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The rate law expressed the relationship of the rate of a reaction to the rate constant and the concentration of the reactants raised to some powers for the general reaction aA + bB → cC + dD
Rate law takes the form r = k [A]X [B]y
where x and y are number that must be determined experimentally k is the rate constant and [A] and [B] are concentration of A & B respectively.
Q.
The initial rate of zero order reaction of the gaseous equation A (g) → 2B (g) is 10-2 M min1 if the initial conc. of A is 0.1 M what would be conc. of B after 60 sec.
- a)0.09 M
- b)0.01 M
- c)0.02 M
- d)0.03 M
Correct answer is option 'C'. Can you explain this answer?
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Solution:
The given reaction is A (g) + 2B (g) → C (g) + D (g)
According to the rate law, the rate of the reaction is given by:
r = k [A]^x [B]^y
Given that the reaction is zero order with respect to A, the rate law becomes:
r = k [A]^0 [B]^y
Since any quantity raised to the power of zero is equal to 1, the rate law simplifies to:
r = k [B]^y
Given that the initial rate of the reaction is 10^-2 M min^-1 and the initial concentration of A is 0.1 M, we can substitute these values into the rate law to find the value of y.
10^-2 = k (0.1)^0 [B]^y
10^-2 = k [B]^y
Now, we need to find the value of y. To do this, we can use the given information that the initial concentration of A is 0.1 M and the initial rate is 10^-2 M min^-1.
Let's consider the time t = 0 and t = 60 seconds. At t = 0, the concentration of A is 0.1 M and the concentration of B is 0 M. After 60 seconds, the concentration of A will remain the same at 0.1 M (since it is a zero-order reaction), and the concentration of B will change.
We can use the integrated rate equation for a zero-order reaction to find the concentration of B at t = 60 seconds:
[B] = [B]0 - kt
Where [B]0 is the initial concentration of B, k is the rate constant, and t is the time.
Since [A]0 = 0.1 M and [B]0 = 0 M, we can substitute these values into the equation:
[B] = 0 - k(60)
[B] = -60k
We need to find the concentration of B after 60 seconds, so we substitute this value into the rate law:
10^-2 = k (-60k)^y
10^-2 = k (-60)^y k^y
10^-2 = k^y (-60)^y
Now we have two equations:
10^-2 = k^y (-60)^y
[B] = -60k
By comparing the two equations, we can find the value of y:
-60k = (-60)^y
Since k is a constant, we can simplify:
-60 = (-60)^y
Now we can solve for y by taking the logarithm of both sides:
log(-60) = y log(-60)
This equation does not have a real solution for y, which means that the given initial rate and concentration values are not consistent with a zero-order reaction. Therefore, the answer cannot be determined.
The given reaction is A (g) + 2B (g) → C (g) + D (g)
According to the rate law, the rate of the reaction is given by:
r = k [A]^x [B]^y
Given that the reaction is zero order with respect to A, the rate law becomes:
r = k [A]^0 [B]^y
Since any quantity raised to the power of zero is equal to 1, the rate law simplifies to:
r = k [B]^y
Given that the initial rate of the reaction is 10^-2 M min^-1 and the initial concentration of A is 0.1 M, we can substitute these values into the rate law to find the value of y.
10^-2 = k (0.1)^0 [B]^y
10^-2 = k [B]^y
Now, we need to find the value of y. To do this, we can use the given information that the initial concentration of A is 0.1 M and the initial rate is 10^-2 M min^-1.
Let's consider the time t = 0 and t = 60 seconds. At t = 0, the concentration of A is 0.1 M and the concentration of B is 0 M. After 60 seconds, the concentration of A will remain the same at 0.1 M (since it is a zero-order reaction), and the concentration of B will change.
We can use the integrated rate equation for a zero-order reaction to find the concentration of B at t = 60 seconds:
[B] = [B]0 - kt
Where [B]0 is the initial concentration of B, k is the rate constant, and t is the time.
Since [A]0 = 0.1 M and [B]0 = 0 M, we can substitute these values into the equation:
[B] = 0 - k(60)
[B] = -60k
We need to find the concentration of B after 60 seconds, so we substitute this value into the rate law:
10^-2 = k (-60k)^y
10^-2 = k (-60)^y k^y
10^-2 = k^y (-60)^y
Now we have two equations:
10^-2 = k^y (-60)^y
[B] = -60k
By comparing the two equations, we can find the value of y:
-60k = (-60)^y
Since k is a constant, we can simplify:
-60 = (-60)^y
Now we can solve for y by taking the logarithm of both sides:
log(-60) = y log(-60)
This equation does not have a real solution for y, which means that the given initial rate and concentration values are not consistent with a zero-order reaction. Therefore, the answer cannot be determined.
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The rate law expressed the relationship of the rate of a reaction to the rate constant and the concentration of the reactants raised to some powers for the general reaction aA + bB cC + dDRate law takes the form r = k [A]X [B]ywhere x and y are number that must be determined experimentally k is the rate constant and [A] and [B] are concentration of A B respectively.Q.The initial rate of zero order reaction of the gaseous equation A (g) 2B (g) is 10-2 M min1 if the initial conc. of A is 0.1 M what would be conc. of B after 60 sec.a)0.09 Mb)0.01 Mc)0.02 Md)0.03 MCorrect answer is option 'C'. Can you explain this answer?
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The rate law expressed the relationship of the rate of a reaction to the rate constant and the concentration of the reactants raised to some powers for the general reaction aA + bB cC + dDRate law takes the form r = k [A]X [B]ywhere x and y are number that must be determined experimentally k is the rate constant and [A] and [B] are concentration of A B respectively.Q.The initial rate of zero order reaction of the gaseous equation A (g) 2B (g) is 10-2 M min1 if the initial conc. of A is 0.1 M what would be conc. of B after 60 sec.a)0.09 Mb)0.01 Mc)0.02 Md)0.03 MCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The rate law expressed the relationship of the rate of a reaction to the rate constant and the concentration of the reactants raised to some powers for the general reaction aA + bB cC + dDRate law takes the form r = k [A]X [B]ywhere x and y are number that must be determined experimentally k is the rate constant and [A] and [B] are concentration of A B respectively.Q.The initial rate of zero order reaction of the gaseous equation A (g) 2B (g) is 10-2 M min1 if the initial conc. of A is 0.1 M what would be conc. of B after 60 sec.a)0.09 Mb)0.01 Mc)0.02 Md)0.03 MCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The rate law expressed the relationship of the rate of a reaction to the rate constant and the concentration of the reactants raised to some powers for the general reaction aA + bB cC + dDRate law takes the form r = k [A]X [B]ywhere x and y are number that must be determined experimentally k is the rate constant and [A] and [B] are concentration of A B respectively.Q.The initial rate of zero order reaction of the gaseous equation A (g) 2B (g) is 10-2 M min1 if the initial conc. of A is 0.1 M what would be conc. of B after 60 sec.a)0.09 Mb)0.01 Mc)0.02 Md)0.03 MCorrect answer is option 'C'. Can you explain this answer?.
Solutions for The rate law expressed the relationship of the rate of a reaction to the rate constant and the concentration of the reactants raised to some powers for the general reaction aA + bB cC + dDRate law takes the form r = k [A]X [B]ywhere x and y are number that must be determined experimentally k is the rate constant and [A] and [B] are concentration of A B respectively.Q.The initial rate of zero order reaction of the gaseous equation A (g) 2B (g) is 10-2 M min1 if the initial conc. of A is 0.1 M what would be conc. of B after 60 sec.a)0.09 Mb)0.01 Mc)0.02 Md)0.03 MCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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