Chapter 16, Problem 48A

Interpretation Introduction

Interpretation:

To express the average rate as mol Mg consumed/min for the reaction,

$\text{Mg}\left(\text{s}\right)\text{+}\text{2HCl}\to {\text{H}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{MgCl}}_{\text{2}}\left(\text{aq}\right)$

Concept introduction:

In a chemical reaction, the reactants are written on the left side of the equation and the product is written on the right side.

The average rate of a reaction can be given by the formula,

Average reaction rate $\text{=}-\frac{\text{Δ}\left[\text{reactant}\right]}{\text{Δt}}$

Here, $\text{Δ}\left[\text{reactant}\right]$ = change in the concentration of the reactant

Answer to Problem 48A

The average rate as mol Mg consumed/min for the reaction, $\text{Mg}\left(\text{s}\right)\text{+}\text{2HCl}\to {\text{H}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{MgCl}}_{\text{2}}\left(\text{aq}\right)$ is $0.0206\frac{\text{mol}}{\text{min}}$

Explanation of Solution

Given, for the reaction,

$\text{Mg}\left(\text{s}\right)\text{+}\text{2HCl}\to {\text{H}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{MgCl}}_{\text{2}}\left(\text{aq}\right)$, 6.00 g of Mg was present at 0.00 min. After 3.00 min, 4.50 g of Mg remained.

So, using the equation,

Average reaction rate $\text{=}-\frac{\text{Δ}\left[\text{reactant}\right]}{\text{Δt}}$, we find,

$\begin{array}{c}\text{Δ}\left[\text{reactant}\right]\text{=Δ}\left[{\text{I}}_{\text{2}}\right]\\ \text{= 4}\text{.50}\text{g}-6.00\text{g}\\ \text{=}-1.50\text{g}\end{array}$

To find the answer in units of moles of Mg consumed over time in minutes, we convert the $\text{Δ}\left[\text{reactant}\right]$ value by multiplying it with the conversion factor $\frac{\text{1}\text{mol}}{\text{24}\text{.305}\text{g}}$

Therefore,

$\begin{array}{c}\text{Δ}\left[\text{reactant}\right]\text{=}-1.50\text{g}\left(\frac{\text{1}\text{mol}}{\text{24}\text{.305}\text{g}}\right)\\ =\left(\frac{-\text{1}.50}{\text{24}\text{.305}}\right)\text{mol}\\ \text{=}-0.0617\text{mol}\end{array}$

The change in time for the reaction is $\Delta t=3.00\min -0.00\min =3.00\min$, $\Delta \left[\text{Mg}\right]=-0.0617\text{mol}$ and $\Delta t=3.00\min$. Using these values, we will calculate the average reaction rate as,

Average reaction rate $\text{=}-\frac{\text{Δ}\left[\text{reactant}\right]}{\text{Δt}}$

$\begin{array}{l}=-\frac{\Delta \left[\text{Mg}\right]}{\Delta t}\\ =-\frac{-0.0617\text{mol}}{3.00\min }\\ =\frac{0.6017}{3.00}\frac{\text{mol}}{\text{min}}\\ =0.0206\frac{\text{mol}}{\text{min}}\end{array}$

Therefore, the reaction rate is $0.0206\frac{\text{mol}}{\text{min}}$

Conclusion

$0.0206\frac{\text{mol}}{\text{min}}$ is the value of average rate as mol Mg consumed/min for the given reaction.