remaining means ratio 10. Time t = (2.303/k)log10 = 2.303/k. Since t½=0.693/k, t/t½=2.303/0.693≈3.32