# Inversion of Sucrose; Kinetics of a Pseudo-First INVERSION OF SUCROSE; KINETICS OF A PSEUDO-FIRST

ORDER REACTION DETERMINED BY POLARIMETRY

DIFFERENTIAL EQUATIONS

Dr. Herman

Nov. 6, 1996

Bonnie Jenkins

Jennifer Williams

Abstract

This is a simple experiment that demonstrates the decomposition of sucrose by dissolving sucrose in hydrochloric acid. The products formed are optically active and their optical rotation can be determined by use of a polarimeter. This first order reaction follows a differential equation.

Introduction

The hydrolysis of sucrose can be considered a pseudo-first order reaction, in which the rate of change at any time is proportional to the concentration of sucrose. The rate of the reaction can be determined by the following equation,

(1)

Integration of equation (1),

, (2)

t, (3)

and therefore,

, (4)

where a is the initial concentration at time t, (a-x) equals the sucrose concentration, and k equals the fractional number of molecules decomposing per unit time.

The reaction rate constant k is calculated using the following equation:

, (5)

where (0) is the initial rotation of pure sucrose, (is the rotation at time (infinite time is the time when the reaction has gone to completion)and (t) is the rotation at some specific time t.

The k value is then used in the following equation to determine the concentration of sucrose at any given time,

, (6)

where c(t) is the calculated concentration for any given time t.

Equipment and Supplies

Hydrochloric acid (4 and 6 Molar), Sucrose, Sodium lamp, Polarimeter, Jacketed polarimeter tube, Volumetric flasks (100 mL), Graduated cylinders (25 mL), Constant temperature bath, Deionized H2O.

Procedure

The sodium lamp should be turned on at the beginning of the lab and positioned on the lab bench. The lamp takes 20-30 minutes to warm up, so this should be done first. Turn on the water bath and adjust the temperature appropriately, connecting a hose from the outlet of the constant temperature bath to one end of the jacketed polarimeter tube. Connect another hose to the other end of the tube so that the flow of the water can return to the constant temperature bath. A 100mL solution of 4 M HCl should be prepared. Place the flask in the constant temperature bath, securing it with a clamp and ring stand. A 100mL solution of 0.200 g/mL sucrose should be prepared. Place the flask in the constant temperature bath, securing it with a clamp and ring stand.

After the apparatus has had time to equilibrate, rinse the polarimeter tube first with deionized water, then with the sucrose solution. Fill the polarimeter tube with the sucrose solution and take an initial reading on the polarimeter. This will be taken as time = 0. Again, rinse the polarimeter tube with deionized water.

Remove the two flasks containing the 4M HCl and the sucrose solution from the constant temperature bath. Separately measure 25mL of each solution into two 50mL graduated cylinders. Mix the solutions in the graduated cylinders, at the same time starting the stopwatch. Mix the solution three or four times, then rinse the polarimeter tube with the mixture. Fill the polarimeter tube with the mixture, being sure to clear all bubbles from the tube. This can be done by slowly tilting the tube back and forth until all the bubbles float out.

Begin taking time readings after ten minutes for the 4 M solution, and continue taking readings every 10 minutes for 30 minutes. After 30 minutes begin taking 5 minute intervals readings as long as time permits. Begin taking readings for the 6M solution after 5 minutes and continue taking readings in 5 minute intervals as long as time permits. Leave the solution in the polarimeter at the end of lab and the final reading will be at time infinity, which should be approximately two days following the experiment.

Data

Pure Sucrose (20g --> 100mL H2O)= 0.56497mol/L or M
4M HCl
Time (Min) / Rotation ( / Concentration of sucrose
(M) / log (C1-C2) (M) / Rate of Reaction (1/min)
10 / 8.00 / 0.36 / -0.88 / -0.01
20 / 4.51 / 0.23 / -1.08 / -0.01
30 / 2.42 / 0.15 / -1.53 / -0.02
35 / 1.76 / 0.12 / -1.63 / -0.02
40 / 0.56 / 0.094 / -1.72 / -0.02
45 / -0.08 / 0.075 / -1.82 / -0.02
50 / -0.68 / 0.060 / -1.92 / -0.02
55 / -0.92 / 0.048 / -2.02 / -0.02
60 / -1.44 / 0.038 / -2.11 / -0.02
65 / -1.52 / 0.031 / -2.21 / -0.02
70 / -1.58 / 0.025 / -2.31 / -0.02
75 / -2.06 / 0.020 / -2.40 / -0.02
80 / -2.88 / 0.016 / -2.50 / -0.02
85 / -2.84 / 0.013 / -2.60 / -0.02
90 / -2.96 / 0.010 / -2.70 / -0.02
95 / -3.18 / 0.0080 / -2.79 / -0.02
100 / -3.18 / 0.0064 / -2.19 / -0.03
 / -3.92 / 0 / -0.03
6M HCl
Time(Min) / Rotation ( / Concentration of sucrose
(M) / log (C1-C2) (M) / Rate of Reaction (1/min)
5 / 5.60 / 0.37 / -0.89 / -0.02
10 / 2.48 / 0.24 / -1.08 / -0.03
15 / -0.02 / 0.16 / -1.26 / -0.04
20 / -1.36 / 0.10 / -1.45 / -0.04
25 / -2.22 / 0.066 / -1.64 / -0.04
30 / -2.36 / 0.043 / -1.82 / -0.04
35 / -2.80 / 0.028 / -2.01 / -0.05
40 / -3.33 / 0.018 / -2.19 / -0.05
45 / -3.52 / 0.012 / -2.38 / -0.05
50 / -3.54 / 0.008 / -2.57 / -0.05
55 / -3.74 / 0.005 / -2.75 / -0.05
60 / -3.98 / 0.0033 / -2.94 / -0.05
65 / -3.96 / 0.0022 / -3.12 / -0.05
70 / -4.02 / 0.0014 / -3.31 / -0.05
75 / -3.98 / 0.00091 / -3.50 / -0.05
80 / -3.86 / 0.00060 / -0.05
 / -3.96 / 0 / -0.05

Calculations

The following k values used to determine the rate of reaction where calculated using equation (5). The k value calculations are shown below.

4M HCl @ 50 min.

k=2.303/50 min*log((26.56+3.92)/(-0.68+3.92))

k=0.04606*log(30.48/3.24)=0.044838026 1/min

6M HCl @ 50 min.

k=2.303/50min*log((26.56+3.96)/(-3.54+3.96))

k=0.04606*log(30.52/0.42)=0.085695874 1/min

C=Coe-kt

C1=Coe-kt

C2=Coe-k(t+t)

C1-C2=C0e-kt(1-e-kt)

log (C1-C2)=-kt/2.303 +log [Co(1-e-kt)]

This is why the slope of the plot, log(C1-C2) vs. Time is -k/2.303.

=ac+b

C1-C2=(1-2)/a

log(1-2)=-kt/2.303+log[a Co(1-e-kt)]

Discussion

The concentration of sucrose as it decomposes over time was determined experimentally by using the first order rate equation as given in the introduction. It was determined that sucrose does indeed decompose over time by hydrolysis giving two products, which are optically active. When plotting log(C1-C2) vs. Time the slope of the line equals -k/2.303, therefore leading to =ac+b where C1-C2=(1-2)/a, log(1-2)=

-kt/2.303+log[aC0(1-exp(-kt)].

Figures  Bibliography

J. H. Reeves and A. M. Halpern, “Experimental Physical Chemistry,” Scott, Foresman/Little, 1988.