tut 1 -q4-5 - ans

# tut 1 -q4-5 - ans - 0 1,110 0.0328 0.169 2,010 0.0292 0.261...

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Reaction Engineering Tutorial 1. 4. The rate of the first order reaction is given by: . Separating the variables and integrating this equation leads to: . Hence: . Recalling the definition of reactant conversion: , we rearrange to find that; , and insert this expression into our integrated rate equation: . Now we know that after 10 minutes at 10 ° C, 10 % of A is consumed, so we can work out the rate constant, k . . Then we use the Arrhenius equation to adjust the rate constant to the temperature of 20 ° C:

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And then we can calculate the conversion at 20 ° C: . 5. We have the reaction: We can calculate the conversion from X A = 1-(C A /C A0 ) Time (sec) C A (M = Molar, mol dm -3 ) X A (conversion C 3 H 7 Br) 0 0.0395

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Unformatted text preview: 0 1,110 0.0328 0.169 2,010 0.0292 0.261 3,192 0.0244 0.382 5,052 0.0195 0.506 7,380 0.0149 0.623 11,232 0.0097 0.754 78,840 0 1.000 For a second order reaction: the integrated form of the rate equation is (See lecture notes): where M = C B / C A . Therefore a straight line plot of against t has slope (M-1)C A k . We know that C A0 = 0.0395 Molar and C B0 = 0.0966 Molar, therefore M = C B / C A = 2.44 The conversion data are manipulated as follows in order to produce the required plot: t (sec) X A 0 0 0 1,110 0.169 0.113 2,010 0.261 0.189 3,192 0.382 0.311 5,052 0.506 0.473 7,380 0.623 0.682 11,232 0.754 1.033 78,840 1.000 - Slope of graph = 9.19 × 10-5 sec-1 Hence ....
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## This note was uploaded on 06/06/2011 for the course CHEM 3040 taught by Professor Reddy during the Spring '10 term at Taylor's.

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tut 1 -q4-5 - ans - 0 1,110 0.0328 0.169 2,010 0.0292 0.261...

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