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The reaction A+OH- =product. obeys rate law expression as -d(A)/dt = k (A) (OH-) if initial conc of (A) and (OH-) are 0.002M and 0.3M and if it takes 30 secs for 1% A to react at 2 centigrade .The rate constant for the reaction is a) 1.05×10^-4 L mol^1 s^1 b) 1.41×10^-3 . c) 1.12 × 10^-3. d) 1.67× 10^-4.?
Most Upvoted Answer
The reaction A+OH- =product. obeys rate law expression as -d(A)/dt = k...
The given reaction is A + OH- → product. The rate law expression for this reaction is -d(A)/dt = k (A) (OH-), where k is the rate constant.
Given:
Initial concentration of A = 0.002 M
Initial concentration of OH- = 0.3 M
Time taken for 1% A to react = 30 seconds
Temperature = 2 °C
To find the rate constant, we can use the integrated rate law for a first-order reaction. The integrated rate law for a first-order reaction is ln(A₀/A) = kt, where A₀ is the initial concentration of A, A is the concentration of A at time t, k is the rate constant, and t is the time.
Let's calculate the concentration of A at the time when 1% of A has reacted:
1% of A = 0.01 * 0.002 M = 2 * 10^-5 M
Using the integrated rate law, we can rearrange it to solve for k:
ln(A₀/A) = kt
ln(0.002 M / (0.002 M - 2 * 10^-5 M)) = k * 30 seconds
Now, let's solve the equation:
ln(0.002 M / (0.002 M - 2 * 10^-5 M)) = k * 30 seconds
Using a calculator, we can find that ln(0.998 / 0.9979998) = 2 * 10^-5 M / (0.002 M - 2 * 10^-5 M) = k * 30 seconds
Simplifying further, we get:
ln(9999 / 9999.9998) = k * 30 seconds
Taking the natural logarithm of both sides, we get:
ln(0.99990001) = 30k
Now, solving for k:
k = ln(0.99990001) / 30
Using a calculator, we find that k ≈ 1.05 × 10^-4 L mol^-1 s^-1
Therefore, the correct answer is option a) 1.05 × 10^-4 L mol^-1 s^-1.
Given:
Initial concentration of A = 0.002 M
Initial concentration of OH- = 0.3 M
Time taken for 1% A to react = 30 seconds
Temperature = 2 °C
To find the rate constant, we can use the integrated rate law for a first-order reaction. The integrated rate law for a first-order reaction is ln(A₀/A) = kt, where A₀ is the initial concentration of A, A is the concentration of A at time t, k is the rate constant, and t is the time.
Let's calculate the concentration of A at the time when 1% of A has reacted:
1% of A = 0.01 * 0.002 M = 2 * 10^-5 M
Using the integrated rate law, we can rearrange it to solve for k:
ln(A₀/A) = kt
ln(0.002 M / (0.002 M - 2 * 10^-5 M)) = k * 30 seconds
Now, let's solve the equation:
ln(0.002 M / (0.002 M - 2 * 10^-5 M)) = k * 30 seconds
Using a calculator, we can find that ln(0.998 / 0.9979998) = 2 * 10^-5 M / (0.002 M - 2 * 10^-5 M) = k * 30 seconds
Simplifying further, we get:
ln(9999 / 9999.9998) = k * 30 seconds
Taking the natural logarithm of both sides, we get:
ln(0.99990001) = 30k
Now, solving for k:
k = ln(0.99990001) / 30
Using a calculator, we find that k ≈ 1.05 × 10^-4 L mol^-1 s^-1
Therefore, the correct answer is option a) 1.05 × 10^-4 L mol^-1 s^-1.
Community Answer
The reaction A+OH- =product. obeys rate law expression as -d(A)/dt = k...
D(A)=1%if A=0.002×(1/100)=2×10^-5
dt=30sec.
(A)=0.002. (B)=0.3.
substitute these values in the given equation.we get option C .
dt=30sec.
(A)=0.002. (B)=0.3.
substitute these values in the given equation.we get option C .
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The reaction A+OH- =product. obeys rate law expression as -d(A)/dt = k (A) (OH-) if initial conc of (A) and (OH-) are 0.002M and 0.3M and if it takes 30 secs for 1% A to react at 2 centigrade .The rate constant for the reaction is a) 1.05×10^-4 L mol^1 s^1 b) 1.41×10^-3 . c) 1.12 × 10^-3. d) 1.67× 10^-4.?
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The reaction A+OH- =product. obeys rate law expression as -d(A)/dt = k (A) (OH-) if initial conc of (A) and (OH-) are 0.002M and 0.3M and if it takes 30 secs for 1% A to react at 2 centigrade .The rate constant for the reaction is a) 1.05×10^-4 L mol^1 s^1 b) 1.41×10^-3 . c) 1.12 × 10^-3. d) 1.67× 10^-4.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The reaction A+OH- =product. obeys rate law expression as -d(A)/dt = k (A) (OH-) if initial conc of (A) and (OH-) are 0.002M and 0.3M and if it takes 30 secs for 1% A to react at 2 centigrade .The rate constant for the reaction is a) 1.05×10^-4 L mol^1 s^1 b) 1.41×10^-3 . c) 1.12 × 10^-3. d) 1.67× 10^-4.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The reaction A+OH- =product. obeys rate law expression as -d(A)/dt = k (A) (OH-) if initial conc of (A) and (OH-) are 0.002M and 0.3M and if it takes 30 secs for 1% A to react at 2 centigrade .The rate constant for the reaction is a) 1.05×10^-4 L mol^1 s^1 b) 1.41×10^-3 . c) 1.12 × 10^-3. d) 1.67× 10^-4.?.
The reaction A+OH- =product. obeys rate law expression as -d(A)/dt = k (A) (OH-) if initial conc of (A) and (OH-) are 0.002M and 0.3M and if it takes 30 secs for 1% A to react at 2 centigrade .The rate constant for the reaction is a) 1.05×10^-4 L mol^1 s^1 b) 1.41×10^-3 . c) 1.12 × 10^-3. d) 1.67× 10^-4.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The reaction A+OH- =product. obeys rate law expression as -d(A)/dt = k (A) (OH-) if initial conc of (A) and (OH-) are 0.002M and 0.3M and if it takes 30 secs for 1% A to react at 2 centigrade .The rate constant for the reaction is a) 1.05×10^-4 L mol^1 s^1 b) 1.41×10^-3 . c) 1.12 × 10^-3. d) 1.67× 10^-4.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The reaction A+OH- =product. obeys rate law expression as -d(A)/dt = k (A) (OH-) if initial conc of (A) and (OH-) are 0.002M and 0.3M and if it takes 30 secs for 1% A to react at 2 centigrade .The rate constant for the reaction is a) 1.05×10^-4 L mol^1 s^1 b) 1.41×10^-3 . c) 1.12 × 10^-3. d) 1.67× 10^-4.?.
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