Question

Question #2 Here are your data for trials 1-4 of Lab 7: [103 initial (S203 Jinitial Initial rate Trial 1 .0023 M 1 .00090 M .000028 M/s Trial 2 .0047 M .00090 M .000069 M/S Trial 3 .0023 M .0018 M .000052 M/S Trial 4 .0047 M .0018 M .000058 M/s The order with respect to 10, should be an integer (either 0, 1, or 2). Let's determine it using two sets of data: Part 1: Use the data from Trials 1 and 2 to determine the order of the reaction with respect to 103. Choose the closest answer Notes you have only one attempt to answer the questo will receive 2 points if your result is accurate and your calculations correct. You will receive point if your calculations are correct but your conclusions accurate will receive no points for an answer not supported by your data you chose o Your data do not support this answer Incorrect. Tries 1/1 Previous Tries a c e but w cada Part 2: Use the data from Trials 3 and 4 to determine the order of the reaction with respect to 10). Choose the dest will receive 2 points of your result is accurate and your calculations correct. You will receive por will receive no points for an answer e d O type here to search

3,4,5!!! please

Answer 1

Part 3 : order of reaction with respect to S₂O₅^2- = 0.89

Part 4 : order of reaction with respect to S₂O₅^2- = -0.25

Part 5 : order with respect to S₂O₅^2- = 0.32

Explanation

Part 3

The general rate law is given as : Rate = k[S₂O₅^2-]ⁿ

where n = order of reaction with respect to S₂O₅^2-

For trial 1, rate = 0.000028 M/s

[S₂O₅^2-] = 0.00090 M

Substituting these values in general rate law, we get

0.000028 M/s = k * (0.00090 M)ⁿ

Similarly from trial 3

0.000052 M/s = k * (0.0018 M)ⁿ

dividing the two equations,

(0.000052 M/s) / (0.000028 M/s) = [k * (0.0018 M)ⁿ] / [k * (0.00090 M)ⁿ]

52 / 28 = (18 / 9)ⁿ

1.857 = (2)ⁿ

n = log(1.857) / log(2)

n = 0.893