Hi CUET aspirants, Welcome to Amans Maths Blogs (AMBIPi). In this post, you will get CUET Chemistry Mock Test Chemical Kinetics AMBIPi. This CUET Chemistry Mock Test is of the Chemistry chapter Chemical Kinetics. Before solving these CUET Chemistry questions, you must read CUET Chemistry Notes, which helps you to revise CUET Chemistry Syllabus and then you must need to solve CUET Previous Years Questions Papers.
CUET Chemistry Mock Test
CUET Chemistry Question No: 1
Rate of which of the following reactions can be determined easily?
Option A : Rusting of iron in the presence of air and moisture
Option B : Hydrolysis of starch
Option C : Reaction of silver nitrate with sodium chloride
Option D : All of the above
Show/Hide Answer Key
Option B: Hydrolysis of starch
Rate of only those chemical reactions can be determined easily which occurs at moderate speed, e.g. hydrolysis of
starch, inversion of cane sugar etc.
CUET Exam Chemistry Question No: 2
For the reaction,
2 X + Y → X2 Y
What will be the expression for instantaneous rate of the reaction?
Option A : −d [X]/ 2dt
Option B : +1/2 d [Y]/dt
Option C : +1/2 d [Y]/dt
Option D : None of these
Show/Hide Answer Key
Option A: −d [X]/ 2dt
For the reaction, 2 X + Y → X2Y
The instantaneous rate of the reaction is given as,
– d [X]/ 2dt = -d[Y]/dt = + d[X2Y/dt
CUET UG Chemistry Question No: 3
Choose the correct value of d[ NH3] / dt for the reaction
N2 + 3 H2 → 2 NH3
Option A : -2/3 d[H2] / dt
Option B : 2/3 d [H2]
Option C : 3/2 d [N2]/ dt
Option D : -3/2 d [N2]/ dt
Show/Hide Answer Key
Option A: -2/3 d[H2] / dt
For the reaction, N2 + 3 H2 → 2 NH3
The instantaneous rate of reaction is given as
d[N2]/ dt = -d [H2]/ 3dt = +d [ NH3]/ 2dt
d [NH3]/dt = 2/3 d[H2]/ dt
CUET Domain Subject Chemistry Question No: 4
In a reaction, 2x → y, the concentration of x decreases from 3.0 M to 1.5 M in 4 min.
Option A : 0.187 M min −1
Option B : 1.87 M min −1
Option C : 3.75 10-1M min −1
Option D : 0.75 M min −1
Show/Hide Answer Key
Option A: 0.187 M min −1
Rate of reaction = – 1/2 d[x]/ dt = -1/2 [1.5 – 3.0]/ 4
= 1.5 /8 = 0.187 M min-1
CUET BSc Chemistry Question No: 5
Rate law cannot be determined from balanced chemical equation, if ……… .
Option A : reverse reaction is involved
Option B : it is an elementary reaction
Option C : it is a sequence of elementary reactions
Option D : Both (a) and (c)
Show/Hide Answer Key
Option B: it is an elementary reaction
CUET Entrance Exam Chemistry Question No: 6
On mixing 1 dm3 of 3M ethanol with 1 dm3 of 2 M ethanoic acid, an ester is formed
C2 H5 OH + CH3 COOH → CH3 COOC2H5 + H2O
If each solution is diluted with an equal volume of water, the decrease in the initial rate would be
Option A : 0.5 times
Option B : 3 times
Option C : 0.25 times
Option D : 2 times
Show/Hide Answer Key
Option C: 0.25 times
For the given reaction,
C2 H5 OH + CH3 COOH → CH3 COOC2 H5 + H20
r = k [CH3 COOH] [C2 H5 OH]
or r = k [n/v] [n/v]
Increasing volume (V) by twice (due to dilution) will reduce the
rate by 4 times.
r = 0.25 k [CH3 COOH[ [C2 H5 OH]
CUET Exam Question No: 7
Which of the following rate expression is correct against its reaction?
