What is the rate law for the reaction between crystal violet and NaOH at 22.2 C?

Write the correct rate law expression for the following reaction at 22.2 C. Replace constants by their values in the equation, and indicate at which temperature it is valid. Concentration of NaOH is 0.100M.

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2
Jun 15, 2017

This is a lab my students have done, where they tracked decreases in absorbance due to the disappearance of ${\text{CV}}^{+}$ (whose molar absorptivity was $\epsilon = {\text{53532 M"^(-1)cdot"cm}}^{- 1}$).

• The temperature of the room was somewhere around $20 - {25}^{\circ} \text{C}$, and the pressure was about $\text{0.918 atm}$ (around $\text{2352 ft}$ above sea level).

• The concentrations were $3.0 \times {10}^{- 5} \text{M}$ ${\text{CV}}^{+}$ and $\text{0.100 M}$ ${\text{OH}}^{-}$, and volumes used were around $1 - 5$ $\text{mL}$.

• Typical rates of disappearance of ${\text{CV}}^{+}$, $r \left(t\right)$, were such that reactant mixing had to be performed within 10 seconds.

• An average rate constant was on the order of $7 \times {10}^{- 2}$ $\text{M/s}$. Keep in mind this is only an estimate from memory.

Based on the various results they've gotten, the ideal orders are likely $1$ and $0$, respectively, for what they did

$r \left(t\right) = k \left[{\text{CV}}^{+}\right]$,

since the concentration of ${\text{OH}}^{-}$ they used was so much larger than that of ${\text{CV}}^{+}$, thus rendering the ${\text{OH}}^{-}$ concentration effectively constant, and the order with respect to ${\text{OH}}^{-}$ effectively zero.

Now of course, YOUR rate law depends on what $\left[{\text{CV}}^{+}\right]$ for you actually was. If it's small compared to $\text{0.100 M}$, then the above rate law should be the ideal one for your experiment. If not... provide the data!

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May 26, 2017

$\text{rate} = {0.1}^{m} \cdot {2.71828}^{- \setminus \frac{E a}{2455.5399}} A e \setminus {\left(C {V}^{+}\right)}^{n}$

Explanation:

$\text{rate} = k {\left[O {H}^{-}\right]}^{m} {\left[C {V}^{+}\right]}^{n}$

Where $\text{rate}$ has the units M/s and sometimes......
$\frac{M}{s} = \frac{m o l}{\mathrm{dm}} ^ 3 \div s = \text{mol} \cdot \frac{1}{{\mathrm{dm}}^{3}} \cdot \frac{1}{s} = m o l \cdot {\mathrm{dm}}^{-} 3 \cdot {s}^{-} 1$

Where OH^-) and CV^+ are the concentration of them in M

We know that the concentration of NaOH is 0.1M or the $O {H}^{-}$concentration is 0.1M as 1 mol of NaOH contributes 1 mole of $O {H}^{-}$ because it is a strong base.

The equation for dissociation in water

${\text{NaOH" + "H"_2"O" rightleftharpoons "Na"^+ + "OH}}^{-}$

Plug in the new found variable

$\text{rate} = k {\left[0.1 M\right]}^{m} {\left[C {V}^{+}\right]}^{2}$

As the temperature is not the room temperature we need to use the Arrhenis equation

$k = A {e}^{-} \left(E \frac{a}{R} T\right)$

Where k is rate constant
A is pre-exponential factor
Ea is activation energy
R is the gas constant
T is temperature in kelvins

Therefore ${22.2}^{\circ} C = {22.2}^{\circ} C + 273.15 K = 295.35 K$

$k = A {e}^{-} \left(\frac{E a}{8.314} \times 295.35\right)$
$k = A {e}^{-} \left(\frac{E a}{2455.5399}\right)$

$\text{rate} = A {e}^{-} \left(\frac{E a}{2455.5399}\right) {\left[0.1 M\right]}^{m} {\left[C {V}^{+}\right]}^{2}$

${0.1}^{m} \cdot {2.71828}^{- \setminus \frac{\left(E a\right)}{2455.5399}} A e \setminus {\left(C {V}^{+}\right)}^{n}$

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