Q.ii. For the reaction, $\mathrm{CH_3Br(aq) + OH^-(aq) \longrightarrow CH_3OH(aq) + Br^-(aq)}$, rate law is rate = $k$[CH$_3$Br][OH$^-$] a. How does reaction rate changes if [OH$^-$] is decreased by a factor of 5 ?
b. What is change in rate if concentrations of both reactants are doubled?

[!NOTE]
The book prints the product methanol as "CH$_3$OH$^{\ominus}$(aq)" -- with a circled-minus charge -- while methanol is a neutral molecule (the reactant OH$^-$ in the same line carries a plain minus). CH$_3$OH is shown uncharged here; everything else is as printed.

A rate constant with units of s$^{-1}$ (per second) indicates a first-order reaction.

The unit of the rate constant $k$ for a reaction of order $n$ is $mol^{1-n}\,L^{n-1}\,s^{-1}$ (concentration$^{1-n}$ time$^{-1}$).

• Zero order: $mol\,L^{-1}\,s^{-1}$
• First order: $s^{-1}$ (concentration terms cancel out)
• Second order: $mol^{-1}\,L\,s^{-1}$

Since the given unit is simply per second (s$^{-1}$), with no concentration term, the reaction is first order.

With $x<0$, Rate $=k[A]^x$ falls as $[A]$ increases — the rate decreases.

Given rate law: $\text{Rate} = k[A]^x$.

If the order $x$ is negative (say $x=-1$), then $\text{Rate} = \dfrac{k}{[A]}$ — the rate is inversely related to $[A]$.

So as the concentration of A is increased, $[A]^x$ (with negative $x$) actually gets smaller, meaning the rate of reaction DECREASES with increasing $[A]$. This behaviour is seen when the reactant itself (or a related species) inhibits the reaction, e.g. by tying up an active catalyst site.

Rate-constant unit $mol\,dm^{-3}s^{-1}$ corresponds to order $n=0$.

The general rate-constant unit is $\text{mol}^{1-n}\,dm^{3(n-1)}\,s^{-1}$ for a reaction of order $n$.

Given unit: $mol\,dm^{-3}\,s^{-1} = mol^{1}\,dm^{-3}\,s^{-1}$, matching $n=0$ (since $1-n=1 \Rightarrow n=0$).

This is the characteristic unit of a zero-order rate constant, where $\text{rate}=k$ (concentration-independent).

$2 = 4^n \Rightarrow n = 0.5$.

For $A \rightarrow$ products, rate $= k[A]^n$. If concentration is increased by a factor of 4 and the rate only doubles:

$\dfrac{\text{rate}_2}{\text{rate}_1} = \left(\dfrac{[A]_2}{[A]_1}\right)^n \Rightarrow 2 = 4^n$

$4^n = 2 \Rightarrow (2^2)^n = 2^1 \Rightarrow 2n = 1 \Rightarrow n = 0.5$.

$k = \text{rate}/[A] = 1.1\times10^{-4}\ s^{-1}$.

For a first order reaction $A \rightarrow B$, rate $= k[A]$.

$k = \dfrac{\text{rate}}{[A]} = \dfrac{6.6\times10^{-5}}{0.6} = 1.1\times10^{-4}\ s^{-1}$