# Question ab272

Feb 25, 2017

WARNING! Long answer! Here's what I get.

#### Explanation:

a). Rate expression

The total volume in each experiment is the same: 50 mL.

Hence, the concentrations of $\text{KI}$ and $\text{Fe"^"3+}$ are directly proportional to their volumes.

We can't get the order with respect to $\text{KI}$ because its concentration is always the same.

However, we can get the order with respect to $\text{Fe"^"3+}$.

We are given

y ∝ ["Fe"^"3+"]^x and

y ∝ 1/t["Fe"^"3+"]^x

y ∝ 1/t or $y = \frac{k}{t}$

From Experiments 1 and 2, it appears that doubling the concentration halves the time (doubles the rate).

Thus the reaction appears to be first order in $\left[\text{Fe"^"3+}\right]$.

Hence, the rate expression is $r = k {\left[\text{Fe"^"3+}\right]}^{1}$ or $r = k \left[\text{Fe"^"3+}\right]$.

(b) Plot of concentration vs $\frac{1}{t}$

If a plot of rate ($\frac{1}{t}$) vs concentration ($x$) is a straight line, the reaction is first order in $\left[\text{Fe"^"3+}\right]$.

We have the following data:

$\boldsymbol{{\text{Expt."color(white)(m)x//"cm}}^{3} \textcolor{w h i t e}{m} t \textcolor{w h i t e}{m m l} 1 / t}$
$\textcolor{w h i t e}{m} 1 \textcolor{w h i t e}{m m m m l} 5 \textcolor{w h i t e}{m m} 92 \textcolor{w h i t e}{m m} 0.0109$
$\textcolor{w h i t e}{m} 2 \textcolor{w h i t e}{m m m l l} 10 \textcolor{w h i t e}{m m} 45.4 \textcolor{w h i t e}{m l} 0.0220$
color(white)(m)3color(white)(mmmll)15color(white)(mmll)?color(white)(mmm)?
$\textcolor{w h i t e}{m} 4 \textcolor{w h i t e}{m m m l l} 20 \textcolor{w h i t e}{m m} 23 \textcolor{w h i t e}{m m l} 0.0435$
$\textcolor{w h i t e}{m} 5 \textcolor{w h i t e}{m m m l l} 25 \textcolor{w h i t e}{m m} 18.1 \textcolor{w h i t e}{m l} 0.0552$

We plot $\frac{1}{t}$ vs $x$ and get a graph that looks like this: You are given that the equation for the line is $y = k x$, so $k$ is the slope of the graph.

The points at ${\text{5 cm}}^{3}$ and ${\text{25 cm}}^{3}$ appear to be on the line, so we will use them to calculate the slope.

$\text{Slope" = (Δy)/(Δx) = (y_2-y_1)/(x_2-x_1) = (0.0552 - 0.0109)/("25 cm"^3 - "5 cm"^3) = 0.0443/("20 cm"^3) = 2.22 × 10^"-3"color(white)(l) "cm"^"-3}$

Thus, the rate expression from the graph is

r = 2.22 × 10^"-3"color(white)(l) "cm"^"-3" × ["Fe"^"3+"].

Since $r$ is directly proportional to $\left[\text{Fe"^"3+}\right]$, the reaction is first order.

The calculated value of $\frac{1}{t}$ for $y = {\text{15 cm}}^{3}$ is

1/t = 2.22 × 10^"-3" color(red)(cancel(color(black)("cm"^"-3"))) × 15 color(red)(cancel(color(black)("cm"^3))) = 0.0333#

(I have shown it as a brown point on the line.)

$t = \frac{1}{0.0333} = 30.0$