Option A : 2 NO (g) + O2 (g) + 2 NO2 (g) → 2 NO2 (g) ; Rate = k [NO]2 [O2]
Option B : CHCI3 + CI2 → CCI4 + HCI; Rate = k [ CH CI3] [CI2]1/2
Option C : CH3 COO C2 H5 + H20 → CH3 COOH + C2 H5 OH; Rate = k [CH3 COOC2 H5]1 [H2O]0
Option D : All of the above
Show/Hide Answer Key
Option D: All of the above
CUET Chemistry Practice Questions No: 8
Consider the following reactions,
O3 ⇔ O2 + 0 (fast)
O + O3 → 2 O2 (slow)
The rate law expression should be
Option A : r = k [O3]2 [O2]-1
Option B : r = k [O3] [O2]
Option C : r = k [O3]2
Option D : r = k [O2]-1
Show/Hide Answer Key
Option A: r = k [O3]2 [O2]-1
Given, O3 ⇔ O2 + O (fast)
O + O3 → 2 O2 (slow)
From Eq, (i) K eq= [O2] [O] / [O3]
or
From Eq, (ii) rate law = k[O3] [O]
or
r = k [O3] [O3]/ [O2] Keq
or
r = k’ [O3]2 [O2]-1 [k’ = k x k eq]
CUET Chemistry Sample Paper Question No: 9
Reaction of chloroform with chlorine is an example of fraction order reaction. Its rate law expression is rate = k[CHCl3 ][Cl 2] 1/ 2 . If both are assumed in gaseous state and pressure is measured in bar, then units of rate and rate constant respectively are
Option A : bar min-1, bar2 min-1
Option B : bar min-1, bar -1/2 , min-1
Option C : bar -1/2, min -1, bar2 , min-1
Option D : bar min-1 , bar -1/2 , min-1
Show/Hide Answer Key
Option D: bar min-1 , bar -1/2 , min-1
Given, rate = k [CH CI3] [CI2]1/2
Order = 1+ 1/2 = 3/2 Units of rate constant = Units of rate / [CH CI3] [CI2]1/2
= bar min-1/ bar 3/2 = bar-1/2 min-1 and units of rate is bar min-1
CUET Chemistry Mock Test Question No: 10
The rate constant is numerically equal for three reactions of first, second and third order respectively.Which of the following is true?
Option A : If [A] > 1 ; r3 > r2 > r1
Option B : If [A] = 1; r1 = r2 = r3
Option C : If [A] < 1; r1 > r2 > r3
Option D : All of the above
Show/Hide Answer Key
Option D: All of the above
CUET Chemistry Question No: 11
For the reaction, I– + OCI– → IO- + CI– in aqueous medium, the rate of a reaction is given by
d[IO–]/ dt = k [I–] [OCI–]/ [OH–] the overall order of reaction is
Option A : -1
Option B : 1
Option C : 0
Option D : 2
Show/Hide Answer Key
Option B: 1
The rate of reaction is given by, d [IO–]/ dt = k [I–] [OCI–]/ [OH–]
Therefore, overall order of reaction = + − 1 1 1= 1
CUET Exam Chemistry Question No: 12
The reaction,
2 N2 O5 ⇔ 2N2 O4 + 02
Option A : bimolecular and first order
Option B : unimolecular and second order
Option C : bimolecular and second order
Option D : unimolecular and first order
Show/Hide Answer Key
Option A: 1.86A
The given reaction,
i.e. 2 N2 O5 ⇔ 2 N2 04 + 02
occurs in two steps and, hence biomolecular reaction and the rate determining step is the slowest step which is of first order.
CUET UG Chemistry Question No: 13
Four reactions are given below. Which one of them is of zero order?
Option A : PCI5 → PCI3 + CI2
Option B : 2 FeCI3 + SNCI2 → 2 Fe CI2 + Sn CI4
Option C : H2 + CI2 → 2 HCI
Option D : N2 O5 → 2 NO2 + 1/2 O2
Show/Hide Answer Key
Option C: H2 + CI2 → 2 HCI
Among the given reactions, reaction between hydrogen and chlorine under influence of UV-light is an example of zero
order reaction,
i.e H2 + CI2 → 2 HCI
CUET Domain Subject Chemistry Question No: 14
Higher order (>3) reactions are rare due to
Option A : low probability of simultaneous collision of all the reacting species
Option B : increase in entropy and activation energy as more molecules are involved
Option C : shifting of equilibrium towards reactants due to elastic collisions
Option D : loss of active species on collision
Show/Hide Answer Key
Option A: low probability of simultaneous collision of all the reacting species
CUET BSc Chemistry Question No: 15
For a zero order reaction, a graph of concentration (along y-axis) and time (along x-axis) is linear with
Option A : a zero intercept and a +ve slope
Option B : a zero intercept and a –ve slope
Option C : a non-zero intercept and a –ve slope
Option D : a non-zero intercept and a +ve slope
Show/Hide Answer Key
Option B: a non-zero intercept and a –ve slope
For a zero order reaction, a graph of concentration and time is linear with a non-zero intercept,[R]0 and a (-)ve slope, -k
CUET Entrance Exam Chemistry Question No: 16
The rate constant of the reaction, A B → is 06 10 -3. ×−mole per litre per second. If the concentration of A is 5 M then, concentration of B after 20 min is
Option A : 1.08 M
Option B : 3.60 M
Option C : 0.36 M
Option D : 0.72 M
Show/Hide Answer Key
Option D: 0.72 M
For a zero order reaction, unit of rate constant is (mol L-1 s-1).
