Final answer:
The expression ln (a/a0) refers to the natural logarithm of the concentration of a reactant at time t divided by its initial concentration. For a first-order reaction, this could be part of the integrated rate law and would result in the right-hand side of the equation being -kt.
Step-by-step explanation:
If the expression on the left-hand side of a derivation step is ln (a/a0), the right-hand side will depend on the context of the problem, specifically on the type of kinetic equation that is being used. However, assuming we are dealing with a first-order reaction, the integrated rate equation is typically expressed as:
ln [A] = ln [A]0 - kt
Where:
- ln [A] is the natural logarithm of the concentration of the reactant A at time t,
- ln [A]0 is the natural logarithm of the initial concentration of reactant A,
- k is the rate constant of the reaction, and
- t is the time.
For a zero-order reaction, the concentration of A at time t is given by:
[A] = -kt + [A0]
Applying the natural logarithm to each side of the zero-order rate equation would not yield the ln(a/a0) form directly because the relationship is not logarithmic.
Therefore, if the expression ln (a/a0) appears on the left, and we know the reaction is first-order, then the rest of the equation would likely resemble the first-order integrated rate law provided, resulting in:
ln (a/a0) = -kt
Here, we subtract ln [A]0 from both sides of the integrated rate law for a first-order reaction to isolate ln (a/a0) on one side.