Lab_report_4

# Lab_report_4 - 8.314 = 38693.356= 38 693 J/mol The...

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Part 1:- 1) rate = ∆S / ∆t = k [S 2 O 3 2- ] m ln ∆S / ∆t = ln k [S 2 O 3 2- ] m ln ∆S – ln ∆t = ln k + ln [S 2 O 3 2- ] m ln ∆S – ln ∆t = ln k + m x ln [S 2 O 3 2- ] – ln ∆t = m x ln [S 2 O 3 2- ] + ln k – ln – ln ∆t = m x ln [S 2 O 3 2- ] + (ln k / ln ∆S) y = mx + b b y = -ln ∆t, x = ln [S 2 O 3 2- ] , b = (ln k / ln ∆S) 2) By using the regression feature in SigmaPlot, the order of the reaction is 1.0179. However, since the order needs to be an integer, this is rounded to 1. Therefore, the reaction is 1 st order with respect to [S 2 O 3 2- ]. y = 1.0179x + 1.2513 -5 -4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 ln t ln S2O3 Effect of Concentration on Reaction Rate

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Part 2:- 3) The slope of the line is 4654. 4) – ln ∆t = ln A – E a / R (1/T) slope = - E a / R E a = - slope x R = 4654 x
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Unformatted text preview: 8.314 = 38693.356= 38 693 J/mol The activation energy for the reaction between two compounds is 38693 J/mol. 5) ln k = ln A – E a / R (1/T) At k 1 , the temperature is 21 C (294 K). The reaction rate doubles at T 2. Therefore, ln [k 1 / k 2 ] = ln A – E a / R (1/T 1 ) – ln A + E a / R (1/T 2 ) ln [1 / 2] = E a / R [1/T 2 – 1/298] -0.6931 = 38693.356 / 8.314 [1/T 2 – 1/294] 3.252 x 10-3 = 1 / T 2 T 2 = 307.462 K = 34.46 C = 34 C To double the reaction rate, the reaction would have to be heated from 25 C to 34 C. y = 4654x - 11.236 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0.00295 0.003 0.00305 0.0031 0.00315 0.0032 0.00325 0.0033 0.00335 0.0034 0.00345 ln t 1/T Effect of Temperature on Reaction Rule...
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Lab_report_4 - 8.314 = 38693.356= 38 693 J/mol The...

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