Question

Chemical reaction in the form 2A → 3B, 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

The average rate of formation of B, given the rate of A's disappearance being –0.15 mol/l-s in the reaction 2A → 3B, is 0.23 mol/l-s when expressed with two significant figures.

Step-by-step explanation:

If a chemical reaction is represented by the equation 2A → 3B, and the rate at which the concentration of A decreases is –0.15 mol/l-s, we can calculate the rate of product formation. The reaction rate should be expressed as a positive number in units of molarity per second (M/s) following convention.

For every 2 moles of A that react, 3 moles of B are produced, maintaining the stoichiometry of the reaction. Therefore, to find the rate at which B forms, we multiply the rate of disappearance of A by the ratio of the stoichiometric coefficients (3/2). The calculation is as follows: (3/2) × 0.15 mol/l-s = 0.225 mol/l-s. This represents the average rate of formation of B in molarity per second. However, rate of reactions are generally expressed to two significant figures which would make the rate 0.23 mol/l-s.