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 MI 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

The rate law for inversion of cane sugar is R=k [C12H22O11][H2O]. Find

1.	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 MI 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» 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.

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