The kinetics of the reaction below were studied. The volumes of 0.040 M KBrO_3,
ID: 512001 • Letter: T
Question
The kinetics of the reaction below were studied. The volumes of 0.040 M KBrO_3, 0.040 M Kl, 0.00050 M Na_2S_2O_3 0.040 M KCI, H_2O and a buffer solution employed in each trial are shown in the table below. One drop of 1% aqueous starch solution was also added in each trial. The buffer solution was a pH = 4.74 solution containing 0.50 M acetic acid and 0.50 M sodium acetate; K_a = 1.8 times 10^-5 for acetic acid. Each trial was performed at 25 degree C. For each trial, the reaction mixture was initially colorless after combining all solutions; appearance of a blue solution color in the reaction mixture signaled that all of the thiosulfate ion initially present had reacted and marked the end of the timing period. The initial reaction rate for each trial was calculated from the change in molar concentration of thiosulfate ion from each trial; the equation for calculating Initial Rate is shown below, where Delta t is the elapsed time between combining all solutions and appearance of the blue solution color. Each solution volume was measured to the nearest 0.1 ml. (ie 5 mL shown in the table is 5.0 mL.) BrO_3 (aq) + 61 (aq) + 6H4^+ (aq) rightarrow 3H_2 O(I) + Br^- (aq) + 3l_2(aq) Initial rate = - 1/3 Delta S_2 O^2-_3/Delta t (a) Which pair of trials could be used to determine the order for l^-? Justify your choice. (b) How would you determine the order for H^+? Be as specific as possible. (c) What would be the value for the initial rate if 5 mL KBr03 solution, 5 mL Kl solution, 5 mL Na_2S_2O_3 solution, 10 mL KCI solution, 15 mL pH = 4.74 buffer solution, and 1 drop of 1% starch solution were used? (d) Why do Trials 1 and 4 have the same initial rate? (e) What time was required for the color change to occur in Trial 1? (f) What was the purpose of adding 5 mL of 0.040 M KCI in Trial 2? (g) How would you experimentally verify that KCI does not cause the color change from colorless to blue? Be as specific as possible.Explanation / Answer
a) First and third expression can be used to determine the order with respect to I-.
As, It has same value for [BrO3]- and [H2O] , so upon cancellation finally, we have I- value.
Rate = k[I ] x [BrO3 ] y [H+ ] z
where k is the rate constant and “x”, “y” and “z” are the exponents associated with the reactant concentrations. Assuming that temperature is constant (and therefore k is constant), by holding [ BrO3 ] and [H+ ] constant, the effect of [i – ] on reaction rate can be determined from two experiments with different values of [I- ] So, we can have order with respect to I-
b) The order with respect to [H+ ] is zero, as in rate reaction 1 and rate 5, though the concentration of H2O increases by 2.5, the rate of reaction remains same. Hence, we can say that the reaction is unaffected by H+ concentration. Hence,The order with respect to [H+ ] is zero.
c) rate would be 1.25 x 10-6 M/s , as the rate of reaction is first order with respect to both [BrO3]-[I ], so if we take half amount of both, then initial rate is decreased by twice.
d) The rate of reaction is first order with respect to both [BrO3]-[I ], and is independependant on other reagents. So, as first and fourth trial have same amount of [BrO3]- and [I ] , the rate also remains same.
e) 33000 seconds
f) KCl act as a catalyst and increase the rate of oxidation.
g) KCl act as only catalyst.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.