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+)

Answer 1

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.