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All questions of Reversible Reactions for Grade 9 Exam
Find the overall order of a reaction whose rate constant is k = 3 x 10 -4 s-1- a)Third
- b)Second
- c)Zero
- d)First
Correct answer is option 'D'. Can you explain this answer?
Find the overall order of a reaction whose rate constant is k = 3 x 10 -4 s-1
a)
Third
b)
Second
c)
Zero
d)
First
|
Aravind Rane answered |
The overall order of a reaction is the sum of the individual orders of the reactants in the rate law. The rate law expresses the relationship between the rate of a chemical reaction and the concentrations of the reactants.
Given that the rate constant (k) is 3 x 10^-4 s^-1, we can determine the order of the reaction by analyzing the units of k.
We know that the overall order of a reaction is determined by adding the individual orders of the reactants. The order of a reactant is the exponent to which its concentration is raised in the rate law.
The units of k for a given reaction can be determined by analyzing the units of the rate law. In this case, we have:
k = rate / [reactant]^n
where n is the order of the reactant in the rate law.
The units of k are s^-1, which means that the units of rate must be in mol/L/s and the units of [reactant] must be in mol/L in order for the units to cancel out and leave only s^-1.
Since we are given only the value of k, we cannot directly determine the individual orders of the reactants. However, we can infer the overall order of the reaction based on the units of k.
Since k has units of s^-1, the units of rate must be mol/L/s. This means that the overall order of the reaction must be 1 (first order).
Therefore, the correct answer is option D: First.
Given that the rate constant (k) is 3 x 10^-4 s^-1, we can determine the order of the reaction by analyzing the units of k.
We know that the overall order of a reaction is determined by adding the individual orders of the reactants. The order of a reactant is the exponent to which its concentration is raised in the rate law.
The units of k for a given reaction can be determined by analyzing the units of the rate law. In this case, we have:
k = rate / [reactant]^n
where n is the order of the reactant in the rate law.
The units of k are s^-1, which means that the units of rate must be in mol/L/s and the units of [reactant] must be in mol/L in order for the units to cancel out and leave only s^-1.
Since we are given only the value of k, we cannot directly determine the individual orders of the reactants. However, we can infer the overall order of the reaction based on the units of k.
Since k has units of s^-1, the units of rate must be mol/L/s. This means that the overall order of the reaction must be 1 (first order).
Therefore, the correct answer is option D: First.
Half-life period of a second order reaction is
a) independent of the concentration
b) inversaly proportional to the initial conc
c) inversely proportional to conc.
d) directly proportional to the conc.
The answer is b.
|
Dr Manju Sen answered |
The correct answer is Option B.
Relation between half-life period of a reaction and initial concentration is as follows:
For first order reaction (Half life α a0 )
For second order reaction (Half life ∝
1/a)
For third order reaction (Half life ∝ 1a2)
Relation between half-life period of a reaction and initial concentration is as follows:
For first order reaction (Half life α a0 )
For second order reaction (Half life ∝
1/a)
For third order reaction (Half life ∝ 1a2)
For a reaction,
hen x, y and z are - a)1:1:1
- b)3:2:3
- c)3:3:2
- d)2:2:3
Correct answer is option 'C'. Can you explain this answer?
For a reaction,
hen x, y and z are
a)
1:1:1
b)
3:2:3
c)
3:3:2
d)
2:2:3
|
Divey Sethi answered |
For the reaction: xA + yB → zC
Given that,
Multiplying equation 2 by 1/3, we get
On comparing equation 1 and 3,
x = 3, y = 3, z = 2.
Given that,
Multiplying equation 2 by 1/3, we get
On comparing equation 1 and 3,
x = 3, y = 3, z = 2.
For a chemical reaction the rate constant is nearly doubled with the rise in temperature by
a)100ºCb)10ºCc)50ºCd)273ºCCorrect answer is option 'B'. Can you explain this answer?
|
Rohit Shah answered |
It has been found that rate constant is nearly doubled for a chemical reaction with a rise in temperature by 10deg. We can explain the dependence of the rate of a chemical reaction on temperature by Arrhenius equation.
The half life of a zero order reaction is equal to:- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The half life of a zero order reaction is equal to:
a)
b)
c)
d)
|
Tanuja Kapoor answered |
[A] = -kt + [A]0
Here, [A] = [A0]/2
[A]0/2 = -kt + [A]0
kt = [A]0 - [A]0/2
t = [A]0/2k
t1/2 = [A]0/2k
Here, [A] = [A0]/2
[A]0/2 = -kt + [A]0
kt = [A]0 - [A]0/2
t = [A]0/2k
t1/2 = [A]0/2k
For the reaction 
. The variation of the concentration of the products is given by curve 
- a)Y
- b)W
- c)Z
- d)X
Correct answer is option 'A'. Can you explain this answer?
For the reaction . The variation of the concentration of the products is given by curve
. The variation of the concentration of the products is given by curvea)
Y
b)
W
c)
Z
d)
X
|
|
Divey Sethi answered |
For the reaction, A+B⟶C+D, the variation of the concentration of the products is given by the curve Y.
Initially, the product concentration is 0, then it gradually rises and reaches a maximum value. After that it remains constant.
