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All questions of Chemical Kinetics for Chemistry Exam
For which the following remarks the concentration of a reactant decreases linearly with time. What is the order of the reaction?
- a)1st order.
- b)Fractional order.
- c)2nd order.
- d)Zero order.
Correct answer is option 'D'. Can you explain this answer?
For which the following remarks the concentration of a reactant decreases linearly with time. What is the order of the reaction?
a)
1st order.
b)
Fractional order.
c)
2nd order.
d)
Zero order.
|
Pooja Choudhury answered |
For Zero order reaction concentration of a reactant decreases linearly with time.
The integral form of zero order reactions can be rewritten as
[A]=–kt+[A0]
Comparing this equation with that of a straight line (y = mx + c), an [A] against t graph can be plotted to get a straight line with slope equal to ‘-k’ and intercept equal to [A]0 as shown below.
Examine the following first order consecutive reactions. The rate constant (in s-1 units) for each step. Is given above the arrow mark:
Steady–state approximation can be applied to.- a)I only
- b)III only
- c)II and III only
- d)I and IV only
Correct answer is 'D'. Can you explain this answer?
Examine the following first order consecutive reactions. The rate constant (in s-1 units) for each step. Is given above the arrow mark:
Steady–state approximation can be applied to.
a)
I only
b)
III only
c)
II and III only
d)
I and IV only
|
Veda Institute answered |
Steady state approximation can be applied to an intermediate that have a transient existence I.e. if the rate of deformation is greater than their rate of formation.
thus answer is d.
thus answer is d.
The decomposition of ozone to oxygen,
Where, M is the catalyst molecule.
K’ are rate constants and Ea1 is the act ivat ion energy for the elementary step.
Q.
Assuming k3[O3]>>k2[O2][M], the activation energy of the overall reaction is:
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The decomposition of ozone to oxygen,
Where, M is the catalyst molecule.
K’ are rate constants and Ea1 is the act ivat ion energy for the elementary step.
K’ are rate constants and Ea1 is the act ivat ion energy for the elementary step.
Q.
Assuming k3[O3]>>k2[O2][M], the activation energy of the overall reaction is:
a)
b)
c)
d)
|
|
Veda Institute answered |
D is the correct answer.
ATQ: B gives the resultant reaction in terms of O2, O3 and O only.
The reaction, 2NO(g) + O2(g) →2NO2(g) proceeds via the following steps:
The rate of this reaction is equal to - a)2kb[NO][O2]
- b)(2kakb[NO]2 [O2]) /(ka + kb [O2])
- c)2kb [NO]2 [O2]
- d)ka[NO]2 [O2]
Correct answer is option 'C'. Can you explain this answer?
The reaction, 2NO(g) + O2(g) →2NO2(g) proceeds via the following steps:
The rate of this reaction is equal to
a)
2kb[NO][O2]
b)
(2kakb[NO]2 [O2]) /(ka + kb [O2])
c)
2kb [NO]2 [O2]
d)
ka[NO]2 [O2]
|
|
Veda Institute answered |
Correct answer is C.
2kb[NO][O2]
2kb[NO][O2]
Calculate order of reaction A → product , from the following data:
- a)–1
- b)–2
- c)1
- d)2
Correct answer is option 'B'. Can you explain this answer?
Calculate order of reaction A → product , from the following data:
a)
–1
b)
–2
c)
1
d)
2
|
Dronacharya Institute answered |
By doubling the concentration the rate becomes half from (i) to (ii) also from (ii) to (iii).
Hence, B is correct.
After two average lives, concentration of the reactant is reduced to:
Consider the unimolecular reaction: A(g) → products, were made:
(I) The reaction is second order at low pressure and become first order at high pressure.
(II) The reaction is first order at low pressure and become second order at high pressure.
(III) The reaction is zero order.Which of the fo llowing is correct?
- a)I and II
- b)II and III
- c)Only III
- d)Only I
Correct answer is option 'D'. Can you explain this answer?
Consider the unimolecular reaction: A(g) → products, were made:
(I) The reaction is second order at low pressure and become first order at high pressure.
(II) The reaction is first order at low pressure and become second order at high pressure.
(III) The reaction is zero order.Which of the fo llowing is correct?
(II) The reaction is first order at low pressure and become second order at high pressure.
(III) The reaction is zero order.Which of the fo llowing is correct?
a)
I and II
b)
II and III
c)
Only III
d)
Only I
|
Edurev.iitjam answered |
A + B → AB* → products
where AB* represents the intermediate complex.
→ At low pressures, the concentration of B is low, so the reaction appears to be second-order in A.
where AB* represents the intermediate complex.
→ At low pressures, the concentration of B is low, so the reaction appears to be second-order in A.
→ At high pressures, the concentration of B is high, so the reaction appears to be first order in A.
The value of the rate constant for the gas phase reaction, 2NO2 +F2 →2NO2F is 38 dm3 mol–1s–1 at 300K. The order of the reaction is:- a)0
- b)1
- c)2
- d)3
Correct answer is option 'C'. Can you explain this answer?
