Rate determining step for parallel reaction?

Question

I have always read that the rate determining step is the slowest step in a reaction (source: Wikipedia).

In chemical kinetics, the overall rate of a reaction is often approximately determined by the slowest step, known as the rate-determining step (RDS) or rate-limiting step.

However on this website (source), it says that the rate determining step for a parallel reaction is the fastest step. I was quite confused about this statement.

For a parallel reaction $\ce{B <- A -> C}$, where $\ce{A -> B}$ has rate constant $k_1$ and $\ce{A -> C}$ has rate constant $k_2$, the equations (from solving the first-order linear differential equations for A, B, and C, give: $A = A_oe^{-(k_1+k_2)t}$, $B = \frac{k_1*A_o}{k_1+k_2}*(1-e^{-(k_1+k_2)*t})$, and $C = \frac{k_2*A_o}{k_1+k_2}*(1-e^{-(k_1+k_2)*t})$.

But I do not know how to proceed from there.

Answer 1

Let's suppose that $\ce{A -> C}$ is the fast step in the parallel reaction scheme. This implies that $k_2 \gg k_1$. Thus, $k_1 + k_2 \approx k_2$, it is therefore (approximately) true to write the rate of consumption of $\ce{A}$ as $A_0 e^{-k_2t}$. This is what is meant by the fast reaction "controlling" rate for parallel reactions.