Initially, the product concentration is 0, then it gradually rises and reaches a maximum value. After that it remains constant.
This behavior is represented by the curve Y.
Can you explain the answer of this question below:For the reaction,
kf = 2.2 x 104 L mol-1s-1 and Kc = 1.00 x 10-4 at 3000 K. Thus, kb is
- A:
2.2 x 108 L2mol-2s-1
- B:
0.45 x 10-8L-2mol2s-1
- C:
2.2 L mol-1s-1
- D:
0.45L-1mol s-1
The answer is a.
For the reaction,
kf = 2.2 x 104 L mol-1s-1 and Kc = 1.00 x 10-4 at 3000 K. Thus, kb is
2.2 x 108 L2mol-2s-1
0.45 x 10-8L-2mol2s-1
2.2 L mol-1s-1
0.45L-1mol s-1
|
Dewansh Tripathi answered |
Given;kf=2.2×10^4Lmol^-1s^-1;KC=1×10^-4at 3000k . we, know that,KC=kf/kb and put the value and solve.
If Ef and Eb are the activation energies of the forward and reverse reactions and the reaction is known to be exothermic, then:- a)Ef = Eb
- b)Ef < EbData insufficient to predict
- c)f < EbData insufficient to predict
- d)Ef > Eb
Correct answer is option 'B'. Can you explain this answer?
If Ef and Eb are the activation energies of the forward and reverse reactions and the reaction is known to be exothermic, then:
a)
Ef = Eb
b)
Ef < EbData insufficient to predict
c)
f < EbData insufficient to predict
d)
Ef > Eb
|
EduRev JEE answered |
The correct answer is Option B.
For exothermic reaction, ΔH<0
Eb = Ef + ∣ΔH∣ or
Ef < Eb
Eb = Ef + ∣ΔH∣ or
Ef < Eb
The term 
in the rate expression refers to the:- a)Increase in concentration of the reactants
- b)Average rate of the reaction
- c)Instantaneous rate of the reaction
- d)Concentration of the reactants
Correct answer is option 'C'. Can you explain this answer?
The term in the rate expression refers to the:
in the rate expression refers to the:a)
Increase in concentration of the reactants
b)
Average rate of the reaction
c)
Instantaneous rate of the reaction
d)
Concentration of the reactants
|
Mira Sharma answered |
The term —dx/dt in the rate expression refers to theinstantaneous rate of the reaction.
Identify A in the given diagram.
- a)Order of reaction
- b)Average rate of reaction
- c)Concentration of the catalyst
- d)Instantaneous rate of reaction
Correct answer is option 'D'. Can you explain this answer?
Identify A in the given diagram.
a)
Order of reaction
b)
Average rate of reaction
c)
Concentration of the catalyst
d)
Instantaneous rate of reaction
|
Siddhi Bhardwaj answered |
D) obviously it is change in conc over time.
The activation energy of a reaction at a given temperature is found to be 2.303 RT J mol–1. The ratio of rate constant to the Arrhenius factor is- a)0.1
- b)0.01
- c)0.001
- d)0.02
Correct answer is option 'A'. Can you explain this answer?
The activation energy of a reaction at a given temperature is found to be 2.303 RT J mol–1. The ratio of rate constant to the Arrhenius factor is
a)
0.1
b)
0.01
c)
0.001
d)
0.02
|
|
Riya Banerjee answered |
The correct answer is option A
Arrhenius equation is,
rate constant, k=Ae−Ea/RT
k=Ae−2.303RT/RT ⇒ k/A = e-2.303
On solving, we get
k/A = 10-1 or 0.1
Arrhenius equation is,
rate constant, k=Ae−Ea/RT
k=Ae−2.303RT/RT ⇒ k/A = e-2.303
On solving, we get
k/A = 10-1 or 0.1
In a reaction, A + B → Product, rate is doubled when the concentration of B is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (Aand B) are doubled, rate law for the reaction can be written as - a)Rate = k[A] [B]
- b)Rate = k[A]2[B]
- c)Rate = k[A][B]2
- d)Rate = k[A]2[B]2
Correct answer is option 'B'. Can you explain this answer?
In a reaction, A + B → Product, rate is doubled when the concentration of B is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (Aand B) are doubled, rate law for the reaction can be written as
a)
Rate = k[A] [B]
b)
Rate = k[A]2[B]
c)
Rate = k[A][B]2
d)
Rate = k[A]2[B]2
|
|
Dr Manju Sen answered |
The correct answer is option B
When conc. of B is doubled, the rate is doubled.
So, Order w.r.t. [B] is 1.
The rate increases by a factor of 8 when conc. of [A] and [B] are doubled.
So, Order w.r.t. [A] is 2.
Overall Rate is
Rate=k[A]2[B]
When conc. of B is doubled, the rate is doubled.
So, Order w.r.t. [B] is 1.
The rate increases by a factor of 8 when conc. of [A] and [B] are doubled.
So, Order w.r.t. [A] is 2.
Overall Rate is
Rate=k[A]2[B]
Which of the following represents the expression for the 3/4th life of a first order reaction?
