OER Lab 5B- Iodine - Clock Reaction-2

CHEM 1412 Lab – 5 5 Sample Problem: Calculate the order for S 2 O 8 2 – and I – using the experimental data. Step 1 : Select two data points for S 2 O 8 2 – in which [ S 2 O 8 2 – ] is changing while [ I – ] remains constant. Only then can you attribute any changes in corresponding rates to be directly because of changing [ S 2 O 8 2 – ]. Here we will select Exp 1 and 3 . Note how [ S 2 O 8 2 – ] changes while [ I – ] remains constant. Step 2 : Write the rate laws for BOTH Exp 1 and 3 . Rate= k [ S 2 O 8 2 – ] x [ I – ] y Rate 1 : 1.10 x 10 -4 = k [0.16] x [0.24 ] y Rate 3 : 2.20 x 10 -4 = k [0.32] x [0.24 ] y Step 3 : Write the rate laws as ratio of each to determine proportions and thus order of S 2 O 8 2 ???? 3/ ???? 1= ? [ S 2 O 8 2 – ] ? [ I – ] ? / ? [ S 2 O 8 2 – ] ? [ I – ] ? (2.20 x 10 − 4 ) /(1.10 ? 10 − 4 ) = ? [0.32].[0.24] ? / ? [0.16].[0.24] ? Notice that k, [ I – ], and y cancel out leaving you with 2 = 2 ? ➔ ? = 1, order of S 2 O 8 2 – is 1 or first order . A theoretical interpretation concludes that as [ S 2 O 8 2 – ] doubles from 0.16M to 0.32M (and [ I – ] remains the same at 0.24M), the rate also doubles (2.20 x10 -4 vs 1.10 x10 -4 M/s). So, this one-to-one exponential relationship is indicative of a first order reaction. Repeat the same for [ I – ] using data from Exp 1 and 2 . ???? 2 / ???? 1 = ? [ S 2 O 8 2 – ] ? [ 𝐼 ] ? / ? [ S 2 O 8 2 – ] ? [ 𝐼 ] ? (4.40 x 10 − 4 ) /(1.10 ? 10 − 4 ) = ? [0.16] ? [0.48] ? / ? [0.16] ? [0.24] ? Notice that k, [ S 2 O 8 2 – ], and x cancel out leaving you with 4 = 2 ? ➔ y = 2, order of I – is 2 or second order . The theoretical interpretation notes that as [ I – ] doubles from 0.24M to 0.48M and [ S 2 O 8 2 – ] remains the same, the rate also quadruples (4.40 x10 -4 vs 1.10 x10 -4 M/s). This means that the effect was exponential to the power of two, indicative of a second order reaction. If there had been no correlation between a change in the reaction rate and a reactant concentration, such that rate was directly attributed to the k, this would indicate a zero-order reaction since any concentration value to the power of zero would be one. If Rate = k [A] 0 whereby [A] 0 = 1, then Rate= k. The overall order of the reaction would be the sum of each reactant order ➔ Overall order = x + y In the case of this reaction, the overall reaction order would be 3 . Once the orders of the reaction are determined, the rate constant, k, can be calculated using any data points because k is a “constant.” The rate law is simply rearranged to solve for k. ???? = ? [ S 2 O 8 2 – ] 1 [ I – ] 2 .