A chemical reaction in the form 2 A --> 3 B, is observed to have the rate of chemical A decrease at 0.15 mol/L-s. How would the rate of this reaction be reported in mol/L-s?

Question

A chemical reaction in the form 2 A --> 3 B, is observed to have the rate of chemical A decrease at 0.15 mol/L-s. How would the rate of this reaction be reported in mol/L-s?

A. 0.075

B. 0.10

C. 0.15

D. 0.23

E. 0.45

Answer 1

2A ==> 3B

For simplicity we can write mol / L-sec as M/sec or Ms^-1

-1/2 ∆[A]/∆t = + 1/3 ∆[B]/∆t

-1/2 (0.15 M/s) = 0.075 M / L-s ANSWER A.