Notes: 1. Carry out the procedure at three separate temperatures at least 15 deg

Question

Notes: 1. Carry out the procedure at three separate temperatures at least 15 degrees apart 2. Use the Rate Law of Rate-k'[I-][H202] 3. Using Excel Plot Ink' (Y-axis) Vs 1/T (x-axis). The Slope of the Graph =-Ea/R and R-8314/(mol/K) 4. Enter your collected data below, The columns that are highlighted red you must calculate using formulas in excels and linking to other cells when appropriate. Determination of the Activation Energy Trial Cold Temperature Medium Temperature High Temperature Time (s)Reaction Rate Calc. k'In(k')Temp (C) Temp (K) 1/Temp(K) 3.58E-03 3.37E-03 3.31E-03 902 302 200 1.11E-06 3.31E-06 5.00E-06 3.7E-03 1.1E-02 1.7E-02 5.6 4.5 4.1 5.8 23.5 29 279 297 302 Plot Graph-> Activation Energy (Ea) (J/Mole) 44724 Value of-Ea/R 5379 Activation Energy (Ea) (KJ/Mole) 44.7

Explanation / Answer

2")

y = -5964x + 15.12

-Ea/R = -5964

ln A = 15.12

ln(K) = -5964 * (1/283) + 15.12

or, k = 2.59*10^-3

[H2O2] = 0.1 *20mL/100mL = 0.02 M

[I-] =0.2M *20mL/100mL = 0.04 M

rate = k [H2O2][I-]

or, rate = 2.59*10^-3 * 0.02 *0.04 = 2.072*10^-6

rate = [H2O2]final-[H2O2]inital/T-0

or, 2.072*10^-6 = 0.02/T

or, T = 0.02/2.072*10^-6 = 9.65.25 *10^3 sec

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