Optical rotation of sucrose
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In eq 5 from here
$$k=left(2.303/tright)logleft[left(alpha(0)-alpha(infty)right)/left(alpha(t)-alpha(0)right)right]$$
What is the logic for difference of the rotations at $t = 0$ and $t = infty$ to represent initial concenteration and similiarly for the reactant at time = $t$. I have seen other quantities like volume or pressure represented in this form but I don't know the reason
kinetics
edited 3 hours ago
Loong♦
31.1k879147
asked 11 hours ago
harambe
1568
add a comment |Â
up vote
1
down vote
favorite
In eq 5 from here
$$k=left(2.303/tright)logleft[left(alpha(0)-alpha(infty)right)/left(alpha(t)-alpha(0)right)right]$$
What is the logic for difference of the rotations at $t = 0$ and $t = infty$ to represent initial concenteration and similiarly for the reactant at time = $t$. I have seen other quantities like volume or pressure represented in this form but I don't know the reason
kinetics
edited 3 hours ago
Loong♦
31.1k879147
asked 11 hours ago
harambe
1568
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In eq 5 from here
$$k=left(2.303/tright)logleft[left(alpha(0)-alpha(infty)right)/left(alpha(t)-alpha(0)right)right]$$
What is the logic for difference of the rotations at $t = 0$ and $t = infty$ to represent initial concenteration and similiarly for the reactant at time = $t$. I have seen other quantities like volume or pressure represented in this form but I don't know the reason
kinetics
edited 3 hours ago
Loong♦
31.1k879147
asked 11 hours ago
harambe
1568
In eq 5 from here
$$k=left(2.303/tright)logleft[left(alpha(0)-alpha(infty)right)/left(alpha(t)-alpha(0)right)right]$$
What is the logic for difference of the rotations at $t = 0$ and $t = infty$ to represent initial concenteration and similiarly for the reactant at time = $t$. I have seen other quantities like volume or pressure represented in this form but I don't know the reason
kinetics
edited 3 hours ago
Loong♦
31.1k879147
asked 11 hours ago
harambe
1568
edited 3 hours ago
Loong♦
31.1k879147
edited 3 hours ago
Loong♦
31.1k879147
edited 3 hours ago
Loong♦
31.1k879147
31.1k879147
asked 11 hours ago
harambe
1568
asked 11 hours ago
harambe
1568
asked 11 hours ago
harambe
1568
1568
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
Not quite sure what your asking, but I think I understand. If I misinterpreted the question please let me know.
The hydrolysis of sucrose to a mixture of the component simple sugars, fructose, and glucose, causes the direction of rotation to "invert" from right to left. This is because sucrose rotates to the right, but the combination of the two simpler sugars rotate to the left. So the rotation at time zero is for the concentration of sucrose, and the rotation expected when the reaction is complete at time infinity can be calculated from the concentration of the sucrose.
answered 10 hours ago
MaxW
13.4k11855
I still can't get the reasoning behind why the difference is directly proportional to the initial concentration and similarly the difference of rotation at infinite time and any time t is directly proportional to concentration at any other time
– harambe
8 hours ago
@harambe - Let's say that sucrose rotates $50^circ$ to the right and the completely reacted product of fructose and glucose rotates $50^circ$ to the left. There is thus a $100^circ$ difference, so each 1% of the reaction will cause a $1^circ$ change in rotation.
– MaxW
8 hours ago
1
Well now I know what "invert sugar" means.
– A.K.
2 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Not quite sure what your asking, but I think I understand. If I misinterpreted the question please let me know.
The hydrolysis of sucrose to a mixture of the component simple sugars, fructose, and glucose, causes the direction of rotation to "invert" from right to left. This is because sucrose rotates to the right, but the combination of the two simpler sugars rotate to the left. So the rotation at time zero is for the concentration of sucrose, and the rotation expected when the reaction is complete at time infinity can be calculated from the concentration of the sucrose.
answered 10 hours ago
MaxW
13.4k11855
I still can't get the reasoning behind why the difference is directly proportional to the initial concentration and similarly the difference of rotation at infinite time and any time t is directly proportional to concentration at any other time
– harambe
8 hours ago
@harambe - Let's say that sucrose rotates $50^circ$ to the right and the completely reacted product of fructose and glucose rotates $50^circ$ to the left. There is thus a $100^circ$ difference, so each 1% of the reaction will cause a $1^circ$ change in rotation.
– MaxW
8 hours ago
1
Well now I know what "invert sugar" means.
– A.K.
