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 → X_{2} 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 → X_{2}Y

The instantaneous rate of the reaction is given as,

– d [X]/ 2dt = -d[Y]/dt = + d[X_{2}Y/dt

**CUET UG Chemistry Question No: 3**

Choose the correct value of d[ NH_{3}] / dt for the reaction

N_{2} + 3 H_{2} → 2 NH_{3}

**Option A** : -2/3 d[H_{2}] / dt

**Option B** : 2/3 d [H_{2}]

**Option C** : 3/2 d [N_{2}]/ dt

**Option D** : -3/2 d [N_{2}]/ dt

**Show/Hide Answer Key**

**Option A: -2/3 d[H _{2}] / dt**

For the reaction, N_{2} + 3 H_{2} → 2 NH_{3}

The instantaneous rate of reaction is given as

d[N_{2}]/ dt = -d [H_{2}]/ 3dt = +d [ NH_{3}]/ 2dt

d [NH_{3}]/dt = 2/3 d[H_{2}]/ 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^{-1}M 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

C_{2} H_{5} OH + CH_{3} COOH → CH_{3} COOC_{2}H_{5} + H_{2}O

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,

C_{2} H_{5} OH + CH_{3} COOH → CH_{3} COOC_{2} H_{5} + H_{2}0

r = k [CH_{3} COOH] [C_{2} H_{5} 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 [CH_{3} COOH[ [C_{2} H_{5} OH]

**CUET Exam Question No: 7**

Which of the following rate expression is correct against its reaction?

**Option A** : 2 NO (g) + O_{2} (g) + 2 NO_{2} (g) → 2 NO_{2} (g) ; Rate = k [NO]_{2} [O_{2}]

**Option B** : CHCI_{3} + CI_{2} → CCI_{4} + HCI; Rate = k [ CH CI_{3}] [CI_{2}]^{1/2}

**Option C** : CH^{3} COO C^{2} H^{5} + H^{2}0 → CH^{3} COOH + C^{2} H^{5} OH; Rate = k [CH^{3} COOC^{2} H^{5}]^{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,

O_{3} ⇔ O_{2} + 0 (fast)

O + O_{3} → 2 O_{2} (slow)

The rate law expression should be

**Option A** : r = k [O_{3}]^{2} [O_{2}]^{-1}

**Option B** : r = k [O_{3}] [O_{2}]

**Option C** : r = k [O3]^{2}

**Option D** : r = k [O_{2}]^{-1}

**Show/Hide Answer Key**

**Option A: r = k [O _{3}]^{2} [O_{2}]^{-1}**

Given, O_{3} ⇔ O_{2} + O (fast)

O + O_{3} → 2 O_{2} (slow)

From Eq, (i) K eq= [O_{2}] [O] / [O_{3}]

or

From Eq, (ii) rate law = k[O^{3}] [O]

or

r = k [O_{3}] [O_{3}]/ [O_{2}] K_{eq }

or

r = k’ [O_{3}]^{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[CHCl_{3} ][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}, bar^{2} min^{-1}

**Option B** : bar min^{-1}, bar ^{-1/2} , min^{-1}

**Option C** : bar ^{-1/2,} min^{ -1}, bar^{2} , 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 CI_{3}] [CI_{2}]^{1/2 }

Order = 1+ 1/2 = 3/2 Units of rate constant = Units of rate / [CH CI_{3}] [CI_{2}]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 ; r_{3} > r_{2} > r_{1}

**Option B** : If [A] = 1; r_{1} = r_{2} = r_{3}

**Option C** : If [A] < 1; r_{1} > r_{2} > r_{3}

**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 N_{2} O_{5} ⇔ 2N_{2} O_{4} + 0_{2}

**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 N_{2} O_{5} ⇔ 2 N_{2} 0_{4 }+ 0_{2 }

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** : PCI_{5} → PCI_{3} + CI_{2}

**Option B** : 2 FeCI_{3} + SNCI_{2} → 2 Fe CI_{2} + Sn CI_{4}

**Option C** : H_{2} + CI_{2} → 2 HCI

**Option D** : N_{2} O_{5} → 2 NO_{2} + 1/2 O_{2}

**Show/Hide Answer Key**

**Option C: H _{2} + CI_{2} → 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 H^{2} + CI^{2} → 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∞ – V_{0} / V∞ – V_{t}

**Show/Hide Answer Key**

**Option D: V∞ – V _{0} / 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 = 10^{3} log 2^{2} = 2 x 10^{3} x log2 = 2 = 10^{3} x 0.3010 = 602 s

**Alternative method **

For first order reaction, t_{1/2} = 0.693/k ⇒ t_{1/2 }(t_{50%}) = 0.693/ 2.303 x 10^{-3} = 301s Also, t_{75%} = 2t_{50% }

t_{75%} = 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. C_{6} H_{5} N^{+} ≣ NCI → C_{6} H_{5} CI + N_{2} ↑

The volume of nitrogen obtained shows the extent of decomposition of benzene diazonium chloride, so

a ∝ V∞ = 162 mL

(a-x) ∝ (V∞ – V_{t}) mL

For first order reaction, k = 2.303/t log^{10} V∞ – V_{0} / V∞ – V_{t} or

k = 2.303/ t _{log10 } V∞ – V_{0} / V∞ – V_{t}

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/ t^{1/2 }

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