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Which of the following represents the expression for the 3/4th life of a first order reaction?
a)
b)
c)
d)
|
Nikita Singh answered |
The correct answer is option A
As we know, for a first order reaction,
As we know, for a first order reaction,
The half life of a first order reaction is equal to:- a)k
- b)Cannot be predicted
- c)0.693/k
- d)1/k
Correct answer is option 'C'. Can you explain this answer?
The half life of a first order reaction is equal to:
a)
k
b)
Cannot be predicted
c)
0.693/k
d)
1/k
|
Kraft Villa answered |
No explation. there is a proof on this topic in ncert books
- a)1
- b)1/2
- c)-1/2
- d)-1
Correct answer is 'D'. Can you explain this answer?
|
Payal Sapra answered |
Hi tausif. ...let dr/dt is proportional to [A]^n is equation 1st.... then 1/2 dr/dt is prportional to [2A]^n.....equation 2nd compare both equations...u'll get 2^n is equal to 1/2....so we get n is equal to minus 1..... hope u get it
Direction (Q. Nos. 14 and 15) This section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)
Passage
The decomposition of NO2 at 400 K proceeds at a of rate of 5.4 x 10 -5 mol L-1 s-1 when [NO2] = 0.01 mol-1
2 NO2(g) → 2NO(g ) + O2(g).
Q. What is the rate law when observed rate is 1.35 x 10-5 mol L-1 s-1 at [NO2] = 0.005 mol L-1?
- a)k[NO2]
- b)k[NO2]0
- c)k[NO2]3
- d)k[NO2]2
Correct answer is option 'D'. Can you explain this answer?
Direction (Q. Nos. 14 and 15) This section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)
Passage
The decomposition of NO2 at 400 K proceeds at a of rate of 5.4 x 10 -5 mol L-1 s-1 when [NO2] = 0.01 mol-1
2 NO2(g) → 2NO(g ) + O2(g).
Q. What is the rate law when observed rate is 1.35 x 10-5 mol L-1 s-1 at [NO2] = 0.005 mol L-1?
a)
k[NO2]
b)
k[NO2]0
c)
k[NO2]3
d)
k[NO2]2
|
Geetika Shah answered |
Given unit of rate = mol L-1 s-1
General formula for unit of rate = mol n-1 Ln-1 s -1
On equating,
1-n = 1 ⇒ n = 0
n-1 = -1 ⇒ n = 0
General formula for unit of rate = mol n-1 Ln-1 s -1
On equating,
1-n = 1 ⇒ n = 0
n-1 = -1 ⇒ n = 0
Hence, it is a zero order reaction.
A first order reaction is 50% completed in 20 minutes at 27°C and in 5 min at 47°C. The energy of activation of the reaction is- a)43.85 kJ/mol
- b)55.3 kJ/mol
- c)11.97 kJ/mol
- d)6.65 kJ/mol
Correct answer is option 'B'. Can you explain this answer?
A first order reaction is 50% completed in 20 minutes at 27°C and in 5 min at 47°C. The energy of activation of the reaction is
a)
43.85 kJ/mol
b)
55.3 kJ/mol
c)
11.97 kJ/mol
d)
6.65 kJ/mol
|
|
Tanuja Kapoor answered |
t1/2 = 20 min at 300 K
t’1/2 = 5 min at 320 K
On solving, Ea = 55303.12 J
= 55.3 KJ
t’1/2 = 5 min at 320 K
On solving, Ea = 55303.12 J
= 55.3 KJ
The ratio of the rate constant of a reaction at two temperatures differing by __________0C is called temperature coefficient of reaction.- a)2
- b)10
- c)100
- d)50
Correct answer is option 'B'. Can you explain this answer?
The ratio of the rate constant of a reaction at two temperatures differing by __________0C is called temperature coefficient of reaction.
a)
2
b)
10
c)
100
d)
50
|
Devika Banerjee answered |
The ratio of the rate constant of a reaction at two temperatures differing by 100C is called temperature coefficient of reaction.
The initial rates of reaction for the equation, 2A + B → Products.Products were determined under various initial concentrations of reactants.
Thus, rate law is equal to- a)k[A][B]
- b)k[A][B]2
- c) k[A]
- d) k[B]
Correct answer is option 'C'. Can you explain this answer?
The initial rates of reaction for the equation, 2A + B → Products.
Products were determined under various initial concentrations of reactants.
Thus, rate law is equal to
a)
k[A][B]
b)
k[A][B]2
c)
k[A]
d)
k[B]
|
|
Ritu Singh answered |
Let order w.r.t. A = a, order w.r.t. B = b
then rate law is
The rate of a chemical reaction doubles for every 10°C rise of temperature. If the temperature is raised by 50°C, the rate of the reaction increases by about [AIEEE 2011]a)64 timesb)32 timesc)24 timesd)10 timesCorrect answer is option 'B'. Can you explain this answer?
|
Vijay Bansal answered |
Rate at 50 degree C/(Rate at T1 degree C)
= (2)^(ΔT/T1)
= (2)^(50/10)
= 2^5
= 32
Hence, the rate of the reaction increases by 32 times.
