# For a reaction whose mechanism is given below with initial rate "0.0030 M/s", (a) what is the new rate if ["A"] is doubled, (b) if ["B"] is doubled, and (c) if both ["A"] and ["B"] are multiplied by 5?

## Overall reaction: $A + C \to D$ Mechanism: $A \implies B$ ($\text{slow}$) $B + C \implies D$ ($\text{fast}$)

Jun 12, 2017

Well, we know that the rate-limiting step is unimolecular with $A$ as the only reactant.

Single mechanistic steps are elementary reactions, for which the stoichiometric coefficients give the order with respect to the reactants involved.

If such a mechanistic step is slow, then it gives you the rate law directly. This rate law has already incorporated the behavior of $B$.

So, the rate law can be written as:

$r \left(t\right) = k \left[A\right]$

a)

Thus, if $\left[A\right]$ doubles, $r \left(t\right)$ doubles...

$\textcolor{red}{2} \times r \left(t\right) = k \times \textcolor{red}{2} \times \left[A\right]$

$\textcolor{b l u e}{r \left(t\right) ' = 2 r \left(t\right) = \text{0.0060 M/s}}$

b)

If $\left[B\right]$ is doubled, we know the rate law has no $B$ in it, i.e. $B$ is in some mechanistic step that is fast, and is the overall reaction is effectively zero order with respect to $B$.

Hence, doubling $\left[B\right]$ does not do anything, and...

$\textcolor{b l u e}{r \left(t\right) ' \approx r \left(t\right) = \text{0.0030 M/s}}$

given the original concentrations.

c)

Again, as in $\left(b\right)$, changing $\left[B\right]$ does nothing, so focus on $\left[A\right]$ and rehash what you did in $\left(a\right)$.

$\textcolor{red}{5} \times r \left(t\right) = k \times \textcolor{red}{5} \times \left[A\right] {\cancel{{\left[B\right]}^{0}}}^{1}$

Evidently the reaction rate then multiplies by $\boldsymbol{5}$ to maintain mathematical equivalence. Thus:

$\textcolor{b l u e}{r \left(t\right) ' = 5 r \left(t\right) = \text{0.0150 M/s}}$