For the reaction -rA = \(\frac{1670[A][Po]}{6+CA} \frac{kmol}{ml.hr}\) write the units for constants 6 and 1670.(a) kmol/ml and 1/hr(b) 1/hr and 1/hr(c) kmol and kmol(d) data insufficientThis question was addressed to me at a job interview.My question is from Rate Laws and Stoichiometry Definitions in section Rate Laws and Stoichiometry of Chemical Reaction Engineering

For the reaction -rA = \(\frac{1670[A][Po]}{6+CA} \frac{kmol}{ml.hr}\)

1.	For the reaction -rA = \(\frac{1670[A][Po]}{6+CA} \frac{kmol}{ml.hr}\) write the units for constants 6 and 1670.(a) kmol/ml and 1/hr(b) 1/hr and 1/hr(c) kmol and kmol(d) data insufficientThis question was addressed to me at a job interview.My question is from Rate Laws and Stoichiometry Definitions in section Rate Laws and Stoichiometry of Chemical Reaction Engineering
Answer» The correct OPTION is (a) kmol/ML and 1/hr The explanation: To GET the final units -rA = \(\frac{1670[\frac{kmol}{ml}][\frac{kmol}{ml}]}{6+\frac{kmol}{ml}} \) The units of CONSTANTS have to be 1/hr for 1670 and kmol/ml for 6.

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