An enzyme catalyzes conversion of sucrose to fructose. The half-life of the reaction depends on the concentration of sucrose. With an initial concentration of 3.0 M sucrose, t1/2 = 1.36x10–4 s. With an initial concentration of 1.0 M sucrose, t1/2 = 4.5x10–5 s.
a. What is the reaction order in sucrose?
b. Calculate the rate constant for the reaction.
Step (1) using the formula
putting the all value in formula
n = [log{(1.36×10-4)/(4.5×10-5)}/log(3.0/1.0)] + 1
n = [log(3)/log(3)] + 1
n = 1+1
n = 2
So
Order of reaction is 2
(b)
t(1/2) = 1/k×an-1
k = 1/t(1/2)×an-1
putting the value in formula
k = 1/(1.36×10-4s)(3.0M)2-1
k = 1/4.08×10-4Ms
k = 2.450×10-3M-1s-1
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