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

2 Answers
May 26, 2017

Answer:

#"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!