Question

If the rate law for the clock reaction is: Rate = k [1] [BrO3 ] [H] A clock reaction is run with the following initial concentrations: [BrO3] 0.008 [H] 0.02 [52032) 0.002 0.0001 The reaction time is 35.6 seconds Calculate k in the rate law: Answer: 1. If the rate law for the clock reaction is: Rate = k [1] [BrO3 ] [H] A clock reaction is run at 20 °C with the following initial concentrations [BrO3] [52032-1 0.002 0.008 0.02 0.0001 [H] Then the experiment is repeated at 33 °C, the rate constant k is found to be 2.5 times larger. Calculate the activation energy Answer:

Answer 1

Rate of the reaction = ∆[I₂]/∆t = [S₂O₃^2-]/(2×∆t) = 0.0001/(2×35.6) = 1.4045×10^-6

= k ×[I^-][H⁺][BrO₃^-] = k × 0.002 × 0.008×0.02

k = 1.4045×10^-6/(0.002×0.008×0.02)

= 4.389

k = 4.4 M^-2s^-1 (Answer)

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ln(k₂/k₁) = E_a/R ( 1/T₁ - 1/T₂)

ln(2.5) = E_a/8.314 × (1/293.15 - 1/306.15)

E_a = 52592.54J/mol

= 52.592 kJ/mol

= 52.6 kJ/mol. (Answer)