->given table -When we mixed solution 1 & 2 together (did this 6 times therefore

Question

->given table

-When we mixed solution 1 & 2 together (did this 6 times therefore had 6 reaction mixtures) to titrate (given in table 1), we waited for a color change to occur and got the results in table 4

Rate Law equation: rate= k [A]m [B]n

[A] & [B] represent concentration of substance A and substance B
k= rate law constant (which is not affected by concentration)
m & n are the "orders of reaction"

Table 1 Solution l Solutio n lI Reaction 0.200 M 0.0200 M Mixture KI (mL) KNO3 (mL) Na2S203 (mL) (mL) (NH4)2S20s (mL) (NH4)2SO4 0.0100 M | 1% starch 0.100 M 0.100 M 20.0 20.0 20.0 10.0 5.0 10.0 10.0 10.0 10.0 10.0 10.0 10.0 1.0 1.0 ol 10.0 1.0 20.0 1.0 1.0 1.0 (mL) 15.0 10.0 0 0 0 10.0 0 5.0 2 3 4 0 10.0 15.0 10.0 20.0 20.0 10.0 6

Explanation / Answer

This reaction involves two steps :

3 I- (aq) + S2O82- (aq) --------> I3- (aq) + 2 SO42- (aq) -------------> (1) slow step

I3- (aq) + 2 S2O32- (aq) --------> S4O62- (aq) + 3I- (aq) ------------> (2) fast step

Slower step is always the rate determining step. so we have to use the step (1) for rate determination. therefore, the calculations you made using the step (2) (concentration of S2O32-) will give wrong results.

A generic rate law would be :

rate = k [I-]m [S2O82-]n

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