What are the orders of #"CV"^(+)# and #"OH"^(-)# for the reaction #"CV"^(+)(aq) + "OH"^(-)(aq) -> "CVOH"(aq)#?
2 Answers
Explanation:
Where
Where
We know that the concentration of NaOH is 0.1M or the
The equation for dissociation in water
Plug in the new found variable
As the temperature is not the room temperature we need to use the Arrhenis equation
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
This is a lab my students have done, where they tracked decreases in absorbance due to the disappearance of
-
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
#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
