Question

Units of average rate of reaction is:
A. ${{se}}{{{c}}^{ - 1}}$
B. ${{mol}}\,\,{{se}}{{{c}}^{ - 1}}$
C. ${{mol}}\,\,{{d}}{{{m}}^{ - 3}}{{se}}{{{c}}^{ - 1}}$
D. unitless

Answer 1

Hint:The rate of a reaction is defined as the change in concentration divided by the change in time. The concentration change is taken in terms of molarity. Time can be taken into seconds, minutes, hours, days or years. Generally, the time is taken into second.

Complete answer:
The average rate is defined as the change in molar concentration of reactant or product concerning per unit time over a specified period.
The formula of average rate is as follows:
${{r}}\,{{ = }}\,\dfrac{{{{\Delta c}}}}{{{{\Delta t}}}}$
Where,
${{r}}\,$is the average rate
${{\Delta c}}$ is the change in concentration
${{\Delta t}}$ is the change in time
The concentration of any reactant or product is defined in terms of molarity. Molarity is defined as the number of moles of solute dissolved in pre litter so the solution. So, the unit of change in concentration ${{\Delta c}}$ is ${{mol}}\,\,{{{L}}^{ - 1}}$.
One liter is equal to $1\,{{d}}{{{m}}^{ - 3}}$.
So, we can substitute the liter in concentration with ${{d}}{{{m}}^{ - 3}}$.
So, the unit of concentration is ${{mol}}\,\,{{d}}{{{m}}^{ - 3}}$.
Generally, the time is taken into seconds.
So, the unit of average rate is,
${{r}}\,{{ = }}\,\dfrac{{{{mol}}\,\,{{d}}{{{m}}^{ - 3}}}}{{{s}}}$
${{r}}\,{{ = }}\,{{mol}}\,\,{{d}}{{{m}}^{ - 3}}{{se}}{{{c}}^{ - 1}}$
So, the unit of average rate is ${{mol}}\,\,{{d}}{{{m}}^{ - 3}}{{se}}{{{c}}^{ - 1}}$.

Therefore, option (C) ${{mol}}\,\,{{d}}{{{m}}^{ - 3}}{{se}}{{{c}}^{ - 1}}$ is correct.

Note:
For the gaseous reactant or product, the concentration term is replaced with pressure. When the rate is defined for reactant, the term $\dfrac{{{{\Delta c}}}}{{{{\Delta t}}}}$ has a negative sign to make the rate positive because the concentration of reactant decreases during the reaction and the rate is known as the rate of disappearance. When the rate is defined for product, the term $\dfrac{{{{\Delta c}}}}{{{{\Delta t}}}}$ has a positive sign because the concentration of product increases during the reaction and the rate is known as rate of formation.
The sign does not affect the unit of rate.