Chapter 6, Problem 6.141QP

Interpretation Introduction

Interpretation:

The composition of the initial gas mixture has to be given.

Concept introduction:

Ideal gas law:

The ideal gas law is given by

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

Explanation of Solution

Given information

$\begin{array}{l}\text{Total}\text{mass}\text{of}\text{hydrogen}\text{and}\text{oxygen}\text{gas}\text{is}\text{2}\text{.500}\text{g}\text{.}\\ \text{Temperature}\text{is}\text{25}{}^{\text{o}}\text{C}\text{and}\text{pressure}\text{is}\text{1}\text{atm}\text{.}\\ \text{Heat}\text{generated}\text{from}\text{the}\text{reaction}\text{is}\text{28}\text{.6}\text{kJ}\text{.}\end{array}$

Let $\text{X}$ is the mass of ${\text{H}}_{\text{2}}$ and $\text{Y}$ is the mass of ${\text{O}}_{\text{2}}\text{.}$  then, $\text{2}\text{.500}\text{g}\text{=}\text{X}\text{+}\text{Y}\text{.}$

Write the balanced equation and calculate the moles of ${\text{H}}_{\text{2}}\text{O}$ formed

${\text{2H}}_{\text{2}}{}_{\left(\text{g}\right)}\text{+}{\text{O}}_{\text{2}}{}_{\left(\text{g}\right)}\stackrel{}{\to }{\text{2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{;}{\text{ΔH}}^{\text{o}}\text{=}\text{-}\text{571}\text{.6}\text{kJ}\text{.}$

Given that, $\text{28}\text{.6}\text{kJ}$ of heat is evolved from the reaction,

$\text{-}\text{28}\text{.6}\text{kJ}\left(\frac{\text{2}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{-}\text{571}\text{.6}\text{kJ}}\right)\text{=}\text{0}\text{.100}{\text{H}}_{\text{2}}\text{O}\text{formed}$

This shows that either $\text{0}\text{.0500}\text{mol}$ of limiting ${\text{O}}_{\text{2}}$ reacted and ${\text{H}}_{\text{2}}$ was in excess amount, or $\text{0}\text{.100}\text{mol}$ limiting ${\text{H}}_{\text{2}}$ reacted and ${\text{O}}_{\text{2}}$ was in excess amount.

Case I, if ${\text{O}}_{\text{2}}$ was the limiting reactant

$\text{0}\text{.0500}\text{mol}{\text{O}}_{\text{2}}\left(\frac{\text{32}\text{.00}\text{g}{\text{O}}_{\text{2}}}{\text{1}\text{.00}\text{mol}{\text{O}}_{\text{2}}}\right)\text{=}\text{1}\text{.600}\text{g}{\text{O}}_{\text{2}}\text{reacted}\text{.}$

If this were true, this would mean $\text{0}\text{.900}\text{g}{\text{H}}_{\text{2}}$ was initially present in the mixture.  If $\text{0}\text{.100}\text{mol}{\text{H}}_{\text{2}}$ reacted $\left(\text{i}\text{.e}\text{.}\text{0}\text{.202}\text{g}\right)$, the mass of ${\text{H}}_{\text{2}}$ present in excess would be $\text{0}\text{.698}\text{g}\text{.}$  although these are feasible quantities, evaluation of the second case with ${\text{H}}_{\text{2}}$ limiting is warranted.

Case II, i.e. with $\text{0}\text{.100}\text{mol}{\text{H}}_{\text{2}}$ limiting the reaction,

$\text{0}\text{.100}\text{mol}{\text{H}}_{\text{2}}\left(\frac{\text{2}\text{.016}\text{g}{\text{H}}_{\text{2}}}{\text{1}\text{mol}{\text{H}}_{\text{2}}}\right)\text{=}\text{0}\text{.202}\text{g}{\text{H}}_{\text{2}}\text{reacted}\text{.}$

If this were true, this would mean $\text{2}\text{.298}\text{g}{\text{O}}_{\text{2}}$ was initially present in the mixture.  If $\text{0}\text{.0500}\text{mol}{\text{O}}_{\text{2}}$ reacted $\left(\text{i}\text{.e}\text{.}\text{1}\text{.600}\text{g}\right)$, the mass of ${\text{O}}_{\text{2}}$ present in excess would be $\text{0}\text{.698}\text{g}\text{.}$  again, feasible quantities result.

Find which one is best.  Had the volume of the initial mixture also been measured, a second independent equation relating the masses of ${\text{H}}_{\text{2}}\text{and}{\text{O}}_{\text{2}}$ would have been available to use:

${\text{n}}_{\text{gas, total}}\text{=}\frac{\text{PV}}{\text{RT}}\text{=}{\text{n}}_{{\text{O}}_{\text{2}}}\text{+}{\text{n}}_{{\text{H}}_{\text{2}}}\text{=}\frac{\text{X}}{\text{32}\text{.00}}\text{+}\frac{\text{Y}}{\text{2}\text{.02}}$

If case I were true, its volume would have been ${\text{V}}_{\text{1}}$;  if case II were true, its volume would have been ${\text{V}}_{\text{2}}$.  These volumes can readily be calculated from the masses shown above and the ideal gas law:

$\begin{array}{l}{\text{V}}_{\text{case}\text{i}}\text{=}\frac{{\text{n}}_{\text{gas,}\text{total}}\text{RT}}{\text{P}}\text{=}\frac{\left({\text{n}}_{\text{case}\text{i,}{\text{O}}_{\text{2}}}\text{+}{\text{n}}_{\text{case}\text{i,}{\text{H}}_{\text{2}}}\right)\text{RT}}{\text{P}}\\ \text{=}\frac{\left(\frac{{\text{X}}_{\text{i}}}{\text{32}\text{.00}\text{g/mol}}\text{+}\frac{{\text{Y}}_{\text{i}}}{\text{2}\text{.016}\text{g/mol}}\right)\text{RT}}{\text{P}}\end{array}$

$\begin{array}{l}{\text{V}}_{\text{case}\text{1}}\text{=}\frac{\left(\frac{\text{1}\text{.600}\text{g}}{\text{32}\text{.00}\text{g/mol}}\text{+}\frac{\text{0}\text{.900}\text{g}}{\text{2}\text{.016}\text{g/mol}}\right)\left(\text{0}\text{.08206}\frac{\text{L}\text{atm}}{\text{mol}\text{K}}\right)\left(\text{298}\text{K}\right)}{\text{1}\text{.00}\text{atm}}\\ \text{=}\text{12}\text{.1}\text{L}\text{if}\text{case}\text{I}\text{were}\text{true}\text{.}\end{array}$

$\begin{array}{l}{\text{V}}_{\text{case}\text{2}}\text{=}\frac{\left(\frac{\text{2}\text{.298}\text{g}}{\text{32}\text{.00}\text{g/mol}}\text{+}\frac{\text{0}\text{.202}\text{g}}{\text{2}\text{.016}\text{g/mol}}\right)\left(\text{0}\text{.08206}\frac{\text{L}\text{atm}}{\text{mol}\text{K}}\right)\left(\text{298}\text{K}\right)}{\text{1}\text{.00}\text{atm}}\\ \text{=}\text{4}\text{.21}\text{L}\text{if}\text{case}\text{II}\text{were}\text{true}\text{.}\end{array}$

In summary, a critical piece of information on the initial gas mixture was omitted.