# What are the orders of "CV"^(+) and "OH"^(-) for the reaction "CV"^(+)(aq) + "OH"^(-)(aq) -> "CVOH"(aq)?

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

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

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

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

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!