2. 3. Thank you in advance!! Kinetics 2018 Experimental Determination of a Rate

Question

2.
3.
Thank you in advance!! Kinetics 2018 Experimental Determination of a Rate Law ConstantsI Periodic Tabke Part A Consider the reaction What is the rate law for this reaction? Express the rate law symbolicaly in terms of k, [AJ, and B). whose rate at 25 C was measured using three different sets of inibai concentrations as isted in the following table: View Available Hint(s) Rate kIAI, M (M/s) 0.15 0.010 27 10 20.15 0.020 54 10 3 0.30 0010 1.1 10 rato. Previous Answers Correct Part B Calculate the initial rane kr the frmation of C at 2s .."(A1-0.50 Mand?-0.075 A. Express your answer to two significant figures and include the appropriate units View Available Hin inital ateValue Units Submit Provide Feedback

Explanation / Answer

1. Part A is correct

Part B

From Trial 1: 2.7*10-4 M/s = k * (0.15 M)2 * (0.010 M)

i.e. k = 1.2 M-2 s-1

Therefore, using the given concentrations in the problem

Rate = 1.2 M-2 s-1 * (0.5 M)2 * (0.075 M)

i.e. Rate = 0.0225 M/s = 2.25*10-2 M/s

2. Part A

ln(k2/k1) = Ea/R (1/T1 - 1/T2)

Here, k2/k1 = 2, Ea = 49.1 KJ/mol = 49100 J/mol

T1 = 28 oC = (28+273) K = 301 K

Therefore, ln(2) = (49100 J mol-1/8.314 J mol-1 K-1) * (1/301 - 1/T2)

i.e. 1/301 - 1/T2 = 1.174*10-4 K-1

i.e. 1/T2 = 0.0032 K-1

i.e. T2 = 312 K

i.e. T2 = (312 - 273) oC = 39 oC

Part B

ln(k2/k1) = Ea/R (1/T1 - 1/T2)

i.e. ln(k2/0.0130 s-1) = (49100 J mol-1/8.314 J mol-1 K-1) * (1/301 - 1/373)

i.e. k2 = 0.5738 s-1

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