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

2 Answers
May 26, 2017

"rate" = 0.1^m* 2.71828^{-\frac{Ea}{2455.5399}}Ae\(CV^+)^n

Explanation:

"rate" = k[OH^-]^m[CV^+]^n

Where "rate" has the units M/s and sometimes......
M/s = (mol)/(dm)^3-:s = "mol"*1/(dm^3)*1/s = mol*dm^-3 *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 OH^- concentration is 0.1M as 1 mol of NaOH contributes 1 mole of OH^- because it is a strong base.

The equation for dissociation in water

"NaOH" + "H"_2"O" rightleftharpoons "Na"^+ + "OH"^-

Plug in the new found variable

"rate" = k[0.1M]^m[CV^+]^2

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

k = Ae^-(Ea/RT)

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^@C = 22.2^@C + 273.15K = 295.35K

k = Ae^-((Ea)/8.314xx295.35)
k = Ae^-((Ea)/2455.5399)

"rate" = Ae^-((Ea)/2455.5399)[0.1M]^m[CV^+]^2

0.1^m * 2.71828^{-\frac{(Ea)}{2455.5399}}Ae\(CV^+)^n

Jun 15, 2017

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

  • The temperature of the room was somewhere around 20 - 25^@ "C", and the pressure was about "0.918 atm" (around "2352 ft" above sea level).

  • The concentrations were 3.0 xx 10^(-5) "M" "CV"^(+) and "0.100 M" "OH"^(-), and volumes used were around 1 - 5 "mL".

  • Typical rates of disappearance of "CV"^(+), r(t), were such that reactant mixing had to be performed within 10 seconds.

  • An average rate constant was on the order of 7 xx 10^(-2) "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(t) = k["CV"^(+)],

since the concentration of "OH"^(-) they used was so much larger than that of "CV"^(+), thus rendering the "OH"^(-) concentration effectively constant, and the order with respect to "OH"^(-) effectively zero.


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