The value of the rate constant for the gas phase reaction, 2NO2 +F2 →2NO2F is 38 dm3 mol–1s–1 at 300K. The order of the reaction is:
a)
0
b)
1
c)
2
d)
3
|
Rajeev Sharma answered |
For the gas phase reaction 2NO2 + F2" 2NO2F, the rate constant is k= 38 dm3/mol-s at 27 oC. The reaction is first-order in NO2 and first-order in F2. A) Calculate the number of moles of NO2, F2, and NO2F after 10.0 s if 2.00 mol of NO2 is mixed with
In the Lineweaver–Burk plot of (initial rate)–1 vs. (initial substrate concentration) –1 for an enzyme catalyzed reaction following Michaelis-Menten mechanism, the y–intercept is 5000 M–1 s. If the initial enzyme concentration is 
the turnover number is: - a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
In the Lineweaver–Burk plot of (initial rate)–1 vs. (initial substrate concentration) –1 for an enzyme catalyzed reaction following Michaelis-Menten mechanism, the y–intercept is 5000 M–1 s. If the initial enzyme concentration is the turnover number is:
the turnover number is: a)
b)
c)
d)
|
Asf Institute answered |
For a second-order reaction, what is the unit of the rate of the reaction?
- a)s-1
- b)mol L-1s-1
- c)Lmol-1s-1
- d)mol-2 L2 s-1
Correct answer is option 'C'. Can you explain this answer?
For a second-order reaction, what is the unit of the rate of the reaction?
a)
s-1
b)
mol L-1s-1
c)
Lmol-1s-1
d)
mol-2 L2 s-1
|
Pioneer Academy answered |
The unit for the rate constant of a second order reaction is Lmol−1s−1.
For a second order reaction, rate=k[A]2
molL−1s−1=k(molL−1)2
k=Lmol−1s−1
For a second order reaction, rate=k[A]2
molL−1s−1=k(molL−1)2
k=Lmol−1s−1
For an enzyme-substrate reaction, a plot between 
yields a slope of 40 s. If the enzyme concentration is 2.5 μM , then the catalytic efficiency of the enzyme is:
- a)40 Lmol–1s–1
- b)10–4 Lmol–1s–1
- c)107 Lmol–1s–1
- d)104 Lmol–1s–1
Correct answer is option 'D'. Can you explain this answer?
For an enzyme-substrate reaction, a plot between yields a slope of 40 s. If the enzyme concentration is 2.5 μM , then the catalytic efficiency of the enzyme is:
yields a slope of 40 s. If the enzyme concentration is 2.5 μM , then the catalytic efficiency of the enzyme is:a)
40 Lmol–1s–1
b)
10–4 Lmol–1s–1
c)
107 Lmol–1s–1
d)
104 Lmol–1s–1
|
|
Asf Institute answered |
The concentration of a reactant undergoing decomposition was 0.1, 0.08 and 0.067 mol L–1 after 1.0, 2.0 and 3.0 hr respectively. The order of the reaction is:- a)0
- b)1
- c)2
- d)3
Correct answer is option 'C'. Can you explain this answer?
The concentration of a reactant undergoing decomposition was 0.1, 0.08 and 0.067 mol L–1 after 1.0, 2.0 and 3.0 hr respectively. The order of the reaction is:
a)
0
b)
1
c)
2
d)
3
|
Vikram Kapoor answered |
Correct Answer :- c
Explanation : A ----->k P
-d[A]/dt = k[A](0 to n)
-{[A]0 - [A]}/(t2 - t1) = k[A](0 to n)
{[A] - [A]0}/(t2 - t1) = k[A](0 to n)
[A]0 = concentration at t1
and [A] = concentration at t2
{[A] - [A]0}/(t2 - t1) = k[A](0 to n)
(0.1-0.08)/(2-1) = k[0.1]n
k[0.1]n = 0.02......(1)
(0.08-0.067)/(3-2) = k[0.08]n
k[0.08]n = 0.013.........(2)
Dividing (1) by (2)
k[0.1]n/k[0.08]n = 0.02/0.013
{[0.1]/[0.08]}n = 1.5385
[1.25]n = 1.5385
[1.25]n = (1.25)2
n = 2
Sucrose is converted to a mixture of glucose and fructose in a pseudo first order process under alkaline conditions. The reaction has a half˜life of 28.4 min. The time required for the reduction of a 8.0 mM sample of sucrose to 1.0 mM is:- a)56.8 min
- b)170.4 min
- c)85.2 min
- d)227.2 min
Correct answer is option 'C'. Can you explain this answer?
Sucrose is converted to a mixture of glucose and fructose in a pseudo first order process under alkaline conditions. The reaction has a half˜life of 28.4 min. The time required for the reduction of a 8.0 mM sample of sucrose to 1.0 mM is:
a)
56.8 min
b)
170.4 min
c)
85.2 min
d)
227.2 min
|
Sneha Menon answered |
Given data:
Sucrose is converted to a mixture of glucose and fructose in a pseudo first order process.
Halflife of reaction = 28.4 min
Initial concentration of sucrose, [sucrose]₀ = 8.0 mM
Final concentration of sucrose, [sucrose] = 1.0 mM
To find: Time required for the reduction of a 8.0 mM sample of sucrose to 1.0 mM
Calculation:
The given reaction is a pseudo first order reaction, which means it can be represented as:
Rate = k [sucrose]
where k is the rate constant of the reaction.
The halflife of the reaction is given as 28.4 min. This means that the concentration of sucrose will reduce to half its initial value in 28.4 min. Therefore, we can use the following equation to find the rate constant:
t1/2 = ln 2 / k
Substituting the given value of halflife in the above equation, we get:
28.4 = ln 2 / k
k = ln 2 / 28.4
Now, we can use the rate constant to find the time required for the concentration of sucrose to reduce from 8.0 mM to 1.0 mM. We can use the following equation for a first order reaction:
ln [sucrose]₀ / [sucrose] = kt
Substituting the given values in the above equation, we get:
ln 8.0 / 1.0 = (ln 2 / 28.4) t
t = (28.4 / ln 2) ln 8.0 / 1.0
t = 85.2 min
Therefore, the time required for the reduction of a 8.0 mM sample of sucrose to 1.0 mM is 85.2 min. Hence, option (c) is the correct answer.
