Practice Start Test

Chemical Kinetics

NEET Chemistry · 91 questions · Page 2 of 10 · Click an option or "Show Solution" to reveal answer

Q11
For a chemical reaction 4A + 3B →\to 6C + 9D Rate of formation of C is 6 ×\times 10 −-2 mol L −-1 s −-1 and rate of disappearance of A is 4 ×\times 10 −-2 mol L −-1 s −-1 . The rate of reaction and amount of B consumed in interval of 10 seconds, respectively will be :
A 10 ×\times 10 −-2 mol L −-1 s −-1 and 30 ×\times 10 −-2 mol L −-1
B 1 ×\times 10 −-2 mol L −-1 s −-1 and 30 ×\times 10 −-2 mol L −-1
C 10 ×\times 10 −-2 mol L −-1 s −-1 and 10 ×\times 10 −-2 mol L −-1
D 1 ×\times 10 −-2 mol L −-1 s −-1 and 10 ×\times 10 −-2 mol L −-1
Correct Answer
Option B
Solution

4A + 3B →\to 6C + 9D Rate of reaction =

−d[A]dt×14=−d[B]dt×13=+d[C]dt×16=+d[D]dt×19{{ - d[A]} \over {dt}} \times {1 \over 4} = {{ - d[B]} \over {dt}} \times {1 \over 3} = {{ + d[C]} \over {dt}} \times {1 \over 6} = {{ + d[D]} \over {dt}} \times {1 \over 9}

Rate of reaction

=+d[C]dt×16=6×10−26=10−2= {{ + d[C]} \over {dt}} \times {1 \over 6} = {{6 \times {{10}^{ - 2}}} \over 6} = {10^{ - 2}}

mol L −-1 s −-1 Rate of reaction

=−13d[B]dt= {{ - 1} \over 3}{{d[B]} \over {dt}}
−d[B]dt=3×{{ - d[B]} \over {dt}} = 3 \times

rate of reaction

=3×10−2= 3 \times {10^{ - 2}}

mol L −-1 s −-1 After interval of 10 sec.

=3×10−2×10= 3 \times {10^{ - 2}} \times 10
=30×10−2= 30 \times {10^{ - 2}}

mol L −-1

Q12
For a first order reaction A →\to Products, initial concentration of A is 0.1 M, which becomes 0.001 M after 5 minutes. Rate constant for the reaction in min −-1 is
A 1.3818
B 0.9212
C 0.4606
D 0.2303
Correct Answer
Option B
Solution

For first order reaction,

K=2.303tlog⁡[A0][A]K = {{2.303} \over t}\log {{[{A_0}]} \over {[A]}}

; where A 0 is the initial concentration of reactant A. A 0 = 0.1 M A = 0.001 M t = 5 minute

K=2.3035log⁡0.10.001=2.3035log⁡102K = {{2.303} \over 5}\log {{0.1} \over {0.001}} = {{2.303} \over 5}\log {10^2}
=2.3035×2= {{2.303} \over 5} \times 2

K = 0.9212 min −-1

Q13
The slope of Arrhenius plot (ln⁡ k v/s1T)\left( {\ln \,k\,v/s{1 \over T}} \right) of first order reactino is −-5 ×\times 10 3 K. The value of E a of the reaction is. Choose the correct option for your answer. [Given R = 8.314 JK −-1 mol −-1 ]
A −-83 kJ mol −-1
B 41.5 kJ mol −-1
C 83.0 kJ mol −-1
D 166 kJ mol −-1
Correct Answer
Option B
Solution

Arrhenius equation

k=Ae−Ea/RTk = A{e^{ - {E_a}/RT}}
ln⁡k=ln⁡A+ln⁡e−Ea/RT\ln k = \ln A + \ln {e^{ - {E_a}/RT}}
ln⁡k=ln⁡A−EaR(1T)\ln k = \ln A - {{{E_a}} \over R}\left( {{1 \over T}} \right)

.... (1) Slope of

ln⁡k \ln k\,

vs

1T{1 \over T}

curve,

m=−EaRm = - {{{E_a}} \over R}
−5×103=−EaR- 5 \times {10^3} = - {{{E_a}} \over R}

E a = 5 ×\times 10 3 ×\times 8.314 J/mol = 41.57 ×\times 10 3 J/mol

≃\simeq

41.5 kJ/mol

Q14
An increase in the concentration of the reactants of a reaction leads to change in :
A heat of reaction
B threshold energy
C collision frequency
D activation energy
Correct Answer
Option C
Solution

The correct answer to this question is collision frequency.

Here's an explanation for each option to understand why: Option A - Heat of Reaction: The heat of reaction, also known as the enthalpy change (ΔH)(\Delta H), is determined by the difference in the energy levels of the reactants and products.

It is a characteristic feature of a particular chemical reaction and is not typically affected by the concentration of reactants except in cases of extremely high or non-ideal concentrations where volume changes might somewhat influence pressure and therefore the heat of reaction in gases.

Under normal conditions, however, changing the concentration of reactants does not alter the heat of reaction.

Option B - Threshold Energy: Threshold energy is the minimum energy that reactant molecules must possess in order to undergo a successful collision, leading to a chemical reaction.

