Chapter 2, Problem 2A.1ST

Interpretation Introduction

Interpretation: The expansion work done when $\text{50 g}$ of water is electrolysed under constant pressure at ${\text{25 }}^{\text{ο}}\text{C}$ has to be calculated.

Concept introduction: Thermodynamics address temperature and heat and their relationship to work, energy and matter.  Work is defined as the form of mechanical energy that is transferred from system to another.  The work done is calculated by the formula,

    $w=-nRT$

Answer to Problem 2A.1ST

The expansion work done when $\text{50 g}$ of water is electrolysed under constant pressure at ${\text{25 }}^{\text{ο}}\text{C}$ is $\underset{\_}{-10.33147kJ}$.

Explanation of Solution

It is given that the mass of water is $\text{50 g}$.

The temperature is ${\text{25 }}^{\text{ο}}\text{C}$.

Pressure is $\text{1 atm}$.

The conversion of Celsius to Kelvin is done as,

    ${\text{0 }}^{\text{ο}}\text{C}=2\text{73 K}$

Therefore, the conversion of ${\text{25 }}^{\text{ο}}\text{C}$ to Kelvin is done as,

    $\begin{array}{c}{\text{25 }}^{\text{ο}}\text{C}=25+2\text{73 K}\\ =2\text{98 K}\end{array}$

The balanced chemical equation for the electrolysis of water is,

    ${\text{2H}}_{\text{2}}\text{O}\left(\text{l}\right)\to {\text{2H}}_{\text{2}}\left(\text{g}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)$

First, the number of moles of oxygen and hydrogen are calculated by the formula,

    $\text{Number of moles}=\frac{\text{Mass}\left(\text{g}\right)}{\text{Molar mass}\left({\text{g mol}}^{-1}\right)}$                                                         (1)

The molar mass of water is $1{\text{8 g mol}}^{-1}$.

For oxygen,

    $\begin{array}{c}\text{Moles of oxygen}=5{\text{0 g H}}_{\text{2}}\text{O}\times \frac{{\text{1 mol H}}_{\text{2}}\text{O}}{{\text{18 g H}}_{\text{2}}\text{O}}\times \frac{{\text{1 mol O}}_2}{{\text{2 mol H}}_{\text{2}}\text{O}}\\ =\text{1}{\text{.39 mol O}}_{\text{2}}\end{array}$

For hydrogen,

    $\begin{array}{c}\text{Moles of oxygen}=5{\text{0 g H}}_{\text{2}}\text{O}\times \frac{{\text{1 mol H}}_{\text{2}}\text{O}}{{\text{18 g H}}_{\text{2}}\text{O}}\times \frac{{\text{2 mol H}}_2}{{\text{2 mol H}}_{\text{2}}\text{O}}\\ =2.78{\text{ mol H}}_{\text{2}}\end{array}$

Now, the initial volume of water is assumed to be negligible when compared to final volume of both hydrogen and oxygen gas.  Therefore,

    $\text{Δ}V=V_{\text{final}}$                                                                                                        (2)

The ideal gas equation is given by the expression,

    $V_{\text{final}}=\frac{nRT}{P_{\text{ext}}}$                                                                                                    (3)

Now, the expansion work is calculated by the formula,

    $w=-P_{\text{ext}}\times V_{\text{final}}$                                                                                               (4)

Substitute the value of ${\text{V}}_{\text{final}}$ from equation (3) to equation (4).

    $\begin{array}{l}w=-P_{\text{ext}}\times \frac{nRT}{P_{\text{ext}}}\\ w=-nRT\end{array}$                                                                                              (5)

Substitute the values of number of moles of oxygen and hydrogen, universal gas constant and temperature in the above equation (5) to calculate the expansion work.

    $\begin{array}{c}w=-\left(\text{1}{\text{.39 mol O}}_{\text{2}}+2.78{\text{ mol H}}_{\text{2}}\right)\times \text{8}{\text{.314 J K}}^{-1}{\text{mol}}^{-1}\times 29\text{8 K}\\ =-4.1\text{7 mol}\times 2477.57{\text{2 J mol}}^{-1}\\ =-10331.4\text{7 J}\end{array}$

The conversion of joule into $\text{kJ}$ is done as,

    $\begin{array}{l}\text{1 kJ}=100\text{0 J}\\ \text{1 J}={10}^{-3}\text{ kJ}\end{array}$

Therefore, the conversion of $-10331.4\text{7 J}$ into $\text{kJ}$ is done as,

    $\begin{array}{c}-10331.4\text{7 J}={10}^{-3}\times -10331.4\text{7 kJ}\\ =\underset{\_}{-10.33147kJ}\end{array}$

Hence, the expansion work done when $\text{50 g}$ of water is electrolysed under constant pressure at ${\text{25 }}^{\text{ο}}\text{C}$ is $\underset{\_}{-10.33147kJ}$.