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Topic: Physical Chemistry Sucrose Inversion laboratory math derivation  (Read 2660 times)

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Offline kinslow135

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Physical Chemistry Sucrose Inversion laboratory math derivation
« on: January 23, 2016, 03:50:51 PM »
So i am working on a physical chemistry lab concerning sucrose inversion and right now my goal is to relate two equations or derive one from the other.  I am using a polarimeter and the observed optical rotation to relate the speed of the acid catalyzed conversion of sucrose to fructose which involves and optical inversion because of the large and negative optical rotation of fructose.  The equation for the optical rotation at any time is :

αt=[α]s*dm*Cs + [α]i*dm*Ci.  (1)
 This equation says the specific rotation of the solution at any time t,αt, is equal to the sum of the specific rotations of sucrose ,[α]s, and of the "inversion sugar" fructose, [α]i, each of them multiplied by their respective concentrations and the path length in decimeters of the machine.

"If the terms for the sucrose and invert sugar rotations are taken as equivalent to the measured initial and equilibrium rotations equation (1) can be rearranged as follows in terms of the fractional concentration of sucrose remaining at time t : "

Cs,t/Cs,0 = [α]t - [α]/[α]0-[α]. (2)

"The fraction of the initial sucrose concentration remaining at time t is equal to the ratio of rotation at time t minus the equilibrium rotation to the initial rotation minus the equilibrium rotation."

n to initial rotation minus the equilibrium rotation.
Now i get conceptually that at the equilibrium the specific rotation associated with new concentration of the invert sugar will be [α] and it makes sense that the optical rotation at the beginning [α]0 will be associated with the starting sucrose concentration but how do i go from (1) to (2)?  It also says that i will need to find the relationship between Cs and Ci to solve this i also understand that there is a 1:1 molar ratio but how does this get represented mathematically as Cs goes down Ci goes up i guess in terms of one quantity?

How can i go from (1) to (2)?

Here is what i thought to start : Cs,0 - Cs,t = Ci or the concentration of fructose is the difference between the starting concentration of sucrose and the concentration of sucrose at time t.  So then Cs,0 = Ci + Cs,t = [α] + [α]t and Cs,t = Cs,0 - Ci = [α]0 - [α] and this gives me definitions of the ratio Cs,t/Cs,0 =(proportional) dunno where to go from here.
« Last Edit: January 23, 2016, 06:20:16 PM by kinslow135 »

Offline mjc123

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Re: Physical Chemistry Sucrose Inversion laboratory math derivation
« Reply #1 on: January 25, 2016, 08:25:29 AM »
First, be clearer with your notation. You don't need square brackets round the αs in equation 2; these are rotation values, not specific rotations. But you do need parentheses to make the maths clear:
Cs,t/Cs,0 = (αt - α)/(α0 - α)
What you wrote suggests something quite different.
Assume that at time 0 Cs = Cs,0 and Ci = 0; α0 = [α]s*dm*Cs,0
At time ∞ reaction is 100% complete, so Ci = Cs,0 and Cs = 0; α = [α]i*dm*Cs,0
At time t αt = [α]s*dm*Cs,t + [α]i*dm*(Cs,0 - Cs,t)
       = α0*Cs,t/Cs,0 + α - α*Cs,t/Cs,0
Can you take it from there?


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