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The decomposition of gaseous acetaldehyde at T(K) follows second-order kinetics. The half-life of this reaction is 400 s when the initial pressure is 250 Torr. What will be the rate constant (in Torr–1s– 1) and half-life (in s) respectively, if the initial pressure of the acetaldehyde is 200 Torr at the same temperature?
- a)105 and 500
- b)10–5 and 400
- c)10–4 and 400
- d)10–5 and 500
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
The decomposition of gaseous acetaldehyde at T(K) follows second-order...
-1s-1) of this reaction at the same temperature if the initial pressure is 500 Torr?
Given:
Half-life (t1/2) = 400 s
Initial pressure (P1) = 250 Torr
New initial pressure (P2) = 500 Torr
The rate constant of a second-order reaction can be calculated using the formula:
k = 1 / (t1/2 * [A]0)
Where:
k = rate constant
t1/2 = half-life
[A]0 = initial concentration of reactant
The initial concentration of acetaldehyde can be related to the initial pressure using the ideal gas law:
P = (nRT) / V
Where:
P = pressure
n = number of moles
R = gas constant
T = temperature
V = volume
Since the volume and number of moles are constant, we can say that:
P1 / [A]0 = P2 / [A]0
[A]0 = P1 * [A]0 / P2
Now we can substitute the values into the rate constant formula:
k = 1 / (400 * (P1 * [A]0 / P2))
k = 1 / (400 * (250 * [A]0 / 500))
k = 1 / (400 * 0.5)
k = 1 / 200
k = 0.005 Torr-1s-1
Therefore, the rate constant of the reaction at the same temperature with an initial pressure of 500 Torr will be 0.005 Torr-1s-1.
Given:
Half-life (t1/2) = 400 s
Initial pressure (P1) = 250 Torr
New initial pressure (P2) = 500 Torr
The rate constant of a second-order reaction can be calculated using the formula:
k = 1 / (t1/2 * [A]0)
Where:
k = rate constant
t1/2 = half-life
[A]0 = initial concentration of reactant
The initial concentration of acetaldehyde can be related to the initial pressure using the ideal gas law:
P = (nRT) / V
Where:
P = pressure
n = number of moles
R = gas constant
T = temperature
V = volume
Since the volume and number of moles are constant, we can say that:
P1 / [A]0 = P2 / [A]0
[A]0 = P1 * [A]0 / P2
Now we can substitute the values into the rate constant formula:
k = 1 / (400 * (P1 * [A]0 / P2))
k = 1 / (400 * (250 * [A]0 / 500))
k = 1 / (400 * 0.5)
k = 1 / 200
k = 0.005 Torr-1s-1
Therefore, the rate constant of the reaction at the same temperature with an initial pressure of 500 Torr will be 0.005 Torr-1s-1.
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Community Answer
The decomposition of gaseous acetaldehyde at T(K) follows second-order...
- Given, the decomposition of gaseous acetaldehyde at T(K) follows second-order kinetics
- For the reaction, Acetaldehyde (CH3CHO)
product
the rate law is given by
rate = k [CH3CHO]2
rate = k [CH3CHO]2
- In terms of partial pressure, rate = k P2[CH3CHO]
- The Integration rate law expression is,
- For a second-order reaction, the half-life will be
- For this reaction, the half-life will be
- Given, t1/2 = 400 sec and PoCH3CHO = 250 torr
- So,
or, k = 10-5 torr-1sec-1 - Now, when PoCH3CHO = 200 torr, t1/2 will be
= 500 sec
Hence, the rate constant (in Torr–1s– 1) and half-life (in s) respectively will be 10–5 and 500.
Hence, the rate constant (in Torr–1s– 1) and half-life (in s) respectively will be 10–5 and 500.
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The decomposition of gaseous acetaldehyde at T(K) follows second-order kinetics. The half-life of this reaction is 400 s when the initial pressure is 250 Torr. What will be the rate constant (in Torr–1s– 1) and half-life (in s) respectively, if the initial pressure of the acetaldehyde is 200 Torr at the same temperature?a)105and 500b)10–5and 400c)10–4and 400d)10–5and 500Correct answer is option 'D'. Can you explain this answer?
Question Description
The decomposition of gaseous acetaldehyde at T(K) follows second-order kinetics. The half-life of this reaction is 400 s when the initial pressure is 250 Torr. What will be the rate constant (in Torr–1s– 1) and half-life (in s) respectively, if the initial pressure of the acetaldehyde is 200 Torr at the same temperature?a)105and 500b)10–5and 400c)10–4and 400d)10–5and 500Correct answer is option 'D'. Can you explain this answer? for UGC NET 2024 is part of UGC NET preparation. The Question and answers have been prepared according to the UGC NET exam syllabus. Information about The decomposition of gaseous acetaldehyde at T(K) follows second-order kinetics. The half-life of this reaction is 400 s when the initial pressure is 250 Torr. What will be the rate constant (in Torr–1s– 1) and half-life (in s) respectively, if the initial pressure of the acetaldehyde is 200 Torr at the same temperature?a)105and 500b)10–5and 400c)10–4and 400d)10–5and 500Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for UGC NET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The decomposition of gaseous acetaldehyde at T(K) follows second-order kinetics. The half-life of this reaction is 400 s when the initial pressure is 250 Torr. What will be the rate constant (in Torr–1s– 1) and half-life (in s) respectively, if the initial pressure of the acetaldehyde is 200 Torr at the same temperature?a)105and 500b)10–5and 400c)10–4and 400d)10–5and 500Correct answer is option 'D'. Can you explain this answer?.
The decomposition of gaseous acetaldehyde at T(K) follows second-order kinetics. The half-life of this reaction is 400 s when the initial pressure is 250 Torr. What will be the rate constant (in Torr–1s– 1) and half-life (in s) respectively, if the initial pressure of the acetaldehyde is 200 Torr at the same temperature?a)105and 500b)10–5and 400c)10–4and 400d)10–5and 500Correct answer is option 'D'. Can you explain this answer? for UGC NET 2024 is part of UGC NET preparation. The Question and answers have been prepared according to the UGC NET exam syllabus. Information about The decomposition of gaseous acetaldehyde at T(K) follows second-order kinetics. The half-life of this reaction is 400 s when the initial pressure is 250 Torr. What will be the rate constant (in Torr–1s– 1) and half-life (in s) respectively, if the initial pressure of the acetaldehyde is 200 Torr at the same temperature?a)105and 500b)10–5and 400c)10–4and 400d)10–5and 500Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for UGC NET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The decomposition of gaseous acetaldehyde at T(K) follows second-order kinetics. The half-life of this reaction is 400 s when the initial pressure is 250 Torr. What will be the rate constant (in Torr–1s– 1) and half-life (in s) respectively, if the initial pressure of the acetaldehyde is 200 Torr at the same temperature?a)105and 500b)10–5and 400c)10–4and 400d)10–5and 500Correct answer is option 'D'. Can you explain this answer?.
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