The time for half life period of a certain reaction A → Products is 1 h. When the initial concentration of the reactant 'A' is 2.0 mol L–1, how much time does it take for its concentration to come from 0.50 to 0.25 mol L–1, if it is a zero order reaction?- a)0.25 h
- b)1 h
- c)4 h
- d)0.5 h
Correct answer is option 'A'. Can you explain this answer?
The time for half life period of a certain reaction A → Products is 1 h. When the initial concentration of the reactant 'A' is 2.0 mol L–1, how much time does it take for its concentration to come from 0.50 to 0.25 mol L–1, if it is a zero order reaction?
a)
0.25 h
b)
1 h
c)
4 h
d)
0.5 h
|
Shashi Kumar answered |
For a zero order reaction,
(T)half = Initial concentration/ 2× reaction constant
from this equation find out k.
also another formula for zero order reaction-
(A)t =(A)o - KT
put the value of k , (A)t=0.25 and (A)o= 0.50
you will get the answer.
(T)half = Initial concentration/ 2× reaction constant
from this equation find out k.
also another formula for zero order reaction-
(A)t =(A)o - KT
put the value of k , (A)t=0.25 and (A)o= 0.50
you will get the answer.
The effect of temperature on reaction rate is given by- a)Arrhenius equation
- b)Kirchoff’s Equation
- c)Clauius Claperyron equation
- d)Gibb’s Helmholtz equation
Correct answer is option 'A'. Can you explain this answer?
The effect of temperature on reaction rate is given by
a)
Arrhenius equation
b)
Kirchoff’s Equation
c)
Clauius Claperyron equation
d)
Gibb’s Helmholtz equation
|
Shanaya Choudhary answered |
Effect of Temperature on Reaction Rate - Arrhenius Equation
The Arrhenius equation explains the effect of temperature on the reaction rate. It is a mathematical formula that relates the rate constant of a chemical reaction to the temperature and activation energy of the reaction.
The equation is given by:
k = A * e^(-Ea/RT)
Where,
k - rate constant
A - pre-exponential factor
Ea - activation energy
R - gas constant
T - temperature
Explanation of the Equation:
Pre-Exponential Factor (A): It is a constant that reflects the frequency at which reactant molecules collide with each other. A higher value of A means that more collisions occur, thus increasing the reaction rate.
Activation Energy (Ea): It is the minimum energy required to initiate a chemical reaction. A higher value of Ea means that more energy is required to initiate the reaction, thus decreasing the reaction rate.
Gas Constant (R): It is a constant that relates the energy of a system to its temperature.
Temperature (T): It is the measure of the average kinetic energy of the molecules in a system. As the temperature increases, the kinetic energy of the molecules increases, leading to more collisions and thus increasing the reaction rate.
Application of the Equation:
The Arrhenius equation is widely used in chemical kinetics to predict the effect of temperature on the reaction rate. It is also used to determine the activation energy of a reaction by measuring the rate constant at different temperatures.
Conclusion:
In conclusion, the Arrhenius equation is a useful tool to understand the effect of temperature on the reaction rate. It helps in predicting the rate of a chemical reaction at different temperatures and determining the activation energy of a reaction.
The Arrhenius equation explains the effect of temperature on the reaction rate. It is a mathematical formula that relates the rate constant of a chemical reaction to the temperature and activation energy of the reaction.
The equation is given by:
k = A * e^(-Ea/RT)
Where,
k - rate constant
A - pre-exponential factor
Ea - activation energy
R - gas constant
T - temperature
Explanation of the Equation:
Pre-Exponential Factor (A): It is a constant that reflects the frequency at which reactant molecules collide with each other. A higher value of A means that more collisions occur, thus increasing the reaction rate.
Activation Energy (Ea): It is the minimum energy required to initiate a chemical reaction. A higher value of Ea means that more energy is required to initiate the reaction, thus decreasing the reaction rate.
Gas Constant (R): It is a constant that relates the energy of a system to its temperature.
Temperature (T): It is the measure of the average kinetic energy of the molecules in a system. As the temperature increases, the kinetic energy of the molecules increases, leading to more collisions and thus increasing the reaction rate.
Application of the Equation:
The Arrhenius equation is widely used in chemical kinetics to predict the effect of temperature on the reaction rate. It is also used to determine the activation energy of a reaction by measuring the rate constant at different temperatures.
Conclusion:
In conclusion, the Arrhenius equation is a useful tool to understand the effect of temperature on the reaction rate. It helps in predicting the rate of a chemical reaction at different temperatures and determining the activation energy of a reaction.
The reactions with low activation energy are- a)Slow
- b)Non-spontaneous
- c)Fast
- d)Always spontaneous
Correct answer is option 'C'. Can you explain this answer?
The reactions with low activation energy are
a)
Slow
b)
Non-spontaneous
c)
Fast
d)
Always spontaneous
|
|
Vivek Godara answered |
Low activation energy mean energy reguire for reaction to occur is low so product making is fast
The rate of a chemical reaction doubles for every 10°C rise of temperature. If the temperature is raised by 50°C, the rate of the reaction increases by about- a)64 times
- b)10 times
- c)24 times
- d)32 times
Correct answer is option 'D'. Can you explain this answer?
