Chapter 4, Problem 82QRT

(a)

Interpretation Introduction

Interpretation:

The reaction enthalpy when $0.054\text{ g}$ of sulfur burns forming $S{O}_2\left(g\right)$ has to be calculated.

Answer to Problem 82QRT

The reaction enthalpy value is $\text{- 0}\text{.50kJ}$.

Explanation of Solution

Combustion reaction of sulfur is $S\left(s\right)+O_2\left(g\right)\to S{O}_2\left(g\right)$.

The ${\Delta }_f{H}^{\circ }$ value of above reaction is $\text{- 1206}\text{.92 kJ/mol}$.

The given grams of sulfur is converted into moles.

    $\begin{array}{c}Moles\text{ = }\frac{Mass}{Molar\text{ mass}}\\ \\ =\frac{0.054\text{ g S}}{32.065\text{ g/mol}}\\ \\ =\text{ 1}\text{.684}\times {\text{10}}^{\text{-3}}\text{ moles}\end{array}$

The reaction enthalpy is as follows,

    $\text{1}\text{.684}\times {\text{10}}^{\text{-3}}\text{ moles}$ of sulfur gives $\text{1}\text{.684}\times {\text{10}}^{\text{-3}}\text{ moles}$ of $S{O}_2\left(g\right)$

    $\begin{array}{l}\text{= 1}\text{.684}\times {\text{10}}^{\text{-3}}{\text{ moles of SO}}_{\text{2}}\times \frac{-296.830\text{ kJ}}{{\text{1 mole of SO}}_{\text{2}}}\\ \\ =\text{ - 0}\text{.50kJ}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The reaction enthalpy when $0.20\text{ mol}$ of $HgO\left(s\right)$ decomposes forming $Hg\left(l\right)+O_2\left(g\right)$ has to be calculated.

Answer to Problem 82QRT

The reaction enthalpy value is $\text{+18 kJ}$.

Explanation of Solution

Decomposition reaction of $HgO\left(s\right)$ is $HgO\left(s\right)\to Hg\left(l\right)+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.O_2\left(g\right)$.

The ${\Delta }_f{H}^{\circ }$ value of above reaction is $\text{- 90}\text{.83 kJ/mol}$ hence, its ${\Delta }_r{H}^{\circ }$ value is $\text{+90}\text{.83 kJ/mol}$

The reaction enthalpy is as follows,

    $\begin{array}{l}\text{= }0.20\text{ moles of HgO}\times \frac{+90.83\text{ kJ}}{\text{1 mole of HgO}}\\ \\ =\text{ +18 kJ}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The reaction enthalpy when $2.40\text{ g}$ of $N{H}_3\left(g\right)$ formed from $N_2\left(g\right)$ and $H_2\left(g\right)$ has to be calculated.

Answer to Problem 82QRT

The reaction enthalpy value is $\text{- 6}\text{.50 kJ}$.

Explanation of Solution

Formation reaction of $N{H}_3\left(g\right)$ is $\frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)\to N{H}_3\left(g\right)$.

The ${\Delta }_f{H}^{\circ }$ value of above reaction is $\text{- 46}\text{.11 kJ/mol}$.

The given grams of $N{H}_3\left(g\right)$ is converted into moles.

    $\begin{array}{c}Moles\text{ = }\frac{Mass}{Molar\text{ mass}}\\ \\ =\frac{2.40\text{ g }}{17.0304\text{ g/mol}}\\ \\ =0.1409\text{ moles}\end{array}$

The reaction enthalpy is as follows,

    $\begin{array}{l}\text{= }0.1409{\text{ moles of NH}}_{\text{3}}\times \frac{-46.11\text{ kJ}}{{\text{1 mole of NH}}_{\text{3}}}\\ \\ =\text{ - 6}\text{.50 kJ}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The reaction enthalpy when $1.05\times {\text{ 10}}^{\text{-2}}$ moles of carbon is oxidized to $C{O}_2\left(g\right)$ has to be calculated.

Answer to Problem 82QRT

The reaction enthalpy value is $\text{- 4}\text{.13 kJ}$.

Explanation of Solution

Formation reaction of $C{O}_2\left(g\right)$ is $C\left(s\right)+O_2\left(g\right)\to C{O}_2\left(g\right)$.

The ${\Delta }_f{H}^{\circ }$ value of above reaction is $\text{- 393}\text{.509 kJ/mol}$.

The reaction enthalpy is as follows,

    $\begin{array}{l}\text{= }1.05\times {10}^{-2}\text{ moles of C}\times \frac{-393.509\text{ kJ}}{\text{1 mole of C}}\\ \\ =\text{ - 4}\text{.13 kJ}\end{array}$