# Advance Study Assignment: Rates of Chemical Reactions, II. A Clock Reaction

1. A student studied the clock reaction described in this experiment. She set up Reaction Mixture 4 by mixing $10 \mathrm{~mL} 0.010 \mathrm{M} \mathrm{KI}, 10 \mathrm{~mL} 0.001 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}, 10 \mathrm{~mL} 0.040 \mathrm{M} \mathrm{KBrO}_{3}$, and $20 \mathrm{~mL} 0.10 \mathrm{M} \mathrm{HCl}$ using the procedure given. It took about 21 seconds for the color to turn blue.
a. She found the concentrations of each reactant in the reacting mixture by realizing that the number of moles of each reactant did not change when that reactant was mixed with the others, but that its concentration did. For any reactant $A$,

$
ext { no. moles } A=M_{A ext { stock }} imes V_{ ext {mix }} \approx M_{A ext { mixture }} imes V_{ ext {mixture }}
$

The volume of the mixture was 50 mL . Revising the above equation, she obtained

$
M_{A ext { mixture }}=M_{A ext { stock }} imes \frac{V_{ ext {mix }}(\mathrm{mL})}{50 \mathrm{~mL}}
$

Find the concentrations of each reactant using the equation above.

$
\left(\mathrm{I}^{-} ight)=\quad \mathrm{M} ;\left(\mathrm{BrO}_{3}^{-} ight)=\quad \mathrm{M} ;\left(\mathrm{H}^{+} ight)=\quad \mathrm{M}
$

b. What was the relative rate of the reaction (1000/r)?

$
\begin{aligned}
& ext { c. Knowing the relative rate of reaction for Mixture } 4 ext { and the concentrations of } \mathrm{I}^{-}, \mathrm{BrO}_{3}^{-} ext {, and } \mathrm{H}^{+} ext {in that } \
& ext { mixture, she was able to set up Equation } 5 ext { for the relative rate of the reaction. The only quantities that } \
& ext { remained unknown were } k^{\prime}, m, n ext {, and } p ext {. Set up Equation } 5 ext { as she did, presuming she did it properly. } \
& ext { Equation } 5 ext { is on page } 5 ext {. }
\end{aligned}
$

2. For Reaction Mixture 1 the student found that 85 seconds were required. On dividing Equation 5 for Reaction Mixture 1 by Equation 5 for Reaction Mixture 4, and after canceling out the common terms $\left[k^{\prime} ight.$ and terms in $\left(\mathrm{I}^{-} ight)$and $\left(\mathrm{BrO}_{3}^{-} ight)$], she got the following equation:

$
\frac{11.8}{48}=\left(\frac{0.020}{0.040} ight)^{p}=\left(\frac{1}{2} ight)^{p}
$

Recognizing that $11.8 / 48$ is about equal to $K$, she obtained an approximate value for $p$. What was that value?

$
p=
$

By taking logarithms of both sides of the equation, she got an exact value for $p$. What was that value?

$
p=
$

Since orders of reactions are often integers, she rounded her value of $p$ to the nearest integer, and reported that value as the order of the reaction with respect to $\mathrm{H}^{+}$.

Question

# Advance Study Assignment: Rates of Chemical Reactions, II. A Clock Reaction

1. A student studied the clock reaction described in this experiment. She set up Reaction Mixture 4 by mixing $10 \mathrm{~mL} 0.010 \mathrm{M} \mathrm{KI}, 10 \mathrm{~mL} 0.001 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}, 10 \mathrm{~mL} 0.040 \mathrm{M} \mathrm{KBrO}_{3}$, and $20 \mathrm{~mL} 0.10 \mathrm{M} \mathrm{HCl}$ using the procedure given. It took about 21 seconds for the color to turn blue.
a. She found the concentrations of each reactant in the reacting mixture by realizing that the number of moles of each reactant did not change when that reactant was mixed with the others, but that its concentration did. For any reactant $A$,

$
	ext { no. moles } A=M_{A 	ext { stock }} 	imes V_{	ext {mix }} \approx M_{A 	ext { mixture }} 	imes V_{	ext {mixture }}
$

The volume of the mixture was 50 mL . Revising the above equation, she obtained

$
M_{A 	ext { mixture }}=M_{A 	ext { stock }} 	imes \frac{V_{	ext {mix }}(\mathrm{mL})}{50 \mathrm{~mL}}
$

Find the concentrations of each reactant using the equation above.

$
\left(\mathrm{I}^{-}ight)=\quad \mathrm{M} ;\left(\mathrm{BrO}_{3}^{-}ight)=\quad \mathrm{M} ;\left(\mathrm{H}^{+}ight)=\quad \mathrm{M}
$

b. What was the relative rate of the reaction (1000/r)?

$
\begin{aligned}
& 	ext { c. Knowing the relative rate of reaction for Mixture } 4 	ext { and the concentrations of } \mathrm{I}^{-}, \mathrm{BrO}_{3}^{-} 	ext {, and } \mathrm{H}^{+} 	ext {in that } \
& 	ext { mixture, she was able to set up Equation } 5 	ext { for the relative rate of the reaction. The only quantities that } \
& 	ext { remained unknown were } k^{\prime}, m, n 	ext {, and } p 	ext {. Set up Equation } 5 	ext { as she did, presuming she did it properly. } \
& 	ext { Equation } 5 	ext { is on page } 5 	ext {. }
\end{aligned}
$

2. For Reaction Mixture 1 the student found that 85 seconds were required. On dividing Equation 5 for Reaction Mixture 1 by Equation 5 for Reaction Mixture 4, and after canceling out the common terms $\left[k^{\prime}ight.$ and terms in $\left(\mathrm{I}^{-}ight)$and $\left(\mathrm{BrO}_{3}^{-}ight)$], she got the following equation:

$
\frac{11.8}{48}=\left(\frac{0.020}{0.040}ight)^{p}=\left(\frac{1}{2}ight)^{p}
$

Recognizing that $11.8 / 48$ is about equal to $K$, she obtained an approximate value for $p$. What was that value?

$
p=
$

By taking logarithms of both sides of the equation, she got an exact value for $p$. What was that value?

$
p=
$

Since orders of reactions are often integers, she rounded her value of $p$ to the nearest integer, and reported that value as the order of the reaction with respect to $\mathrm{H}^{+}$.

StudyXY · Accepted Answer

: Find the concentrations of each reactant in the reacting mixture. Using the given equation, we can calculate the concentrations of each reactant in the reacting mixture.

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