The rate of a chemical reaction doubles for every 10°C rise of temperature. If the temperature is raised by 50°C, the rate of the reaction increases by about
a)
64 times
b)
10 times
c)
24 times
d)
32 times
|
Mohit Rajpoot answered |
For every 10°C rise of temperature, the rate is doubled. Thus, the temperature coefficient of the reaction = 2
When temperature is increased by 50°, rate becomes
=2(50/10) = 25 times = 32 times
When temperature is increased by 50°, rate becomes
=2(50/10) = 25 times = 32 times
A foreign substance that increase the speed of a chemical reaction is called- a)promotor
- b)catalyst
- c)moderator
- d)inhibitor
Correct answer is option 'B'. Can you explain this answer?
A foreign substance that increase the speed of a chemical reaction is called
a)
promotor
b)
catalyst
c)
moderator
d)
inhibitor
|
|
Nandini Patel answered |
Catalyst: Substances which alter the rate of a chemical reaction and themselves remain chemically and quantitatively unchanged after the reaction are known as catalysts and the phenomenon is known as catalysis.
If 75% of a first order reaction was completed in 32 min, then 50% of the reaction was completed in- a)24 min
- b)4 min
- c)16 min
- d)8 min
Correct answer is option 'C'. Can you explain this answer?
If 75% of a first order reaction was completed in 32 min, then 50% of the reaction was completed in
a)
24 min
b)
4 min
c)
16 min
d)
8 min
|
Avi Chawla answered |
75% completion means 2 half lifes so 50% completion means only one half life.
Read the passage given below and answer the following questions:The half-life of a reaction is the time required for the concentration of reactant to decrease by half, i.e.,[A]t = [A]/2For first order reaction,t1/2 = 0.693/kThis means t1/2 is independent of initial concentration. Figure shows that typical variation of concentration of reactant exhibiting first order kinetics. It may be noted that though the major portion of the first order kinetics may be over in a finite time, but the reaction will never cease as the concentration of reactant will be zero only at infinite time.
The following questions are multiple choice questions. Choose the most appropriate answer:Q. A first order reaction has a rate constant k = 3.01 x 10-3 /s. How long it will take to decompose half of the reactant?- a)2.303 s
- b)23.03 s
- c)230.3 s
- d)2303 s
Correct answer is option 'C'. Can you explain this answer?
Read the passage given below and answer the following questions:
The half-life of a reaction is the time required for the concentration of reactant to decrease by half, i.e.,
[A]t = [A]/2
For first order reaction,
t1/2 = 0.693/k
This means t1/2 is independent of initial concentration. Figure shows that typical variation of concentration of reactant exhibiting first order kinetics. It may be noted that though the major portion of the first order kinetics may be over in a finite time, but the reaction will never cease as the concentration of reactant will be zero only at infinite time.
The following questions are multiple choice questions. Choose the most appropriate answer:
Q. A first order reaction has a rate constant k = 3.01 x 10-3 /s. How long it will take to decompose half of the reactant?
a)
2.303 s
b)
23.03 s
c)
230.3 s
d)
2303 s
|
Ambition Institute answered |
The following mechanism has been proposed for the exothermic catalyzed complex reaction.
If k1 is much smaller than k2. The most suitable qualitative plot of potential energy (P.E.) versus reaction coordinate for the above reaction.- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
The following mechanism has been proposed for the exothermic catalyzed complex reaction.
If k1 is much smaller than k2. The most suitable qualitative plot of potential energy (P.E.) versus reaction coordinate for the above reaction.
a)
b)
c)
d)
|
Imk Pathsala answered |
Since
K1 <<< K2 = most Imp. peack will be higher
The rate law for a reaction between the substances A and 8 is given byrate = k[A]n [B]mIf concentration of A is doubled and that of 8 is halved, the new rate as compared to the earlier rate would be - a)
- b)(m + n)
- c)(n - m)
- d)2(n-m)
Correct answer is option 'D'. Can you explain this answer?
The rate law for a reaction between the substances A and 8 is given by
rate = k[A]n [B]m
If concentration of A is doubled and that of 8 is halved, the new rate as compared to the earlier rate would be
a)
b)
(m + n)
c)
(n - m)
d)
2(n-m)
|
Top Rankers answered |
The unit of rate constant for a first order reaction is- a)Mol/L
- b)Mol2 / L2 / S2
- c)S-1
- d)Mol/L/S
Correct answer is option 'C'. Can you explain this answer?
The unit of rate constant for a first order reaction is
a)
Mol/L
b)
Mol2 / L2 / S2
c)
S-1
d)
Mol/L/S
|
|
Nikita Singh answered |
The correct answer is Option C.
Let R be the rate of reaction.
For first order reaction,
R=K[A]1
⇒K=R[A]-1
Whereas, K and [A] are rate constant and initial concentration of reactant respectively.
Therefore,
Unit of rate constant =(mol L-1)1-nsec-1
For first order reaction, n=1
Unit of rate constant = sec-1
Hence the unit of rate constant for first order reaction is sec-1.
For first order reaction,
R=K[A]1
⇒K=R[A]-1
Whereas, K and [A] are rate constant and initial concentration of reactant respectively.
