1) Write the correct rate law expression for reaction 2 (from the lab manual) at
ID: 501003 • Letter: 1
Question
1) Write the correct rate law expression for reaction 2 (from the lab manual) at temperature T1. Replace constants by their values in the equation, and indicate at which temperature it is valid.
t1 = 294 K and the order is first (1)
We can simplify equation 1 to a less itimidating form: CV+(aq) + OH(aq) CVOH(aq) (2)
Here, CV+is crystal violet in its protonated form. The rate law for this reaction is: where: rate = k[OH]m[CV+]n (3) k : rate constant for the reaction m : is the order with respect to the hydroxide ion (OH) n: is the order with respect to cyrstal violet(CV+)
Explanation / Answer
1) reaction of crystal violet CV with base NaOH written as OH-, is done.
The reaction is,
CV(aq) + OH-(aq) --> CVOH(aq)
The correct rate law with using large excess of OH- would thus be writtent as,
rate = k[CV]
with order with respect to [CV] being 1, that is first-order kinetics.
k is the rate constant with units inverse of time.
order with respect to [OH-] is zero.
Molecularity of the reaction is 2, with both CV and OH- acting as reactant.
the overall reaction thus becomes a first order.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.