Chapter 16, Problem 6STP

Interpretation Introduction

Interpretation:

The correct option which is not acceptable unit for expressing a reaction rate needs to be determined.

Concept introduction:

Rate of a reaction is the change in the amount of products or reactants per unit time.

$\text{Rate}=\frac{-\text{Change}\text{in}\text{reactants}}{\text{Change}\text{in}\text{time}}$

Answer to Problem 6STP

B. $\text{L}/\text{s}$

Explanation of Solution

Reaction rate refers speed that the reaction takes place. Average rate is the concentration of reactant over a period of time.

$\text{Average}\text{reaction}\text{rate}=\frac{-\Delta \left[\text{reactants}\right]}{\Delta \text{t}}$

Negative sign out in front line which corrects for decreases in concentration of reactant to yield a +ve average reaction rate.

From the above formula denominator refers time thus unit will be hours (h), seconds(s) and minutes (m). Numerator refers concentration, thus unit of concentration will be number of moles per milliliter $\left(\text{mol}/\text{mL}\right)$ or liter $\left(\text{mol}/\text{L}\right)$ or molarity (M).

In option B numerator is not correct unit for concentration. It gives unit of time in denominator (s) and unit of time in numerator (L)

Conclusion

The correct options are as follows:

$\text{M}/\min$

In numerator it has unit of concentration (M) and denominator it has unit of time (min).

C. $\text{mol}/\left(\text{mL}\cdot \text{h}\right)$

In numerator it has unit of concentration $\left(\text{mol}/\text{mL}\right)$ and denominator it has unit of time (h).

D. $\text{mol}/\left(\text{L}\cdot \min \right)$

In numerator it has unit of concentration $\left(\text{mol}/\text{L}\right)$ and denominator it has unit of time (min).