Concentration of the limiting reactant (with initial concentration of a moles/litre) after time t is (a-x). Then 't' for a first order reaction is given by____________________?
Correct answer: A. k. t = ln a/(a - x)
- A. k. t = ln a/(a - x)
- B. k. t = x/a (a - x)
- C. k. t = ln (a - x)/a
- D. k. t = ln a (a - x)/x
Explanation
For a first-order reaction, integration of −dC/dt = kC from C = a to C = a−x gives kt = ln[a/(a−x)]. This is option a; option c has the logarithm inverted and would give a negative value.
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