Chapter 6, Problem 6.134QP

Interpretation Introduction

Interpretation:

The volume of oxygen was collected has to be calculated.

Concept introduction:

Ideal gas law:

The ideal gas law is given by

$\begin{array}{l}\text{PV}\text{=}\text{nRT}\\ \text{where,}\\ \text{P-}\text{pressure}\\ \text{V}\text{-}\text{volume}\\ \text{n-}\text{number}\text{of}\text{moles}\text{of}\text{gas}\\ \text{R}\text{-}\text{ideal}\text{gas}\text{constant}\text{and}\text{T-}\text{temperature}\end{array}$

Answer to Problem 6.134QP

• The volume of oxygen collected is $\text{17}\text{.6}\text{L}\text{.}$

Explanation of Solution

Given information,

$\begin{array}{l}{\text{4KO}}_{\text{2}}{}_{\left(\text{s}\right)}\text{+}{\text{2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\stackrel{}{\to }{\text{4KOH}}_{\left(\text{s}\right)}\text{+}{\text{3O}}_{\text{2}}{}_{\left(\text{g}\right)}\\ {\text{ΔH}}^{\text{o}}{}_{\text{f}}\text{:}\text{4}\text{×}\text{-}\text{284}\text{.5}\text{2}\text{×}\text{-285}\text{.8}\text{4}\text{×}\text{-424}\text{.7}\text{3}\text{×}\text{0}\text{kJ/mol}\\ \text{-1138}\text{.0}\text{-571}\text{.6}\text{-1698}\text{.8}\text{0}\text{kJ/mol}\end{array}$

Calculate the enthalpy change of the given reaction

$\begin{array}{l}{\text{ΔH}}^{\text{o}}\text{=}{\displaystyle \sum {\text{nΔH}}^{\text{o}}{}_{\text{f}}}\left(\text{products}\right)\text{-}{\displaystyle \sum {\text{mΔH}}^{\text{o}}{}_{\text{f}}}\left(\text{reactants}\right)\\ \text{=}\left[\left(\text{-1698}\text{.8}\right)\text{ + }\left(\text{0}\right)\right]\text{ - }\left[\left(\text{-}\text{1138}\text{.0}\right)\text{+}\left(\text{-}\text{571}\text{.6}\right)\right]\text{ kJ}\\ \text{=}\text{10}\text{.8 kJ}\text{.}\end{array}$

Calculate the number of moles of ${\text{O}}_{\text{2}}$ formed when $\text{2}\text{.70}\text{kJ}$ heat is absorbed

$\begin{array}{l}\text{mol}{\text{O}}_{\text{2}}\text{formed}\text{=}\text{2}\text{.70}\text{kJ}\text{×}\frac{\text{3}\text{mol}{\text{O}}_{\text{2}}}{\text{10}\text{.8}\text{kJ}}\\ \text{=}\text{0}\text{.750}\text{mol}{\text{O}}_{\text{2}}\text{.}\end{array}$

Collection conditions for oxygen is $\text{23}{}^{\text{o}}\text{C}\text{and}\text{789}\text{mmHg}$

Calculate the volume of oxygen by using ideal gas equation

$\begin{array}{l}\text{V}\text{=}\frac{\text{nRT}}{\text{P}}\text{=}\frac{\left(\text{0}\text{.750}\text{mol}\right)\left(\frac{\text{0}\text{.08206}\text{L}·\text{atm}}{\text{mol}·\text{K}}\right)\left(\text{296}\text{.2}\text{K}\right)}{\left(\text{789}\text{mmHg}\right)\left(\frac{\text{1}\text{atm}}{\text{760}\text{mmHg}}\right)}\\ \text{=}\text{17}\text{.56}\text{=}\text{17}\text{.6}\text{L}\text{.}\end{array}$

The volume of oxygen was collected is $\text{17}\text{.6}\text{L}\text{.}$

Conclusion

By using ideal gas equation the volume of oxygen collected was calculated.