Question

8. The following data were determined during an experiment to determine the kinetics of the removal of glucose from a water source. Time Glucose (hr) (mg/L) 250 3 90 6 30 9 10 123 a) Balance the following oxidation equation C6H12O6 + O2 + CO2 + H20. b) Determine the order of the reaction (make graphs for 0 order, 1st order, and 2nd order) and conclude which plot is the best fit to the data. 1st 2nd Time (hr) glucose (mg/L) Time 2500 903 306 10 9 In glucose 5.52 4.50 3.40 3 2.30 Time 1/glucose 0 0.0040 0.0111 0.0333 0.1000 0.3333 1.0000 -0.1826 0.0573 0.8264 1 Intercept Slope Correl 173 -14.5333 -0.84101 1.10 0.00 5.58319487 -0.370566753 -0.999738091

Answer 1

a) The balanced reaction is :

C6H12O6 + 6O2 $\rightarrow$ 6 CO2 + 6H2O

b) Let concentration of glucose be represented as A. Then the following graphs are drawn.

i) A vs t

Note that It is not a good fit for straight line.

Avst y=-14.53x + 173 R = 0.707 Avst — Linear (A vs t) . 15 time

ii) ln A vs t

Note that It is a very good fit for straight line.

In Avst y-0.370x+5.581 R=0.999 In A In A vst — Linear (In A vs t) ME time

iii) (1/A) vs t

Note that It is also not a good fit for straight line.

(1/A) vs t Y=0.057x -0.182 R2=0.682 (1/A) (1/A) vst NINH - Linear ((1/A) vs t) 10 1520

Now note that for a first order reaction, ln A vs t gives a straight line.

So, the reaction is first order in A. From the equation ;

y = -0.370x + 5.581

Slope = -k, or k = 0.370 hr-1

c) So, rate (r) = 0.370 [A] or = 0.370[C6H12O6]

d) For a first order reaction: [A] = [Ao] e-kt

   So, for 99 % removal, means [A] = 0.01[Ao] = [Ao] e-kt

Note that no need here for initial concentration, as final concentration is given as fraction of the original concentration.

Solving,   ln (0.01) = -kt = -(0.370 hr-1 ) x t (in hr)

$\Rightarrow$      t (in hr) = ( -4.605 / -0.370 ) = 12.45 hr

e) half life of a first order reaction (t1/2) is given by:

t1/2 = 0.693 / k = 0.693 / (0.370 hr-1 ) = 1.87 hr

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