Sucrose is converted to a mixture of glucose and fructose in a pseudo first order process.
Halflife of reaction = 28.4 min
Initial concentration of sucrose, [sucrose]₀ = 8.0 mM
Final concentration of sucrose, [sucrose] = 1.0 mM
To find: Time required for the reduction of a 8.0 mM sample of sucrose to 1.0 mM
Calculation:
The given reaction is a pseudo first order reaction, which means it can be represented as:
Rate = k [sucrose]
where k is the rate constant of the reaction.
The halflife of the reaction is given as 28.4 min. This means that the concentration of sucrose will reduce to half its initial value in 28.4 min. Therefore, we can use the following equation to find the rate constant:
t1/2 = ln 2 / k
Substituting the given value of halflife in the above equation, we get:
28.4 = ln 2 / k
k = ln 2 / 28.4
Now, we can use the rate constant to find the time required for the concentration of sucrose to reduce from 8.0 mM to 1.0 mM. We can use the following equation for a first order reaction:
ln [sucrose]₀ / [sucrose] = kt
Substituting the given values in the above equation, we get:
ln 8.0 / 1.0 = (ln 2 / 28.4) t
t = (28.4 / ln 2) ln 8.0 / 1.0
t = 85.2 min
Therefore, the time required for the reduction of a 8.0 mM sample of sucrose to 1.0 mM is 85.2 min. Hence, option (c) is the correct answer.
The t1/2 of a reaction is doubled as the initial concentration of a reactant is increased 4 times. The order of the reaction is:
- a)0.5
- b)1
- c)0
- d)3
Correct answer is option 'C'. Can you explain this answer?
The t1/2 of a reaction is doubled as the initial concentration of a reactant is increased 4 times. The order of the reaction is:
a)
0.5
b)
1
c)
0
d)
3
|
Chirag Verma answered |
Correct Answer :- c
Explanation : A → P . For zero order reaction, rate= k. So, unit of k is
moles/sec.
- d[A]/dt = k
Upon integration between limits, [ A ] − [ A0 ] = − kt
At t1/2, [A]=[A0]/2.
Hence, t1/2 = [A0]/2k
Increased by 4 times, so it will become double
A zero-order reaction is 25% complete in 30seconds. What time does it take for 50% completion?- a)40s
- b)70s
- c)50s
- d)60s
Correct answer is option 'D'. Can you explain this answer?
A zero-order reaction is 25% complete in 30seconds. What time does it take for 50% completion?
a)
40s
b)
70s
c)
50s
d)
60s
|
|
Pioneer Academy answered |
t25% = (1-0.75)/K
K = 0.25/30
t50% = (1 x 30)/(2 x 0.25)
t50% = 15/0.25
t50% = 60s.
K = 0.25/30
t50% = (1 x 30)/(2 x 0.25)
t50% = 15/0.25
t50% = 60s.
A reaction follows second order rate law,
- a)A plot of [A] versus t is a straight line
- b)A plot of 1/[A] versus t is a straight line.
- c)A plot of ln[A] versus t is a straight line.
- d)A plot of e[A] versus t is a straight line.
Correct answer is option 'B'. Can you explain this answer?
A reaction follows second order rate law,
a)
A plot of [A] versus t is a straight line
b)
A plot of 1/[A] versus t is a straight line.
c)
A plot of ln[A] versus t is a straight line.
d)
A plot of e[A] versus t is a straight line.
|
|
Shruti Datta answered |
The Arrhenius parameters for the thermal decomposition of NaCl, 2NOCl(g)→2NO(g)+CI2 are A = 1013 M–1s–1, E0 = 105kJ mol–1 and RT = 2.5 kJ mol–1. The enthalpy (in kJ mol–1) of the activated complex will be.- a)110
- b)105
- c)102.5
- d)100
Correct answer is option 'D'. Can you explain this answer?
The Arrhenius parameters for the thermal decomposition of NaCl, 2NOCl(g)→2NO(g)+CI2 are A = 1013 M–1s–1, E0 = 105kJ mol–1 and RT = 2.5 kJ mol–1. The enthalpy (in kJ mol–1) of the activated complex will be.
a)
110
b)
105
c)
102.5
d)
100
|
Niharika Kulkarni answered |
Arrhenius Parameters for Thermal Decomposition of NaCl
- The Arrhenius parameters for the thermal decomposition of NaCl, 2NOCl(g)2NO(g) CI2 are:
- A = 1013 M1s1
- E0 = 105 kJ mol1
- RT = 2.5 kJ mol1
Enthalpy of the Activated Complex
- The enthalpy (in kJ mol1) of the activated complex can be calculated using the Arrhenius equation:
k = Ae(-E0/RT)
- where k is the rate constant, A is the pre-exponential factor, E0 is the activation energy, R is the gas constant (8.314 J K-1 mol-1), and T is the temperature in Kelvin.
- Taking the natural logarithm of both sides of the equation, we get:
ln(k) = ln(A) - (E0/RT)
- Rearranging the equation, we get:
(E0/RT) = ln(A/k) + ln(k)
- Substituting the given values, we get:
(E0/RT) = ln(1013 M1s1/2) + ln(2.5 kJ mol1)
- Simplifying the equation, we get:
(E0/RT) = 20.83 + 1.10
- Therefore, the enthalpy (in kJ mol1) of the activated complex is:
(E0/RT) = 21.93 kJ mol1
- Rounded to the nearest whole number, the answer is:
(E0/RT) = 22 kJ mol1
- Therefore, the correct answer is option 'D' (100 kJ mol1).