2 hours ago
add a comment |Â
up vote
6
down vote
accepted
Not quite sure what your asking, but I think I understand. If I misinterpreted the question please let me know.
The hydrolysis of sucrose to a mixture of the component simple sugars, fructose, and glucose, causes the direction of rotation to "invert" from right to left. This is because sucrose rotates to the right, but the combination of the two simpler sugars rotate to the left. So the rotation at time zero is for the concentration of sucrose, and the rotation expected when the reaction is complete at time infinity can be calculated from the concentration of the sucrose.
answered 10 hours ago
MaxW
13.4k11855
I still can't get the reasoning behind why the difference is directly proportional to the initial concentration and similarly the difference of rotation at infinite time and any time t is directly proportional to concentration at any other time
– harambe
8 hours ago
@harambe - Let's say that sucrose rotates $50^circ$ to the right and the completely reacted product of fructose and glucose rotates $50^circ$ to the left. There is thus a $100^circ$ difference, so each 1% of the reaction will cause a $1^circ$ change in rotation.
– MaxW
8 hours ago
1
Well now I know what "invert sugar" means.
– A.K.
2 hours ago
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Not quite sure what your asking, but I think I understand. If I misinterpreted the question please let me know.
The hydrolysis of sucrose to a mixture of the component simple sugars, fructose, and glucose, causes the direction of rotation to "invert" from right to left. This is because sucrose rotates to the right, but the combination of the two simpler sugars rotate to the left. So the rotation at time zero is for the concentration of sucrose, and the rotation expected when the reaction is complete at time infinity can be calculated from the concentration of the sucrose.
answered 10 hours ago
MaxW
13.4k11855
Not quite sure what your asking, but I think I understand. If I misinterpreted the question please let me know.
The hydrolysis of sucrose to a mixture of the component simple sugars, fructose, and glucose, causes the direction of rotation to "invert" from right to left. This is because sucrose rotates to the right, but the combination of the two simpler sugars rotate to the left. So the rotation at time zero is for the concentration of sucrose, and the rotation expected when the reaction is complete at time infinity can be calculated from the concentration of the sucrose.
answered 10 hours ago
MaxW
13.4k11855
answered 10 hours ago
MaxW
13.4k11855
answered 10 hours ago
MaxW
13.4k11855
answered 10 hours ago
MaxW
13.4k11855
13.4k11855
I still can't get the reasoning behind why the difference is directly proportional to the initial concentration and similarly the difference of rotation at infinite time and any time t is directly proportional to concentration at any other time
– harambe
8 hours ago
@harambe - Let's say that sucrose rotates $50^circ$ to the right and the completely reacted product of fructose and glucose rotates $50^circ$ to the left. There is thus a $100^circ$ difference, so each 1% of the reaction will cause a $1^circ$ change in rotation.
– MaxW
8 hours ago
1
Well now I know what "invert sugar" means.
– A.K.
2 hours ago
add a comment |Â
I still can't get the reasoning behind why the difference is directly proportional to the initial concentration and similarly the difference of rotation at infinite time and any time t is directly proportional to concentration at any other time
– harambe
8 hours ago
@harambe - Let's say that sucrose rotates $50^circ$ to the right and the completely reacted product of fructose and glucose rotates $50^circ$ to the left. There is thus a $100^circ$ difference, so each 1% of the reaction will cause a $1^circ$ change in rotation.
– MaxW
8 hours ago
1
Well now I know what "invert sugar" means.
– A.K.
2 hours ago
I still can't get the reasoning behind why the difference is directly proportional to the initial concentration and similarly the difference of rotation at infinite time and any time t is directly proportional to concentration at any other time
– harambe
8 hours ago
I still can't get the reasoning behind why the difference is directly proportional to the initial concentration and similarly the difference of rotation at infinite time and any time t is directly proportional to concentration at any other time
– harambe
8 hours ago
@harambe - Let's say that sucrose rotates $50^circ$ to the right and the completely reacted product of fructose and glucose rotates $50^circ$ to the left. There is thus a $100^circ$ difference, so each 1% of the reaction will cause a $1^circ$ change in rotation.
– MaxW
8 hours ago
@harambe - Let's say that sucrose rotates $50^circ$ to the right and the completely reacted product of fructose and glucose rotates $50^circ$ to the left. There is thus a $100^circ$ difference, so each 1% of the reaction will cause a $1^circ$ change in rotation.
– MaxW
8 hours ago
1
1
Well now I know what "invert sugar" means.
– A.K.
2 hours ago
Well now I know what "invert sugar" means.
– A.K.
2 hours ago
add a comment |Â
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