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For the reaction given below A to B , relaxation time is 10^-6 s. Given that 10% of A remains at equilibrium the value of k1 is a)9*10^5 b)10^-5 c) 10^5 d) 9*10^-5?
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For the reaction given below A to B , relaxation time is 10^-6 s. Give...
Solution:
Given that the relaxation time is 10^-6 s, we can use the following equation to determine the rate constant k1.
k1 = 1/relaxation time
k1 = 1/10^-6
k1 = 10^6 s^-1
Now, let's use the following equation to determine the percentage of A that remains at equilibrium.
% A remaining = 100/(1 + k1/k-1)
Given that % A remaining is 10%, we can plug in the values and solve for k-1.
10 = 100/(1 + (10^6)/k-1)
1 + (10^6)/k-1 = 10
(10^6)/k-1 = 9
k-1 = (10^6)/9
k-1 = 1.11*10^5 s^-1
Therefore, the value of k1 is given by option (c) 10^5 s^-1.
Explanation:
- The relaxation time is used to determine the rate constant k1 using the equation k1 = 1/relaxation time.
- The percentage of A remaining at equilibrium can be determined using the equation % A remaining = 100/(1 + k1/k-1).
- The given value of 10% for % A remaining can be used to solve for k-1.
- The value of k1 is then calculated using the equation k-1 = (10^6)/9 and is given by option (c) 10^5 s^-1.
Given that the relaxation time is 10^-6 s, we can use the following equation to determine the rate constant k1.
k1 = 1/relaxation time
k1 = 1/10^-6
k1 = 10^6 s^-1
Now, let's use the following equation to determine the percentage of A that remains at equilibrium.
% A remaining = 100/(1 + k1/k-1)
Given that % A remaining is 10%, we can plug in the values and solve for k-1.
10 = 100/(1 + (10^6)/k-1)
1 + (10^6)/k-1 = 10
(10^6)/k-1 = 9
k-1 = (10^6)/9
k-1 = 1.11*10^5 s^-1
Therefore, the value of k1 is given by option (c) 10^5 s^-1.
Explanation:
- The relaxation time is used to determine the rate constant k1 using the equation k1 = 1/relaxation time.
- The percentage of A remaining at equilibrium can be determined using the equation % A remaining = 100/(1 + k1/k-1).
- The given value of 10% for % A remaining can be used to solve for k-1.
- The value of k1 is then calculated using the equation k-1 = (10^6)/9 and is given by option (c) 10^5 s^-1.
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For the reaction given below A to B , relaxation time is 10^-6 s. Given that 10% of A remains at equilibrium the value of k1 is a)9*10^5 b)10^-5 c) 10^5 d) 9*10^-5?
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For the reaction given below A to B , relaxation time is 10^-6 s. Given that 10% of A remains at equilibrium the value of k1 is a)9*10^5 b)10^-5 c) 10^5 d) 9*10^-5? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about For the reaction given below A to B , relaxation time is 10^-6 s. Given that 10% of A remains at equilibrium the value of k1 is a)9*10^5 b)10^-5 c) 10^5 d) 9*10^-5? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For the reaction given below A to B , relaxation time is 10^-6 s. Given that 10% of A remains at equilibrium the value of k1 is a)9*10^5 b)10^-5 c) 10^5 d) 9*10^-5?.
For the reaction given below A to B , relaxation time is 10^-6 s. Given that 10% of A remains at equilibrium the value of k1 is a)9*10^5 b)10^-5 c) 10^5 d) 9*10^-5? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about For the reaction given below A to B , relaxation time is 10^-6 s. Given that 10% of A remains at equilibrium the value of k1 is a)9*10^5 b)10^-5 c) 10^5 d) 9*10^-5? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For the reaction given below A to B , relaxation time is 10^-6 s. Given that 10% of A remains at equilibrium the value of k1 is a)9*10^5 b)10^-5 c) 10^5 d) 9*10^-5?.
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