- The Arrhenius parameters for the thermal decomposition of NaCl, 2NOCl(g)2NO(g) CI2 are:
- A = 1013 M1s1
- E0 = 105 kJ mol1
- RT = 2.5 kJ mol1
Enthalpy of the Activated Complex
- The enthalpy (in kJ mol1) of the activated complex can be calculated using the Arrhenius equation:
k = Ae(-E0/RT)
- where k is the rate constant, A is the pre-exponential factor, E0 is the activation energy, R is the gas constant (8.314 J K-1 mol-1), and T is the temperature in Kelvin.
- Taking the natural logarithm of both sides of the equation, we get:
ln(k) = ln(A) - (E0/RT)
- Rearranging the equation, we get:
(E0/RT) = ln(A/k) + ln(k)
- Substituting the given values, we get:
(E0/RT) = ln(1013 M1s1/2) + ln(2.5 kJ mol1)
- Simplifying the equation, we get:
(E0/RT) = 20.83 + 1.10
- Therefore, the enthalpy (in kJ mol1) of the activated complex is:
(E0/RT) = 21.93 kJ mol1
- Rounded to the nearest whole number, the answer is:
(E0/RT) = 22 kJ mol1
- Therefore, the correct answer is option 'D' (100 kJ mol1).
How many times will the rate of the elementary reaction 3X + Y → X2Y change if the concentration of the substance X is doubled and that of Y is halved?- a)r2 = 4r1
- b)r2 = 4.5r1
- c)r2 = 5r1
- d)r2 = 2r1
Correct answer is option 'A'. Can you explain this answer?
How many times will the rate of the elementary reaction 3X + Y → X2Y change if the concentration of the substance X is doubled and that of Y is halved?
a)
r2 = 4r1
b)
r2 = 4.5r1
c)
r2 = 5r1
d)
r2 = 2r1
|
|
Edurev.iitjam answered |
Since it is an elementary reaction, its rate law r1 = k [A] 3[B]
When the concentrations are changed the new rate will be r2 = k (2[A])3([B]/2) = 4k[A]3[B]
So, r2 = 4r1.
When the concentrations are changed the new rate will be r2 = k (2[A])3([B]/2) = 4k[A]3[B]
So, r2 = 4r1.
In a zero-order reaction for every 10° rise of temperature, the rate is doubled. If the temperature is increased from 10°C to 100°C, the rate of the reaction will become - a)64 times
- b)128 times
- c)256 times
- d)512 times
Correct answer is option 'D'. Can you explain this answer?
In a zero-order reaction for every 10° rise of temperature, the rate is doubled. If the temperature is increased from 10°C to 100°C, the rate of the reaction will become
a)
64 times
b)
128 times
c)
256 times
d)
512 times
|
Kaavya Sengupta answered |
For 10^o rise in temperature, n = 1
so rate = 2^n = 2^1 = 2
When temperature is increased from 10^o C to 100^o C change in temperature = 100 -10 = 90^oC i.e n=9 so, rate = 29 = 512 times.
Alternate method: With every 10^o C rise in temperature, rate becomes double, so
r^r/r = 2 (100 - 10 /10) = 2^9 = 512 times
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
|
Arnab Pillai answered |
For every 10o C rise of temperature, the rate is doubled. Thus, temperature coefficient of the reaction = 2 when temperature is increased by 50o rate becomes
What time does it take for reactants to reduce to 3/4 of initial concentration if the rate constant is 7.5 x 10-3 s-1?- a)38.4s
- b)40.2s
- c)39.3s
- d)36.8s
Correct answer is option 'A'. Can you explain this answer?
What time does it take for reactants to reduce to 3/4 of initial concentration if the rate constant is 7.5 x 10-3 s-1?
a)
38.4s
b)
40.2s
c)
39.3s
d)
36.8s
|
|
Asf Institute answered |
Given,
K=7.5 x 10-3 s-1
K = (2.303/t) x log([R]0/[R])
t = (2.303/7.5 x 10-3) x log([100]/[25])
t = (2.303/7.5 x 10-3) x 0.6
t = 38.4s.
K=7.5 x 10-3 s-1
K = (2.303/t) x log([R]0/[R])
t = (2.303/7.5 x 10-3) x log([100]/[25])
t = (2.303/7.5 x 10-3) x 0.6
t = 38.4s.
respectively. The rate of the reaction when the substrate concentration is 
is
In a consecutive first order reaction,(where k1 and k2 are the respective rate constants) species B has transient existence. Therefore,- a)k1 ≈ k2
- b)k1 = 2 k2
- c)k1 >> k2
- d)k1 << k2
Correct answer is option 'D'. Can you explain this answer?
In a consecutive first order reaction,
(where k1 and k2 are the respective rate constants) species B has transient existence. Therefore,
a)
k1 ≈ k2
b)
k1 = 2 k2
c)
k1 >> k2
d)
k1 << k2
|
|
Anisha Pillai answered |
We cannot determine the values of k1 and k2 based on the given information.
However, we can make some observations based on the fact that species B has a transient existence.
In a consecutive first order reaction, the reaction mechanism involves a series of consecutive steps, each with its own rate constant. Species B is likely an intermediate that is formed in the first step and consumed in the second step.
If species B has a very short lifespan, it means that the second step is much faster than the first step. This implies that k2 is much larger than k1.
We can also say that the concentration of species B at any given time is proportional to k1/k2. This is because the rate of formation of B is proportional to k1[A], while the rate of consumption of B is proportional to k2[B]. At equilibrium, the rate of formation of B equals the rate of consumption of B, so k1[A] = k2[B]. Solving for [B] gives [B] = k1/k2 [A].