Therefore,
Unit of rate constant =(mol L-1)1-nsec-1
For first order reaction, n=1
Unit of rate constant = sec-1
Hence the unit of rate constant for first order reaction is sec-1.
The rate is independent of the concentration of the reactants in- a)First order
- b)Second order
- c)Third order
- d)Zero order
Correct answer is option 'D'. Can you explain this answer?
The rate is independent of the concentration of the reactants in
a)
First order
b)
Second order
c)
Third order
d)
Zero order
|
|
Naiham Difoesa answered |
Bcz rate is proportional to the zeroth power of the concentration of reactant.
Direction (Q. Nos. 1-13) This section contains multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correctQ. In the following reaction, which has maximum rate w.r.t. rate of disappearance of NH3?4NH3 + 502 → 4NO + 6H2O- a)O2
- b)NO
- c)H2O
- d)Equal
Correct answer is option 'C'. Can you explain this answer?
Direction (Q. Nos. 1-13) This section contains multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct
Q. In the following reaction, which has maximum rate w.r.t. rate of disappearance of NH3?
4NH3 + 502 → 4NO + 6H2O
a)
O2
b)
NO
c)
H2O
d)
Equal
|
Nabanita Basu answered |
Understanding the Reaction
The given reaction is:
4NH3 + 5O2 → 4NO + 6H2O
This reaction involves the disappearance of ammonia (NH3) and the appearance of the products NO and H2O.
Rate of Reaction
The rate of a chemical reaction can be expressed in terms of the rate of disappearance of reactants or the rate of appearance of products.
Stoichiometry of the Reaction
- According to the stoichiometry:
- 4 moles of NH3 produce 4 moles of NO and 6 moles of H2O.
- The coefficients in the balanced equation indicate the relative rates of disappearance and appearance.
Rate of Disappearance
- The rate of disappearance of NH3 is given by:
Rate = - (1/4) * d[NH3]/dt
- The rates for O2, NO, and H2O can be expressed similarly:
- O2: Rate = - (1/5) * d[O2]/dt
- NO: Rate = (1/4) * d[NO]/dt
- H2O: Rate = (1/6) * d[H2O]/dt
Comparison of Rates
To find the maximum rate of disappearance, we can compare the rates derived from the balanced equation:
- NH3: - (1/4) (for every 1 mole of disappearance)
- O2: - (1/5) (for every 1 mole of disappearance)
- NO: (1/4) (for every 1 mole of appearance)
- H2O: (1/6) (for every 1 mole of appearance)
The fractions reveal how many moles of each substance are involved in the reaction. The lower the denominator, the higher the rate of disappearance or appearance.
Conclusion
- Among the reactants and products, H2O has the highest coefficient when considering the rate of disappearance of NH3.
- Therefore, the maximum rate of disappearance is related to H2O's formation.
Thus, the correct answer is option 'C' (H2O).
The given reaction is:
4NH3 + 5O2 → 4NO + 6H2O
This reaction involves the disappearance of ammonia (NH3) and the appearance of the products NO and H2O.
Rate of Reaction
The rate of a chemical reaction can be expressed in terms of the rate of disappearance of reactants or the rate of appearance of products.
Stoichiometry of the Reaction
- According to the stoichiometry:
- 4 moles of NH3 produce 4 moles of NO and 6 moles of H2O.
- The coefficients in the balanced equation indicate the relative rates of disappearance and appearance.
Rate of Disappearance
- The rate of disappearance of NH3 is given by:
Rate = - (1/4) * d[NH3]/dt
- The rates for O2, NO, and H2O can be expressed similarly:
- O2: Rate = - (1/5) * d[O2]/dt
- NO: Rate = (1/4) * d[NO]/dt
- H2O: Rate = (1/6) * d[H2O]/dt
Comparison of Rates
To find the maximum rate of disappearance, we can compare the rates derived from the balanced equation:
- NH3: - (1/4) (for every 1 mole of disappearance)
- O2: - (1/5) (for every 1 mole of disappearance)
- NO: (1/4) (for every 1 mole of appearance)
- H2O: (1/6) (for every 1 mole of appearance)
The fractions reveal how many moles of each substance are involved in the reaction. The lower the denominator, the higher the rate of disappearance or appearance.
Conclusion
- Among the reactants and products, H2O has the highest coefficient when considering the rate of disappearance of NH3.
- Therefore, the maximum rate of disappearance is related to H2O's formation.
Thus, the correct answer is option 'C' (H2O).
Reaction kinetics deals with the study of- a)Rate of reaction
- b)Mechanism of reaction
- c)Factors which affects the rate of reaction
- d)All of the mentioned
Correct answer is option 'D'. Can you explain this answer?
Reaction kinetics deals with the study of
a)
Rate of reaction
b)
Mechanism of reaction
c)
Factors which affects the rate of reaction
d)
All of the mentioned
|
Om Desai answered |
Reaction kinetics deals with the study of rate of reaction, their mechanism and the factors which affects the rate of reaction. It specifies all the general characteristics of a chemical reaction.
