Answer:
(a)
Rate of appearance of sucrose = - d[C12H22O11] / dt = - ( 0.808 - 1.002 ) / ( 60.0 - 0.0) = 0.00323 M/s
(b)
Rate of appearance of fructose = d[C6H12O6] / dt = (1.002 - 0.808) / (60.0 - 0.0) = 0.00323 M/s
(c)
k = (1 / t ) * ln[A]/[A]t
k = ( 1 / 60.0 ) * ln[1.002 / 0.808]
k = 0.00359 min-1
(d)
0.00359 = ( 1 / t ) * ln[1.002 / 0.212]
t = 432.6 min
(e)
Half life time = 0.693 / k = 0.693 / 0.00359 = 193 min
Explanation:
First-order reactions are defined as the chemical reactions in which rate of the reaction is linearly dependent on the concentration of only one reactant.
The answers can be explained as:
(a) Rate of appearance of the sucrose from the chemical reaction is:
Rate = [tex]\dfrac{\text d [\text C_{12}\text H_{22}\text O_{11}]}{\text {dt}}[/tex] = [tex]\dfrac{0.808 - 1.002}{60.0 -0.0}[/tex]
Rate = 0.00323 m/s
(b) Rate of appearance of Fructose from the given chemical reaction is:
Rate = [tex]\dfrac{\text d [\text C_{6}\text H_{12}\text O_{6}]}{\text {dt}}[/tex] = [tex]\dfrac{1.002 - 0.808 }{60.0 -0.0}[/tex]
Rate = 0.00323 m/s
(C) Rate constant for the reaction is:
[tex]\text k &= \dfrac{1}{\text t}\times \dfrac {\text{ln [A]}}{\text {[A]} \text t}[/tex]
[tex]\text k &= \dfrac{1}{60}\times \dfrac {\text{ln} (1.002)}{(0.808)}[/tex]
k = 0.00359 minute⁻¹
(d) Time required for the concentration of sucrose to drop from 1.002 to 0.212 M is:
[tex]0.00359 &= \dfrac{1}{t} \times {\text{ln}\dfrac{[1.002]}{[0.212]}[/tex]
t = 432.6 minutes
(e) The half-life of the decomposition of sucrose at 25°C is:
Half-life = [tex]\dfrac{0.693}{\text k} = \dfrac{0.693}{0.00359}[/tex]
Half-life = 193 minutes.
To know more about first-order reaction, refer to the following link:
https://brainly.com/question/12446045