In summary, we cannot determine the values of k1 and k2, but we can say that k2 is much larger than k1 and that the concentration of species B is proportional to k1/k2.
However, we can make some observations based on the fact that species B has a transient existence.
In a consecutive first order reaction, the reaction mechanism involves a series of consecutive steps, each with its own rate constant. Species B is likely an intermediate that is formed in the first step and consumed in the second step.
If species B has a very short lifespan, it means that the second step is much faster than the first step. This implies that k2 is much larger than k1.
We can also say that the concentration of species B at any given time is proportional to k1/k2. This is because the rate of formation of B is proportional to k1[A], while the rate of consumption of B is proportional to k2[B]. At equilibrium, the rate of formation of B equals the rate of consumption of B, so k1[A] = k2[B]. Solving for [B] gives [B] = k1/k2 [A].
In summary, we cannot determine the values of k1 and k2, but we can say that k2 is much larger than k1 and that the concentration of species B is proportional to k1/k2.
the rate of formation of X2 is:
Examine the following first order consecutive reactions. The rate constant (in s-1 units) for each step. Is given above the arrow mark:
Steady–state approximation can be applied to.- a)I only
- b)III only
- c)II and III only
- d)I and IV only
Correct answer is option 'D'. Can you explain this answer?
Examine the following first order consecutive reactions. The rate constant (in s-1 units) for each step. Is given above the arrow mark:
Steady–state approximation can be applied to.
a)
I only
b)
III only
c)
II and III only
d)
I and IV only
|
|
Veda Institute answered |
Steady state approximation can be applied to an intermediate that have a transient existence I.e. if the rate of deformation is greater than their rate of formation.
thus answer is d.
thus answer is d.
Which of the following assumption was not made in rapid equilibrium model?- a)Flexible nature of enzymes
- b)[S] >> [E]
- c)[E]+[S]⇌[ES]
- d)It is the initial substrate concentration [So] that determines efficacy of an enzyme to mediate a reaction
Correct answer is option 'A'. Can you explain this answer?
Which of the following assumption was not made in rapid equilibrium model?
a)
Flexible nature of enzymes
b)
[S] >> [E]
c)
[E]+[S]⇌[ES]
d)
It is the initial substrate concentration [So] that determines efficacy of an enzyme to mediate a reaction
|
|
Pioneer Academy answered |
The flexible nature of enzymes was not dealt by lock and key model, but the induced fit model. This model dealt with change in shape after enzyme substrate interaction. The following assumptions were made in rapid equilibrium model:
* The substrate concentration will always be higher than enzyme concentration. [E] >> [S]
* An equilibrium state is assumed between reactants and products. [E]+[S]⇌[ES]
* It is the initial substrate concentration [So] that determines the efficacy of an enzyme to mediate a reaction.
* The substrate concentration will always be higher than enzyme concentration. [E] >> [S]
* An equilibrium state is assumed between reactants and products. [E]+[S]⇌[ES]
* It is the initial substrate concentration [So] that determines the efficacy of an enzyme to mediate a reaction.
State true or false.
Gas phase decomposition of N2O follows first order mechanism for low concentrations of N2O and second order mechanism for high concentrations of N2O.- a)True
- b)False
Correct answer is option 'A'. Can you explain this answer?
State true or false.
Gas phase decomposition of N2O follows first order mechanism for low concentrations of N2O and second order mechanism for high concentrations of N2O.
Gas phase decomposition of N2O follows first order mechanism for low concentrations of N2O and second order mechanism for high concentrations of N2O.
a)
True
b)
False
|
|
Pioneer Academy answered |
The rate expression is, rN2 = 
At low concentrations of N2O, kNO2 << 1 and the reactions follows second order. At high concentrations of N2O, kNO2 >> 1 and the reaction follows first order.
At low concentrations of N2O, kNO2 << 1 and the reactions follows second order. At high concentrations of N2O, kNO2 >> 1 and the reaction follows first order.
What is the rate law for acid hydrolysis of an ester such as CH3COOC2H5 in aqueous solution?- a)k [CH3COOC2H5]
- b)k [CH3COOC2H5] [H2O]
- c)k [CH3COOC2H5]2
- d)k
Correct answer is option 'A'. Can you explain this answer?
What is the rate law for acid hydrolysis of an ester such as CH3COOC2H5 in aqueous solution?
a)
k [CH3COOC2H5]
b)
k [CH3COOC2H5] [H2O]
c)
k [CH3COOC2H5]2
d)
k
|
|
Vivek Khatri answered |
Acid hydrolysis of ester, CH3COOC2H5 + H2O → CH3COOH + C2H5OH
The order of the reaction may be altered sometimes by taking reactant in excess compared to the other.
The rate law R= k [CH3COOC2H5] [H2O] however water is present in excess.
So, R= k [CH3COOC2H5].
The order of the reaction may be altered sometimes by taking reactant in excess compared to the other.
The rate law R= k [CH3COOC2H5] [H2O] however water is present in excess.
So, R= k [CH3COOC2H5].
Experimentally determined rate law for the chemical reaction,
The rate determining step consistent with the rate law is:- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Experimentally determined rate law for the chemical reaction,
The rate determining step consistent with the rate law is:
a)
b)
c)
d)
|
|
Sona Harnathka answered |
Bcz...slowest step is rate determining step
The specific rate constant of decomposition of a compound is represented by:
The activation energy of decomposition for this compound at 300 K is
- a)24 kcal/mol
- b)12 kcal/mol
- c)24 cal/mol
- d)12 cal/mol
Correct answer is option 'A'. Can you explain this answer?
