Question

The rate law for inversion of cane sugar is R=k [C12H22O11][H2O]. Find the concentration of sucrose if the rate of the reaction is 0.032 s^-1 and rate constant k=0.005.

(a) 5.8 M

(b) 6 M

(c) 6.2 M

(d) 6.4 M

I have been asked this question at a job interview.

I need to ask this question from Chemical Kinetics in section Chemical Kinetics of Chemistry – Class 12

Answer 1

The correct answer is (d) 6.4 M

Easiest explanation: Given, rate constant k=0.005, rate of the reaction R=0.032 s^-1.

 Inversion of cane sugar is a pseudo first order reaction and water is present in large excess so the rate law becomes R=k [C12H22O11]

0.032= 0.005 [C12H22O11]

[C12H22O11]=6.4 M.