A Certain Zeroth Order reaction has k = 0.025 Ms-1 for the disappearance of A. What will be the concentration of A after 15s, if the intial concentration is 0.50 M?- a)0.50 M
- b)0.375 M
- c)0.125 M
- d)0.060 M
Correct answer is option 'C'. Can you explain this answer?
A Certain Zeroth Order reaction has k = 0.025 Ms-1 for the disappearance of A. What will be the concentration of A after 15s, if the intial concentration is 0.50 M?
a)
0.50 M
b)
0.375 M
c)
0.125 M
d)
0.060 M
|
Sinjini Tiwari answered |
x (product) formed after 15 s = 0.025 Ms-1 x 15s= 0.375 M Then, reactant left = 0.500 - 0.375= 0.125 M
Direction (Q. Nos. 1-13) This section contains multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correctQ. In the following reaction, which has maximum rate w.r.t. rate of disappearance of NH3?4NH3 + 502 → 4NO + 6H2O- a)O2
- b)NO
- c)H2O
- d)Equal
Correct answer is 'C'. Can you explain this answer?
Direction (Q. Nos. 1-13) This section contains multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct
Q. In the following reaction, which has maximum rate w.r.t. rate of disappearance of NH3?
4NH3 + 502 → 4NO + 6H2O
a)
O2
b)
NO
c)
H2O
d)
Equal
|
Manoj Chauhan answered |
By stoichiometry of the reaction
similarly,
Study of rate of chemical reaction is known as- a)Chemical thermodynamics
- b)Chemical kinetics
- c)Chemical equilibrium
- d)Chemical change
Correct answer is option 'B'. Can you explain this answer?
Study of rate of chemical reaction is known as
a)
Chemical thermodynamics
b)
Chemical kinetics
c)
Chemical equilibrium
d)
Chemical change
|
Sarthak Khanna answered |
Chemical Kinetics deals with rate of a Chemical reactions.
for the given reaction is 
Two reactions with different activation energies have the same rate at room temperature. Which statement correctly describes the rates of these two reactions at the same higher temperature?- a)The reaction with the greater activation energy will be faster
- b)The reaction with the smaller activation energy will be faster
- c)The two reactions will have the same rate at higher temperature also
- d)Temperature range is also required
Correct answer is option 'B'. Can you explain this answer?
Two reactions with different activation energies have the same rate at room temperature. Which statement correctly describes the rates of these two reactions at the same higher temperature?
a)
The reaction with the greater activation energy will be faster
b)
The reaction with the smaller activation energy will be faster
c)
The two reactions will have the same rate at higher temperature also
d)
Temperature range is also required
|
Pranjal Pillai answered |
If temperature is increased to T`,
If 
Thus, reaction with smaller activation energy will be faster on increasing temperature.
Thus, reaction with smaller activation energy will be faster on increasing temperature.
For the reaction, 
Mechanism is given by
Rate law is true, if - a)step I is the slow step
- b)step II is the slow step
- c)step III is the slow step
- d)step I and step II are the slow steps
Correct answer is option 'B'. Can you explain this answer?
For the reaction,
Mechanism is given by
Rate law is true, if
a)
step I is the slow step
b)
step II is the slow step
c)
step III is the slow step
d)
step I and step II are the slow steps
|
Thedot Starry answered |
Yes b is correct because the rate law is written using the slowest step and it matches with second option.
Decomposition of H2O2 was studied by titration against KMnO4 solution. It was found that 0.4 mol of H2O2 was reduced to 0.2 mol in 20 min and to 0.1 mol in 40 min and to 0.05 mol after 1 hr, the order of reaction must be- a)3
- b)1
- c)0
- d)2
Correct answer is option 'B'. Can you explain this answer?
Decomposition of H2O2 was studied by titration against KMnO4 solution. It was found that 0.4 mol of H2O2 was reduced to 0.2 mol in 20 min and to 0.1 mol in 40 min and to 0.05 mol after 1 hr, the order of reaction must be
a)
3
b)
1
c)
0
d)
2
|
Om Kumar answered |
Decomposition of nH2O2 is 1st order.
If a reaction proceeds with a uniform rate throughout, the reaction is- a)Third order
- b)Second order
- c)First order
- d)Zero order
Correct answer is option 'D'. Can you explain this answer?
If a reaction proceeds with a uniform rate throughout, the reaction is
a)
Third order
b)
Second order
c)
First order
d)
Zero order
|
|
Rohan Singh answered |
A reaction with uniform rate means dC/dt is a constant, where C is the concentration of the reactant being consumed.
For an n-th order reaction,
dC/dt = kC^n
As dC/dt is independent of C
dC/dt = k C^0
Thus, n = 0, and the reaction is a zero order reaction.
Which among the following is an example of first order reaction?- a)Acid catalysed hydrolysis of ethyl acetate
- b)Formation and dissociation of ozone
- c)Inversion of cane sugar
- d)Decomposition of nitrogen pentoxide
Correct answer is option 'D'. Can you explain this answer?