The specific rate constant of decomposition of a compound is represented by:
The activation energy of decomposition for this compound at 300 K is
a)
24 kcal/mol
b)
12 kcal/mol
c)
24 cal/mol
d)
12 cal/mol
|
|
Asf Institute answered |
Concept:
The relationship between reaction rate and temperature:
- According to Arrhenius, the molecules in a reaction must possess a certain amount of energy in order to undergo a chemical reaction.
- This means that before the reaction occurs, the energy of the molecules must be raised in order to bring any transformation.
- This energy-rich state is called the activated state of the molecules.
- The extra energy required to attain this minimum energy has to be supplied externally and is known as the activation energy
- Reactions with low requirements of activation energy have faster rates.
- The higher the activation energy, the slower is the reaction rate.
- The relation between activation energy and temperature and the rate of reaction is given by the Arrhenius equation:
The half–line of a order reaction varies with temperature according to:- a) In t1/2α1 / T
- b)In t1/ 2 αT
- c)t1/ 2α1 / T2
- d)t1/ 2αT2
Correct answer is option 'A'. Can you explain this answer?
The half–line of a order reaction varies with temperature according to:
a)
In t1/2α1 / T
b)
In t1/ 2 αT
c)
t1/ 2α1 / T2
d)
t1/ 2αT2
|
|
Vedika Singh answered |
k = ln2/t1/2 ----(i)
k = Ae(-Ea/RT) ------(ii)
Taking ln on both sides
lnk = ln((ln2)/ t1/2)
lnk = lnA + (-Ea/R) x 1/T
Hence comparing the two equations, we get:
-ln(t1/2) α 1/T
Therefore A is the correct answer.
What is the time taken to complete 75 percent of the reaction if the rate of the first-order reaction is 0.023 min-1?- a)60.28 minutes
- b)69.28 minutes
- c)50.37 minutes
- d)65.97 minutes
Correct answer is option 'A'. Can you explain this answer?
What is the time taken to complete 75 percent of the reaction if the rate of the first-order reaction is 0.023 min-1?
a)
60.28 minutes
b)
69.28 minutes
c)
50.37 minutes
d)
65.97 minutes
|
|
Ishani Dasgupta answered |
Given:
- Rate of the first-order reaction = 0.023 min-1
- We need to find the time taken to complete 75 percent of the reaction
Formula:
The integrated rate law for a first-order reaction is given by the equation:
ln([A]t/[A]0) = -kt
Where:
[A]t is the concentration of reactant at time t
[A]0 is the initial concentration of reactant
k is the rate constant
t is the time
Explanation:
To find the time taken to complete 75 percent of the reaction, we need to find the time at which the concentration of the reactant is reduced to 25 percent of its initial concentration.
Let's assume the initial concentration of the reactant is 100 units (for simplicity).
Step 1: Calculate the half-life of the reaction
The half-life (t1/2) of a first-order reaction can be calculated using the equation:
t1/2 = (0.693/k)
Given that k = 0.023 min-1, we can substitute this value into the equation to find the half-life:
t1/2 = (0.693/0.023) = 30.13 minutes
The half-life represents the time it takes for the concentration of the reactant to reduce to half its initial value.
Step 2: Calculate the time taken to reach 25 percent of the initial concentration
Since the half-life is 30.13 minutes, it represents the time taken to reach 50 percent of the initial concentration. We need to find the time taken to reach 25 percent.
To do this, we can multiply the half-life by the natural logarithm of 2 (ln(2)) and divide it by ln(4) to find the time taken to reach 25 percent.
t25% = (t1/2 * ln(2))/ln(4)
t25% = (30.13 * 0.693)/1.386
t25% ≈ 15.07 minutes
Step 3: Calculate the time taken to complete 75 percent of the reaction
Since we are looking for the time taken to complete 75 percent of the reaction, we can subtract the time taken to reach 25 percent from the total time taken to complete the reaction.
t75% = total reaction time - t25%
t75% = 30.13 - 15.07
t75% ≈ 15.06 minutes
Therefore, the time taken to complete 75 percent of the reaction is approximately 15.06 minutes.
Conclusion:
The correct answer is option A) 15.06 minutes.
- Rate of the first-order reaction = 0.023 min-1
- We need to find the time taken to complete 75 percent of the reaction
Formula:
The integrated rate law for a first-order reaction is given by the equation:
ln([A]t/[A]0) = -kt
Where:
[A]t is the concentration of reactant at time t
[A]0 is the initial concentration of reactant
k is the rate constant
t is the time
Explanation:
To find the time taken to complete 75 percent of the reaction, we need to find the time at which the concentration of the reactant is reduced to 25 percent of its initial concentration.
Let's assume the initial concentration of the reactant is 100 units (for simplicity).
Step 1: Calculate the half-life of the reaction
The half-life (t1/2) of a first-order reaction can be calculated using the equation:
t1/2 = (0.693/k)
Given that k = 0.023 min-1, we can substitute this value into the equation to find the half-life:
t1/2 = (0.693/0.023) = 30.13 minutes
The half-life represents the time it takes for the concentration of the reactant to reduce to half its initial value.
Step 2: Calculate the time taken to reach 25 percent of the initial concentration
Since the half-life is 30.13 minutes, it represents the time taken to reach 50 percent of the initial concentration. We need to find the time taken to reach 25 percent.
