Question

When sucrose, C₁₂H₂₂O₁₁, is dissolved in H₂O in the presence of an acid catalyst it reacts according to the equation C₁₂H₂₂O₁₁ + H₂O → 2 C₆H₁₂O₆with a rate law of Rate = k[C₁₂H₂₂O₁₁]. If 3.00 g of sucrose decreases to 2.70 g in 2.50 hours in the presence of a certain concentration of an acid catalyst, what is the half-life for this reaction under these same conditions?

12.5 hours

16.4 hours

23.7 hours

37.9 hours

Answer 1

Solution- the reaction is-

C12H22O11+ H2O=====> 2C6H12O6.  

And rate of law for the reaction

Rate k(C12H22O11).

It is the first order reaction,

for first order reaction rate constant k is

k =( 2.303/t) log ( initial concentration/ final concentration) Given data:- Initial concentration = 3.00g Final concentration = 2.70g Time (t) = 2.50 h

Now , . k = (2.303/2.50) log ( 3.0/2.7)

k = 0.04215 h^-1

For the 1st order reaction half life = 0.693/k = (0.693/0.04215)

= 16.4 hours

Option B is correct