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KINETICS_SOLUTION

# KINETICS_SOLUTION - ENV 6015 Solutions to Kinetics Problem...

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1 ENV 6015 Solutions to Kinetics Problem Set 1. The inversion of sucrose proceeded as shown at 25 o C: The initial concentration of sucrose was 1.0022 moles/liter. Calculate the first order rate constant and the half-life of the reaction. Why does this reaction follow a first order law despite the fact water enters into the stoichiometric equation? How long would it take to invert 95% of a pound of sugar? C 12 H 22 O 11 + H 2 O C 6 H 12 O 6 + C 6 H 12 O 6 Time, min. 0 30 60 90 130 180 Sucrose inverted-M/L 0 0.1001 0.1946 0.2770 0.3726 0.4626 Solution: The solution is diluted, [H 2 O] = 55.5 M/L vs [C 12 H 12 O 11 ] = 1.0 M/L. The reaction is limited by the concentration of C 12 H 12 O 11 and follows first-order kinetics. Sucrose + H 2 O Glucose A A C * k dt dC = after integration: t * k C C ln Ai A = or ln C A = -k*t + ln C Ai Thus, if we plot ln [C A ] vs time, the resulting slope is equal to the rate constant k (min -1 ). Time, min. 0 30 60 90 130 180 Sucrose inverted- M/L 0 0.1001 0.1946 0.2770 0.3726 0.4626 Sucrose remaining M/L 1.0022 0.9021 0.8076 0.7252 0.6296 0.5396 Ln [C A ] 0.0022 -0.1030 -0.2137 -0.3213 -0.4627 -0.6169

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2 Plot of ln C vs time ln C A = -0.0035*time - 0.0025 R 2 = 1.0 -0.8 -0.6 -0.4 -0.2 0 0.2 0 20 40 60 80 100 120 140 160 180 200 Time (min) ln C A (sucrose remaining) From the equation: k = 0.0035 min -1 At half-life of the reaction: 2 C C Ai A = 2 1 Ai Ai t * 0.0035 C C * 0.5 ln = () min 198 0035 . 0 5 . 0 ln t 2 1 = = To invert 95% of a pound of sugar: 2 1 Ai Ai t * 0.0035 C C * 0.05 ln = min 856 0.0035 0.05 ln t 95% = =
3 2. Find the reaction order and rate constant for the decomposition of diazobenzenechloride at 50 o C if the initial diazobenzenechloride concentration is 10 grams/liter. The reaction and nitrogen evolution are shown. C 6 H 5 N 2 Cl C 6 H 5 Cl + N 2 Time, min. 6 9 12 14 18 22 24 26 39 N 2 cm 3 evolved 19.3 26.0 32.6 36 41.3 45.0 46.5 48.4 50.4 58.3 Solution: For t = , V N2 = 58.3 cm 3 ; using the ideal gas law: () moles 0.0022 K 323 K * mol cm * atm 82 cm 58.3 atm 1 T * R V * P n o o 3 3 N N 2 2 = = = by stoichiometry, 0.0022 moles of C 6 H 5 N 2 Cl produce 0.0022 moles of N 2 . Assuming complete reaction: Initial mass of C 6 H 5 N 2 Cl = (0.0022moles)*(140 g/mole) = 0.309 grams Volume of solution = 3 cm 1000 L 1 * L g 10 g 0.309 = 30.9 cm 3 To determine the order of the reaction and the rate constant, we need to graph: a) C vs time Æ for zero-order reaction b) Ln C vs time Æ for first-order reaction c) 1/C vs time Æ for second-order reaction Note: the graphs can also be made by using mass or moles of C 6 H 5 N 2 Cl in lieu of concentration Time (min) N 2 cm 3 evolved N 2 moles evolved

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KINETICS_SOLUTION - ENV 6015 Solutions to Kinetics Problem...

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