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For the reaction given below A B the relaxation time is 10 -4 S with the 20%. Of A remains at equilibrium, the value of K(S-1) is?
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For the reaction given below A B the relaxation time is 10 -4 S with t...
Understanding the Reaction and Parameters
The reaction given is A ⇌ B. In this equilibrium reaction, we need to determine the equilibrium constant (K) in terms of the relaxation time.
Relaxation Time
- The relaxation time (τ) is given as 10^-4 seconds.
- This time constant indicates how quickly the system approaches equilibrium.
Equilibrium Concentration
- It is stated that 20% of A remains at equilibrium.
- Thus, 80% of A has converted to B during the reaction.
Calculation of Equilibrium Constant (K)
- If we denote the initial concentration of A as [A]₀, then at equilibrium:
- [A] = 0.2[A]₀ (20% remains)
- [B] = 0.8[A]₀ (80% converted)
- The equilibrium constant K can be defined as:
K = [B]/[A]
- Substituting the equilibrium concentrations:
K = (0.8[A]₀) / (0.2[A]₀)
- Simplifying gives:
K = 4
Units of K
- Since K is a rate constant with units of S^-1 (per second), the value of K is simply 4 S^-1.
Conclusion
- The equilibrium constant K for the reaction A ⇌ B, given a relaxation time of 10^-4 S and 20% of A remaining, is 4 S^-1.
This analysis shows how equilibrium concentrations and relaxation time influence the equilibrium constant in a reaction.
The reaction given is A ⇌ B. In this equilibrium reaction, we need to determine the equilibrium constant (K) in terms of the relaxation time.
Relaxation Time
- The relaxation time (τ) is given as 10^-4 seconds.
- This time constant indicates how quickly the system approaches equilibrium.
Equilibrium Concentration
- It is stated that 20% of A remains at equilibrium.
- Thus, 80% of A has converted to B during the reaction.
Calculation of Equilibrium Constant (K)
- If we denote the initial concentration of A as [A]₀, then at equilibrium:
- [A] = 0.2[A]₀ (20% remains)
- [B] = 0.8[A]₀ (80% converted)
- The equilibrium constant K can be defined as:
K = [B]/[A]
- Substituting the equilibrium concentrations:
K = (0.8[A]₀) / (0.2[A]₀)
- Simplifying gives:
K = 4
Units of K
- Since K is a rate constant with units of S^-1 (per second), the value of K is simply 4 S^-1.
Conclusion
- The equilibrium constant K for the reaction A ⇌ B, given a relaxation time of 10^-4 S and 20% of A remaining, is 4 S^-1.
This analysis shows how equilibrium concentrations and relaxation time influence the equilibrium constant in a reaction.
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For the reaction given below A B the relaxation time is 10 -4 S with the 20%. Of A remains at equilibrium, the value of K(S-1) is?
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For the reaction given below A B the relaxation time is 10 -4 S with the 20%. Of A remains at equilibrium, the value of K(S-1) is? for UPSC 2025 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about For the reaction given below A B the relaxation time is 10 -4 S with the 20%. Of A remains at equilibrium, the value of K(S-1) is? covers all topics & solutions for UPSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For the reaction given below A B the relaxation time is 10 -4 S with the 20%. Of A remains at equilibrium, the value of K(S-1) is?.
For the reaction given below A B the relaxation time is 10 -4 S with the 20%. Of A remains at equilibrium, the value of K(S-1) is? for UPSC 2025 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about For the reaction given below A B the relaxation time is 10 -4 S with the 20%. Of A remains at equilibrium, the value of K(S-1) is? covers all topics & solutions for UPSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For the reaction given below A B the relaxation time is 10 -4 S with the 20%. Of A remains at equilibrium, the value of K(S-1) is?.
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