Hence, we can easily calculate the concentration of B after 20 min by the following formula, X = kt = 0.6 x 10-3 x 20 x 60 = 0.72 M
CUET Exam Question No: 17
A first order has a rate constant of 2.303 x 10-3 s-1. The time required for 40 g of this reactant to reduce to 10 g will be
[Given that log10 2 = 0.3010]
Option A : 230.3 s
Option B : 301 s
Option C : 2000 s
Option D : V∞ – V0 / V∞ – Vt
Show/Hide Answer Key
Option D: V∞ – V0 / V∞ – Vt
For first order reaction, t = 2.303/ k log a/a-x
Given k = 2.303 x 10-3 s-1 a = 40g , a-x= 10g,
On substituting the given values in Eq. (i), we get
t = 2.303/ 2.303 x 10-3 log 40/10 = 103 log 22 = 2 x 103 x log2 = 2 = 103 x 0.3010 = 602 s
Alternative method
For first order reaction, t1/2 = 0.693/k ⇒ t1/2 (t50%) = 0.693/ 2.303 x 10-3 = 301s Also, t75% = 2t50%
t75% = 2x 301 = 602 s
CUET Chemistry Practice Questions No: 18
For a first order reaction,
Option A : t 1/2 = k/a
Option B : t 3/4 = 2t 1/2
Option C : t 1/2 = 0.693/ 2k
Option D : t 1/2 = k x 0.693
Show/Hide Answer Key
Option B: t 3/4 = 2t 1/2
Decomposition of benzene diazonium chloride occur the following manner. C6 H5 N+ ≣ NCI → C6 H5 CI + N2 ↑
The volume of nitrogen obtained shows the extent of decomposition of benzene diazonium chloride, so
a ∝ V∞ = 162 mL
(a-x) ∝ (V∞ – Vt) mL
For first order reaction, k = 2.303/t log10 V∞ – V0 / V∞ – Vt or
k = 2.303/ t log10 V∞ – V0 / V∞ – Vt
At 20 min,
k = 2.303/20 log10 162/ 162-10 = 2.303/20 log10 162/152
= 3.18 x 10-3 min-1 = 3.2 x10-3 min-1
CUET Chemistry Sample Paper Question No: 19
If half-life of a substance is 36 minutes. The amount left after 2 hrs will be [Given (A0) = 10 g)
Option A : 1 g
Option B : 2 g
Option C : 3 g
Option D : 4 g
Show/Hide Answer Key
Option A: 1 g
Given, t 1/2 = 36 min , t = 2 hours = 120 min, [A]0 = 10 g
For first order reaction,
k = 0.693/ t1/2
k = 0.693/36 = 0.01925
As, we know that, t = 2.303/ k log [A]0/ [A]
where [A]0 = initial amount [A] = final amount
120 = 2.303/ 0.019 log [A]0 / [A] log 10 – log [A] = 1 ⇒ 1 – log [A] =1
Thus, [A] = 1 g
CUET Chemistry Mock Test Question No: 20
99% completion of a first order reaction takes place in 32 min. The time taken in 99.9% completion of the reaction will be
Option A : 48 min
Option B : 52 min
Option C : 56 min
Option D :44 min
Show/Hide Answer Key
Option A: 48 min
For first order reaction, k = 2 303/ t log a / a-x
k = 2.303/ 32 min log a/ a- 0.99a = 0.1439 min -1
⇒ t = 2.303/ 0.1439 log a / a- 0.999 a = 48 min