To do this, we can multiply the half-life by the natural logarithm of 2 (ln(2)) and divide it by ln(4) to find the time taken to reach 25 percent.
t25% = (t1/2 * ln(2))/ln(4)
t25% = (30.13 * 0.693)/1.386
t25% ≈ 15.07 minutes
Step 3: Calculate the time taken to complete 75 percent of the reaction
Since we are looking for the time taken to complete 75 percent of the reaction, we can subtract the time taken to reach 25 percent from the total time taken to complete the reaction.
t75% = total reaction time - t25%
t75% = 30.13 - 15.07
t75% ≈ 15.06 minutes
Therefore, the time taken to complete 75 percent of the reaction is approximately 15.06 minutes.
Conclusion:
The correct answer is option A) 15.06 minutes.
If the fermentation of sugar in an enzymatic solution that in 0.12 M. the concentration of the sugar is reduced to 0.06 M in 10 h and to 0.03 M in 20 h. What is the order of the reaction:- a)1
- b)2
- c)3
- d)0
Correct answer is option 'A'. Can you explain this answer?
If the fermentation of sugar in an enzymatic solution that in 0.12 M. the concentration of the sugar is reduced to 0.06 M in 10 h and to 0.03 M in 20 h. What is the order of the reaction:
a)
1
b)
2
c)
3
d)
0
|
|
Aashna Shah answered |
Ans.
Option (a)
Half life is independent to the initial concentration of the reactant.
If the concept of half-life is generalized to quarter–life of a first order chemical reaction, it will e equal to:- a)In 2/k
- b)In 4/k
- c)4/k
- d)1/4k
Correct answer is option 'B'. Can you explain this answer?
If the concept of half-life is generalized to quarter–life of a first order chemical reaction, it will e equal to:
a)
In 2/k
b)
In 4/k
c)
4/k
d)
1/4k
|
Bijoy Kapoor answered |
The half-life of a reaction is the time required for the reactant concentration to decrease to one-half its initial value. The half-life of a first-order reaction is a constant that is related to the rate constant for the reaction: t1/2 = 0.693/k. Radioactive decay reactions are first-order reactions.
In radical chain polymerization, the quantity given by the rate of monomer depletion, divided by the rate of propagating radical formation is called:
- a)Kinetic chain length.
- b)Propagation efficiency.
- c)Propagation rate constant
- d)Polymerization time.
Correct answer is option 'A'. Can you explain this answer?
In radical chain polymerization, the quantity given by the rate of monomer depletion, divided by the rate of propagating radical formation is called:
a)
Kinetic chain length.
b)
Propagation efficiency.
c)
Propagation rate constant
d)
Polymerization time.
|
|
Edurev.iitjam answered |
What is the formula to calculate the time taken for the completion of a zero-order reaction?- a)t100% = [A]0/k
- b)t100% = [A]0/2k
- c)t100% = 2[A]0/k
- d)t100% = [A]0/3k
Correct answer is option 'A'. Can you explain this answer?
What is the formula to calculate the time taken for the completion of a zero-order reaction?
a)
t100% = [A]0/k
b)
t100% = [A]0/2k
c)
t100% = 2[A]0/k
d)
t100% = [A]0/3k
|
|
Ishani Dasgupta answered |
Calculation of time taken for the completion of a zero-order reaction:
Explanation:
A zero-order reaction is a reaction where the rate of the reaction is independent of the concentration of the reactant. In this type of reaction, the rate constant (k) remains constant throughout the reaction. The formula to calculate the time taken for the completion of a zero-order reaction is:
t100% = [A]0/k
Where:
t100% = time taken for the completion of the reaction
[A]0 = initial concentration of the reactant
k = rate constant of the reaction
Therefore, option 'A' is the correct formula to calculate the time taken for the completion of a zero-order reaction.
Explanation:
A zero-order reaction is a reaction where the rate of the reaction is independent of the concentration of the reactant. In this type of reaction, the rate constant (k) remains constant throughout the reaction. The formula to calculate the time taken for the completion of a zero-order reaction is:
t100% = [A]0/k
Where:
t100% = time taken for the completion of the reaction
[A]0 = initial concentration of the reactant
k = rate constant of the reaction
Therefore, option 'A' is the correct formula to calculate the time taken for the completion of a zero-order reaction.
A reaction A → D , involves following mechanism:
The rate law of the reaction may be given as:
- a)Rate = k1[A]
- b)Rate = k2[B]
- c)Rate = k3[C]
- d)Rate = k1 k2 k3[B] [C]
Correct answer is option 'B'. Can you explain this answer?
A reaction A → D , involves following mechanism:
The rate law of the reaction may be given as:
a)
Rate = k1[A]
b)
Rate = k2[B]
c)
Rate = k3[C]
d)
Rate = k1 k2 k3[B] [C]
|
Pie Academy answered |
Correct option is B.
Rate always depends on the slowest step.
So, v=k2[B] is true.
Rate always depends on the slowest step.
So, v=k2[B] is true.
For the reaction A + H2O → products, find the rate of the reaction when [A] = 0.75 M, k= 0.02.- a)0.077 s-1
- b)0.085 s-1
- c)0.015 s-1
- d)0.026 s-1
Correct answer is option 'C'. Can you explain this answer?
For the reaction A + H2O → products, find the rate of the reaction when [A] = 0.75 M, k= 0.02.
a)
0.077 s-1
b)
0.085 s-1
c)
0.015 s-1
d)
0.026 s-1
|
|
Vivek Khatri answered |
Given,
[A] = 0.75 M, k= 0.02
The reaction belongs to pseudo first order reaction so, the unit is s-1
R= k [A]= 0.02 × 0.75= 0.015 s-1.
[A] = 0.75 M, k= 0.02
The reaction belongs to pseudo first order reaction so, the unit is s-1
R= k [A]= 0.02 × 0.75= 0.015 s-1.
