Chapter 4, Problem 101QRT

(a)

Interpretation Introduction

Interpretation:

The standard reaction enthalpy for following reaction has to be calculated.

    $N{H}_{4}N{O}_3\left(s\right)\to N_{2}O\left(g\right)+2H_{2}O\left(g\right)$

Concept Introduction:

Enthalpy is the amount energy absorbed or released in a process.

The value of standard enthalpy change ${\text{ΔH}}^{\text{ο}}$ of the given reaction is calculated by the formula,

  ${\text{ΔH}}^{\text{ο}}\text{=}{{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{products}\right)}-{{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{ο}}}}_{\left(\text{reactants}\right)}$

Answer to Problem 101QRT

The standard reaction enthalpy of given reaction is $\text{-36}\text{.03 kJ}$.

Explanation of Solution

Given: $N{H}_{4}N{O}_3\left(s\right)\to N_{2}O\left(g\right)+2H_{2}O\left(g\right)$

Using the above given reaction enthalpy is calculated by subtracting the summation of formation of products from the summation of formation enthalpy of reactants.

  $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}\text{ = }\left[{\text{Δ}}_{\text{f}}{\text{H}}^{\text{o}}{N}_{2}O\left(g\right)+2{\text{Δ}}_{\text{f}}{\text{H}}^{\text{o}}{H}_{2}O\left(g\right)\right]-\left[{\text{Δ}}_{\text{f}}{\text{H}}^{\text{o}}N{H}_{4}N{O}_3\left(s\right)\right]\\ \text{= }\left[\left(\text{82}\text{.05kJ/mol}\right)\text{+2}\left(\text{-241}\text{.818kJ/mol}\right)\right]\text{-}\left(\text{-365}\text{.56kJ/mol}\right)\\ =\text{ -36}\text{.03kJ}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The amount of heat transferred when eight kilogram of ammonium nitrate explodes has to be calculated.

Answer to Problem 101QRT

The amount of heat energy evolved is $-1.18\times {10}^4\text{ kJ }$.

Explanation of Solution

The minus sign in reaction enthalpy change value shows that heat is evolved in the given reaction.

    $N{H}_{4}N{O}_3\left(s\right)\to 2N_2\left(g\right)+4H_{2}O\left(g\right)+O_2\left(g\right)$

    $\begin{array}{c}{\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}\text{ = }\left[{\text{4Δ}}_{\text{f}}{\text{H}}^{\text{o}}{H}_{2}O\left(g\right)\right]-\left[{\text{2 Δ}}_{\text{f}}{\text{H}}^{\text{o}}N{H}_{4}N{O}_3\left(s\right)\right]\\ \text{= }\left[4\left(\text{-241}\text{.818kJ/mol}\right)\right]\text{-2}\left(\text{-365}\text{.56kJ/mol}\right)\\ =\text{- 236}\text{.15kJ}\end{array}$

Molar mass of $N{H}_{4}N{O}_3\left(s\right)$ is as follows,

    $\begin{array}{l}=2\left(14.0067\text{ g/mol}\text{}\text{N}\right)+3\left(15.9994\text{ g/mol O}\right)+4\left(1.0079\text{ g/mol H}\right)\\ =\text{ 80}\text{.0432 g/mol }N{H}_{4}N{O}_3\end{array}$

The amount of heat transferred when eight kilograms of $N{H}_{4}N{O}_3\left(s\right)$ explodes is indicated as shown below,

  $\begin{array}{l}\text{=8 kg }N{H}_{4}N{O}_3\left(s\right)\text{×}\frac{1000\text{ g}}{1\text{ kg}}\frac{1\text{ mol }N{H}_{4}N{O}_3\left(s\right)}{80.0432\text{ g }N{H}_{4}N{O}_3\left(s\right)}\times \frac{-236.15\text{ kJ}}{\text{2 mol }N{H}_{4}N{O}_3\left(s\right)}\\ =-1.18\times {10}^4\text{ kJ }\end{array}$