# Question #1f772

##### 1 Answer

Here's what I got.

#### Explanation:

First thing first, you mistyped the **complex ion** listed for option **(2)**. More specifically, it's not *bromopentaamminecobalt(III) complex ion*.

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

#log(k/k_0) = -4 * 0.51 * sqrt(I)" " " "color(purple)((1))#

with this equation right here

#color(blue)(logk = logk_0 + 2 * Q * Z_AZ_B * sqrt(I))" "# , where

**rate constant** for the reaction

**all the activity coefficients** are equal to

**solvent**

**net charges** of the ions

**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.

#Q = 0.509 ~~0.51#

So, rearrange the second equation to get

#logk - logk_0 = 2 * 0.51 * Z_AZ_B * sqrt(I)#

This is of course equivalent to

#log(k/k_0) = 2 * 0.51 * Z_AZ_B * sqrt(I)" " " "color(purple)((2))#

Equations

#-4 * color(red)(cancel(color(black)(0.51))) * color(red)(cancel(color(black)(sqrt(I)))) = 2 * color(red)(cancel(color(black)(0.51))) * Z_AZ_B * color(red)(cancel(color(black)(sqrt(I))))#

Therefore, you can say that

#Z_AZ_B = ((-4))/2 = -2#

Simply put, the product of the net charges carried by your ions **must** be equal to

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_AZ_B = 0 -># rate constant isindependentof ionic strength

By comparison, you can say that if

#Z_BZ_B > 0 -># the rate of the reaction willincreasewith an increase in the solution's ionic strength

#Z_AZ_B < 0 -># the rate of the reaction willdecreasewith 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*,

For option **(1)**, you have the *peroxydisulfate anion*, *iodide anion*,

#Z_AZ_B = (2-) xx (1-) = +2 != -2#

Option **(3)** features a neutral molecule as well, *ethyl acetate*,

This of course leaves you with option **(2)**, for which you have a

#Z_AZ_B = (2+) xx (1-) = -2 color(white)(x)color(green)(sqrt())#