For the reaction shown below,
The value of k1 is 
. If the reaction starts from X, the ratio of the concentration of Y and Z at any given time during the course of the reaction is found to be 
The value of k2 is- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
For the reaction shown below,
The value of k1 is . If the reaction starts from X, the ratio of the concentration of Y and Z at any given time during the course of the reaction is found to be The value of k2 is
. If the reaction starts from X, the ratio of the concentration of Y and Z at any given time during the course of the reaction is found to be The value of k2 isa)
b)
c)
d)
|
|
Edurev.iitjam answered |
Option C: 4×10−44×10−4 s⁻¹
and 
The activation energy for the bimolecular reaction A+BC→AB+C is E0 in the gas phase. If the reaction is carried out in a confined volume of λ3 , the activation energy is expected to:- a)Remain unchanged.
- b)Increase with decreasing λ
- c)Decrease with decreasing λ
- d)Oscillate with decreasing λ
Correct answer is option 'A'. Can you explain this answer?
The activation energy for the bimolecular reaction A+BC→AB+C is E0 in the gas phase. If the reaction is carried out in a confined volume of λ3 , the activation energy is expected to:
a)
Remain unchanged.
b)
Increase with decreasing λ
c)
Decrease with decreasing λ
d)
Oscillate with decreasing λ
|
|
Vikram Kapoor answered |
The minimum energy requirement that must be met for a chemical reaction to occur is called the activation energy. The activation energy depends on nature of the reacting species or reactants. It does not depends on volume that is it will remain constant irrespective of the volume.
So, (A) remain unchanged -- is the correct option
Consider the reaction H2 + C2 H2 → C2 H6 the molecular diameters of H2 and C2H4 
The pre-exponential factor in the rate constant calculated using collision theory in m3 (mol) –1 s–1 is approximately (For this reaction at 300K, 
where the symbols have their usual meanings):- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Consider the reaction H2 + C2 H2 → C2 H6 the molecular diameters of H2 and C2H4 The pre-exponential factor in the rate constant calculated using collision theory in m3 (mol) –1 s–1 is approximately (For this reaction at 300K, where the symbols have their usual meanings):
The pre-exponential factor in the rate constant calculated using collision theory in m3 (mol) –1 s–1 is approximately (For this reaction at 300K, where the symbols have their usual meanings):a)
b)
c)
d)
|
Aparna Ramesh answered |
what actually is the question?
Acid hydrolysis of ester in first order reaction and rate constant is given by,
Where, V0, Vt and V∞ , are the volume of standard NaOH required to neutralize acid present as a given time, if ester is 50% neutralized then - a)V∞ = Vt
- b)V∞ = (Vt - V0)
- c)V∞ = 2Vt - V0
- d)V∞ = 2Vt + V0
Correct answer is option 'C'. Can you explain this answer?
Acid hydrolysis of ester in first order reaction and rate constant is given by,
Where, V0, Vt and V∞ , are the volume of standard NaOH required to neutralize acid present as a given time, if ester is 50% neutralized then
a)
V∞ = Vt
b)
V∞ = (Vt - V0)
c)
V∞ = 2Vt - V0
d)
V∞ = 2Vt + V0
|
|
Edurev.iitjam answered |
For the parallel reaction A → B and A → C, of rate constants k1 and k2 respectively, both reactions of order 1, the rate expression is given as ____- a)(-rA) = k1CA + k2CA
- b)(-rA) = k1CA – k2CA
- c)(-rA) = k2CA
- d)(-rA) = k1CA
Correct answer is option 'A'. Can you explain this answer?
For the parallel reaction A → B and A → C, of rate constants k1 and k2 respectively, both reactions of order 1, the rate expression is given as ____
a)
(-rA) = k1CA + k2CA
b)
(-rA) = k1CA – k2CA
c)
(-rA) = k2CA
d)
(-rA) = k1CA
|
|
Vivek Khatri answered |
For a zero order reaction,
A forms two products: B and C. The concentration of A decreases as the reaction progresses.
A forms two products: B and C. The concentration of A decreases as the reaction progresses.
State true or false.
Decomposition of nitrogen pentoxide follows second order mechanism, assuming that the kinetics is controlled by a single step mechanism.- a)True
- b)False
Correct answer is option 'B'. Can you explain this answer?
State true or false.
Decomposition of nitrogen pentoxide follows second order mechanism, assuming that the kinetics is controlled by a single step mechanism.
Decomposition of nitrogen pentoxide follows second order mechanism, assuming that the kinetics is controlled by a single step mechanism.
a)
True
b)
False
|
|
Vivek Khatri answered |
The rate of N2O5 decomposition is given by, r = k[N2O5].
Hence, the reaction follows first order.
Hence, the reaction follows first order.
The half-life of a given reaction is doubled if the initial concentration of the reactant is doubled. What is the order of the reaction?- a)0
- b)1
- c)2
- d)3
Correct answer is option 'A'. Can you explain this answer?
The half-life of a given reaction is doubled if the initial concentration of the reactant is doubled. What is the order of the reaction?
a)
0
b)
1
c)
2
d)
3
|
|
Pioneer Academy answered |
Half-life (t1⁄2) is the time required for a quantity to reduce to half of its initial value. The half-life of a zero-order reaction is directly proportional to its initial concentration. They are related as:
t1/2 = [R]0/2k.
t1/2 = [R]0/2k.
Chapter doubts & questions for Chemical Kinetics - Physical Chemistry 2025 is part of Chemistry exam preparation. The chapters have been prepared according to the Chemistry exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Chemistry 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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