Which among the following is an example of first order reaction?
a)
Acid catalysed hydrolysis of ethyl acetate
b)
Formation and dissociation of ozone
c)
Inversion of cane sugar
d)
Decomposition of nitrogen pentoxide
|
Gauri Datta answered |
First Order Reaction:
A first-order reaction is a type of chemical reaction in which the rate of reaction is directly proportional to the concentration of only one reactant. The reaction follows first-order kinetics, which means that the rate equation can be expressed as:
Rate = k[A]
Where:
- Rate is the rate of the reaction
- k is the rate constant
- [A] is the concentration of the reactant A
The given options are:
a) Acid catalysed hydrolysis of ethyl acetate
b) Formation and dissociation of ozone
c) Inversion of cane sugar
d) Decomposition of nitrogen pentoxide
Explanation:
To determine which reaction is an example of a first-order reaction, we need to analyze the rate equation and identify the reactant concentration that affects the rate of the reaction.
a) Acid catalysed hydrolysis of ethyl acetate:
The hydrolysis of ethyl acetate is a second-order reaction because it involves the simultaneous collision of two molecules (ethyl acetate and water). Therefore, it does not follow first-order kinetics.
b) Formation and dissociation of ozone:
The formation and dissociation of ozone involve multiple steps and the concentration of ozone does not directly affect the rate of the reaction. Hence, it is not a first-order reaction.
c) Inversion of cane sugar:
The inversion of cane sugar is a hydrolysis reaction that occurs in the presence of an acid or an enzyme. It follows pseudo-first-order kinetics because the concentration of the acid or enzyme is much higher than the concentration of the cane sugar. Therefore, it is not a first-order reaction.
d) Decomposition of nitrogen pentoxide:
The decomposition of nitrogen pentoxide is a first-order reaction. The rate of reaction is directly proportional to the concentration of nitrogen pentoxide. As the concentration of nitrogen pentoxide decreases, the rate of reaction also decreases. Therefore, option d is the correct answer.
In conclusion, among the given options, the decomposition of nitrogen pentoxide is an example of a first-order reaction.
A first-order reaction is a type of chemical reaction in which the rate of reaction is directly proportional to the concentration of only one reactant. The reaction follows first-order kinetics, which means that the rate equation can be expressed as:
Rate = k[A]
Where:
- Rate is the rate of the reaction
- k is the rate constant
- [A] is the concentration of the reactant A
The given options are:
a) Acid catalysed hydrolysis of ethyl acetate
b) Formation and dissociation of ozone
c) Inversion of cane sugar
d) Decomposition of nitrogen pentoxide
Explanation:
To determine which reaction is an example of a first-order reaction, we need to analyze the rate equation and identify the reactant concentration that affects the rate of the reaction.
a) Acid catalysed hydrolysis of ethyl acetate:
The hydrolysis of ethyl acetate is a second-order reaction because it involves the simultaneous collision of two molecules (ethyl acetate and water). Therefore, it does not follow first-order kinetics.
b) Formation and dissociation of ozone:
The formation and dissociation of ozone involve multiple steps and the concentration of ozone does not directly affect the rate of the reaction. Hence, it is not a first-order reaction.
c) Inversion of cane sugar:
The inversion of cane sugar is a hydrolysis reaction that occurs in the presence of an acid or an enzyme. It follows pseudo-first-order kinetics because the concentration of the acid or enzyme is much higher than the concentration of the cane sugar. Therefore, it is not a first-order reaction.
d) Decomposition of nitrogen pentoxide:
The decomposition of nitrogen pentoxide is a first-order reaction. The rate of reaction is directly proportional to the concentration of nitrogen pentoxide. As the concentration of nitrogen pentoxide decreases, the rate of reaction also decreases. Therefore, option d is the correct answer.
In conclusion, among the given options, the decomposition of nitrogen pentoxide is an example of a first-order reaction.
. What will be the value of instantaneous rate of reaction from the graph?
- a)rinst = 1/ Slope
- b)rinst = Slope
- c)rinst = – Slope
- d)rinst = – 1/ Slope
Correct answer is option 'B'. Can you explain this answer?
. What will be the value of instantaneous rate of reaction from the graph?
a)
rinst = 1/ Slope
b)
rinst = Slope
c)
rinst = – Slope
d)
rinst = – 1/ Slope
|
Preeti Iyer answered |
If you make a graph of concentration of reactant vs time, the instantaneous reaction rate at a given time is the slope of the tangent line at that point in time. It is also the value of the rate law at a specific concentration (dA/dt is rate = k [A]).
For a reaction A + B → C + D if the concentration of A is doubled without altering the concentration of B, the rate gets doubled. If the concentration of B is increased by nine times without altering the concentration of A, the rate gets tripled. The order of the reaction is - a)2
- b)1
- c)3/2
- d)4/3
Correct answer is option 'C'. Can you explain this answer?
For a reaction A + B → C + D if the concentration of A is doubled without altering the concentration of B, the rate gets doubled. If the concentration of B is increased by nine times without altering the concentration of A, the rate gets tripled. The order of the reaction is
a)
2
b)
1
c)
3/2
d)
4/3
|
|
Madhurima Mishra answered |
The reaction A → B represents a chemical reaction where substance A is converted into substance B. This is often referred to as a reactant(s) being converted into a product(s). The reaction can be written in a chemical equation format as:
A → B
A → B
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