This value is inherent to the specific reaction and is related to the strength of the bonds in the reactants as well as the energy required to form the activated complex during the reaction.

The threshold energy does not change simply because the concentration of reactants is altered.

Option C - Collision Frequency: Collision frequency corresponds to how often reacting particles collide in a given time interval.

When the concentration of reactants is increased, there are more reactant particles per unit volume.

This statistically leads to an increased number of collisions per unit time, thereby raising the collision frequency.

This is in accordance with the collision theory of chemical kinetics.

Option D - Activation Energy: Activation energy is the minimum energy barrier that must be overcome for reactants to be converted into products.

It is a property of a particular reaction and is related to the nature of the reactants and the reaction pathway.

Altering the concentration of reactants does not change the activation energy for that reaction.

Thus, the only factor among the provided options that changes with an increase in the concentration of reactants is indeed the collision frequency.

Q15
The rate constant for a first order reaction is 4.606 × \times 10 -3 s -1 . The time required to reduce 2.0 g of the reactant to 0.2 g is :
A 200 s
B 500 s
C 1000 s
D 100 s
Correct Answer
Option B
Solution
k=(2.303t)k = \left( {{{2.303} \over t}} \right)

log

(A0A)\left( {{{{A_0}} \over A}} \right)

for first order reaction

4.606×10−3=(2.303t)4.606 \times {10^{ - 3}} = \left( {{{2.303} \over t}} \right)

log

(20.2)\left( {{2 \over {0.2}}} \right)

⇒\Rightarrow t = 500 s

Q16
If the rate constant for a first order reaction is k, the time (t) required for the completion of 99 % of the reaction is given by :
A t=4.606kt = {{4.606} \over k}
B t=2.303kt = {{2.303} \over k}
C t=0.693kt = {{0.693} \over k}
D t=6.909kt = {{6.909} \over k}
Correct Answer
Option A
Solution

For a first order reaction, t =

2.303klog⁡[A0][A]{{2.303} \over k}\log {{\left[ {{A_0}} \right]} \over {\left[ A \right]}}

for 99% completion of the reaction,

t=2.303klog⁡100100−99t = {{2.303} \over k}\log {{100} \over {100 - 99}}

⇒\Rightarrow

t=4.606kt = {{4.606} \over k}
Q17
For the chemical reaction N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g) the correct option is :
A −d[N2]dt=2d[NH3]dt - {{d\left[ {{N_2}} \right]} \over {dt}} = 2{{d\left[ {N{H_3}} \right]} \over {dt}}
B −d[N2]dt=12d[NH3]dt - {{d\left[ {{N_2}} \right]} \over {dt}} = {1 \over 2}{{d\left[ {N{H_3}} \right]} \over {dt}}
C 3d[H2]dt=2d[NH3]dt3{{d\left[ {{H_2}} \right]} \over {dt}} = 2{{d\left[ {N{H_3}} \right]} \over {dt}}
D −13d[H2]dt=−12d[NH3]dt - {1 \over 3}{{d\left[ {{H_2}} \right]} \over {dt}} = -{1 \over 2}{{d\left[ {N{H_3}} \right]} \over {dt}}
Correct Answer
Option B
Solution

N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g Rate of the reaction can be written as :

−d[N2]dt=−13d[H2]dt=+12d[NH3]dt- {{d\left[ {{N_2}} \right]} \over {dt}} = - {1 \over 3}{{d\left[ {{H_2}} \right]} \over {dt}} = + {1 \over 2}{{d\left[ {N{H_3}} \right]} \over {dt}}
Q18
The correct difference between first and second order reactions is that
A the rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations
B the half-life of a first-order reaction does not depend on [A] 0 ; the half-life of a second order reaction does depend on [A] 0
C a first-order reaction can be catalysed; a second-order reaction cannot be catalysed
D the rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations
Correct Answer
Option B
Solution

For the first order reaction, t 1/2 =

0.693k{{0.693} \over k}

which is independent of initial concentration [A] 0 . For second order reaction, t 1/2 =

1k[A]0{1 \over {k{{\left[ A \right]}_0}}}

which depends on initial concentration [A] 0 .

Q19
When initial concentration of the reactant is doubled, the half-life period of a zero order reaction
A is halved
B is doubled
C is tripled
D remains unchanged.
Correct Answer
Option B
Solution

Half life of zero order t 1/2 =

[A]02k{{{{{\left[ A \right]}_0}} \over {2k}}}

∴\therefore t 1/2 ∞\infty [A] 0 ∴\therefore If [A] 0 = doubled then, t 1/2 = doubled.

Q20
A first order reaction has a specific reaction rate of 10 −-2 sec −-1 . How much time will it take for 20 g of the reactant to reduce to 5 g?
A 138.6 sec
B 346.5 sec
C 693.0 sec
D 238.6 sec
Correct Answer
Option A
Solution

For a first order reaction, k =

2.303tlog⁡[A]0[A]t{{2.303} \over t}\log {{{{\left[ A \right]}_0}} \over {{{\left[ A \right]}_t}}}

⇒\Rightarrow 10 -2 =

2.303tlog⁡205{{2.303} \over t}\log {{20} \over 5}

⇒\Rightarrow t = 138.6 sec

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