KVPY2020ChemistryChemical Kinetics
Consider a reaction that is first order in both directions A ⇌ k b k f B Initially only A is present, and its concentration is A 0 . Assume A t and A eq are the concentrations of A at time " t " and at equilibrium, respectively. The time " t " at which A t = A 0 + A eq / 2 is:
Options
- At = ln 3 2 k f + k b
- Bt = ln 3 2 k f - k b
- Ct = ln 2 k f + k b
- Dt = ln 2 k f - k b
Correct answer
C. t = ln 2 k f + k b
Step-by-step solution
A ⇌ B t = 0      A 0     0 t = t    A 0 - x    x = A t t = t eq    A 0 - X eq    X eq = A eq Given at time t = t   A t = A 0 + A aq 2 and x eq . = A 0 - A eq Now, t = 1 k f + k b ln x c x c - x = ln 2 k f + k b