# Question #1f772

Dec 28, 2015

Here's what I got.

#### Explanation:

First thing first, you mistyped the complex ion listed for option (2). More specifically, it's not ${\text{Co"("NH"_3)"Br}}^{2 +}$, it's actually ${\text{Co"("NH"_3)_5"Br}}^{2 +}$, the bromopentaamminecobalt(III) complex ion.

Now, the idea here is that you need to compare the equation given to you

$\log \left(\frac{k}{k} _ 0\right) = - 4 \cdot 0.51 \cdot \sqrt{I} \text{ " " } \textcolor{p u r p \le}{\left(1\right)}$

with this equation right here

$\textcolor{b l u e}{\log k = \log {k}_{0} + 2 \cdot Q \cdot {Z}_{A} {Z}_{B} \cdot \sqrt{I}} \text{ }$, where

$k$ - the rate constant for the reaction
${k}_{0}$- the rate constant for the reaction when all the activity coefficients are equal to $1$
$Q$ - a constant specific to the solvent
${Z}_{A} {Z}_{B}$ - the product between the net charges of the ions
$I$ - the ionic strength of the solution

SIDE NOTE I will not go into the derivation for this particular equation, since I assume that you're familiar with the salt effect (both primary and secondary).

Before moving forward, it's important to note that for water at room temperature, i.3. $\text{298 K}$, the value of $Q$ is

$Q = 0.509 \approx 0.51$

So, rearrange the second equation to get

$\log k - \log {k}_{0} = 2 \cdot 0.51 \cdot {Z}_{A} {Z}_{B} \cdot \sqrt{I}$

This is of course equivalent to

$\log \left(\frac{k}{k} _ 0\right) = 2 \cdot 0.51 \cdot {Z}_{A} {Z}_{B} \cdot \sqrt{I} \text{ " " } \textcolor{p u r p \le}{\left(2\right)}$

Equations $\textcolor{p u r p \le}{\left(1\right)}$ and $\textcolor{p u r p \le}{\left(2\right)}$ will give you

$- 4 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{0.51}}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{\sqrt{I}}}} = 2 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{0.51}}} \cdot {Z}_{A} {Z}_{B} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{\sqrt{I}}}}$

Therefore, you can say that

${Z}_{A} {Z}_{B} = \frac{\left(- 4\right)}{2} = - 2$

Simply put, the product of the net charges carried by your ions must be equal to $- 2$.

Right from the start, you can conclude that your solution must feature both positively charged ions, or cations, and an odd number of negatively charged ions, or anions.

Any reaction that features a neutral molecule will have

${Z}_{A} {Z}_{B} = 0 \to$ rate constant is independent of ionic strength

By comparison, you can say that if

${Z}_{B} {Z}_{B} > 0 \to$ the rate of the reaction will increase with an increase in the solution's ionic strength

${Z}_{A} {Z}_{B} < 0 \to$ the rate of the reaction will decrease with an increase in the solution's ionic strength

In your case, the rate constant decreases upon the addition of an electrolyte.

So, take a look at you options. Option (4) features hydrogen peroxide, ${\text{H"_2"O}}_{2}$, a neutral molecule, so you know that ${Z}_{A} {Z}_{B} = 0$.

For option (1), you have the peroxydisulfate anion, ${\text{S"_2"O}}_{8}^{2 -}$, and the iodide anion, ${\text{I}}^{-}$, so

${Z}_{A} {Z}_{B} = \left(2 -\right) \times \left(1 -\right) = + 2 \ne - 2$

Option (3) features a neutral molecule as well, ethyl acetate, ${\text{CH"_3"COOC"_2"H}}_{5}$, so this will not be the correct option.

This of course leaves you with option (2), for which you have a $2 +$ charge on the complex ion and a $1 -$ charge on the hydroxide anion

${Z}_{A} {Z}_{B} = \left(2 +\right) \times \left(1 -\right) = - 2 \textcolor{w h i t e}{x} \textcolor{g r e e n}{\sqrt{}}$