Question

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, t_1/2 = 1.36x10^–4 s. With an initial concentration of 1.0 M sucrose, t_1/2 = 4.5x10^–5 s.

a. What is the reaction order in sucrose?

b. Calculate the rate constant for the reaction.

Answer 1

Step (1) using the formula

putting the all value in formula

HALF-LIFE DETERMINATION METHOD In general the half-life of a reaction in which the concentration of all reactants are identical, the relationship is; where n is the order of reaction. . Thus if two reactions are run at different initial concentrations, a, & a2 with their respective half-lives and putting them in above equation in logarithmic form we finally get nlog (tiam/tina)+ l log (a2011)

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×a^n-1

k = 1/t_(1/2)×a^n-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