Question

I-
2H2O2(aq) --> O2(g) + 2H2O(l)

If you use 0.20 M of KI instead of 0.10 M of KI, How would this affect
(a) the slopes of your curves,
(b) the rates of the reactions, and
(c) the numerical value of k for the reaction?

Answer 1

KI is a catalyst?

Answer 2

@aaronq my lab manual days that iodide is the catalyst so i would assume KI is

Answer 3

says*

Answer 4

(a) the slopes of your curves

what kind of curves are these?

Answer 5

let me show you my graphs..

Answer 6

okay

Answer 7

These are all V of O2 vs time?

Answer 8

yes! There were three trials performed for 3 solutions

Answer 9

would you need to see the rate data?

Answer 10

okay, soo assuming that the reaction wasnt saturated with the catalyst \(I^-\) at 0.1 M KI, 0.2 M Ki would increase the slope of the line (because there would a higher rate of O2 production).

Answer 11

b) in other words, it would increase the overall rate of both consumption of reactants and production O2 and H2O

Answer 12

c) you know that k is called the rate constant, it is dependent on the rate.

If you were to plot \(rate=k[O_2]\) ..which is just like \(y=mx+b\)

you can see that the slope is the rate constant, k.

Answer 13

If you increase the rate without changing the concentrations the graph would increase in slope

Answer 14

This is all assuming that the reaction isnt saturated with the catalyst - in other words, there are reactants free to MAKE USE of the catalyst

Answer 15

@aaronq so using 0.20 M KI would cause numerical value of K to increase as the slope increases?

Answer 16

yes

Answer 17

@aaronq Thank you :)

Answer 18

no problem!