Chapter 4, Problem 4D.6BST

Interpretation Introduction

Interpretation:

The standard enthalpy of formation of urea has to be determined.

Concept Introduction:

The difference in enthalpies of product and reactant species in a reaction is called enthalpy change $\left(\Delta H\right)$.  If the difference is negative, the reaction is said to be exothermic.  The value of $\Delta H_{\text{rxn}}^{\text{ο}}$ is determined by using,

  $\Delta H_{\text{rxn}}^{\text{ο}}={\displaystyle \sum m\cdot H_{\text{f}}{}^{\text{ο}}}\left(\text{products}\right)-{\displaystyle \sum n\cdot H_{\text{f}}{}^{\text{ο}}}\left(\text{reactants}\right)$

Explanation of Solution

The thermochemical equation for the combustion of one mole of urea is shown below.

    $CO{\left(N{H}_2\right)}_2\left(s\right)+\frac{3}{2}{O}_{\text{2}}\left(g\right)\to C{O}_{\text{2}}\left(g\right)+N_{\text{2}}\left(g\right)+2H_2\text{O}\left(l\right),\Delta {\text{H}}_c^{\circ }=-\text{632 kJ}\cdot {\text{mol}}^{-\text{1}}$

The standard enthalpy of formation of products that are $C{O}_{\text{2}}\left(g\right)$, $N_{\text{2}}\left(g\right)$ and $H_2\text{O}\left(l\right)$ is $-\text{393}\text{.51}\text{kJ}\cdot {\text{mol}}^{-1}$, $0.0\text{kJ}\cdot {\text{mol}}^{-1}$ and $-\text{285}\text{.83}\text{kJ}\cdot {\text{mol}}^{-1}$ respectively.

The total enthalpy of formation of products is calculated by the expression shown below.

  ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}={\text{n}}_{C{O}_2}\Delta {\text{H}}_f^{\circ }\left(C{O}_2,g\right)+{\text{n}}_{N_2}\Delta {\text{H}}_f^{\circ }\left(N_2,g\right)+{\text{n}}_{H_{2}O}\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$        (1)

Where,

• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is total enthalpy of formation of products.
• ${\text{n}}_{C{O}_2}$ is the number of moles of $C{O}_{\text{2}}\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(C{O}_2,g\right)$ is the standard enthalpy of formation of $C{O}_{\text{2}}\left(g\right)$.
• ${\text{n}}_{N_2}$ is the number of moles of $N_{\text{2}}\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(N_2,g\right)$ is the standard enthalpy of formation of $N_{\text{2}}\left(g\right)$.
• ${\text{n}}_{H_{2}O}$ is the number of moles of $H_2\text{O}\left(l\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$ is the standard enthalpy of formation of $H_2\text{O}\left(l\right)$.

The value of ${\text{n}}_{C{O}_2}$ is $1mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(C{O}_2,g\right)$ is $-\text{393}\text{.51}\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of ${\text{n}}_{N_2}$ is $1mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(N_2,g\right)$ is $0.0\text{kJ}\cdot {\text{mol}}^{-1}$.

The value of ${\text{n}}_{H_{2}O}$ is $2mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$ is $-\text{285}\text{.83}\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of ${\text{n}}_{C{O}_2}$, $\Delta {\text{H}}_f^{\circ }\left(C{O}_2,g\right)$, ${\text{n}}_{N_2}$, $\Delta {\text{H}}_f^{\circ }\left(N_2,g\right)$, ${\text{n}}_{H_{2}O}$ and $\Delta {\text{H}}_f^{\circ }\left(H_{2}O,l\right)$ in equation (1).

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}=\left(\begin{array}{c}\left(1mol\right)\times \left(-\text{393}\text{.51}\text{kJ}\cdot {\text{mol}}^{-1}\right)+\left(1mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ +\left(2mol\right)\times \left(-\text{285}\text{.83}\text{kJ}\cdot {\text{mol}}^{-1}\right)\end{array}\right)\\ =-\text{393}\text{.51}\text{kJ}+0.0\text{kJ}+\left(-571.66\text{kJ}\right)\\ =-\text{393}\text{.51}\text{kJ}-571.66\text{kJ}\\ =-965.17\text{kJ}\end{array}$

The standard enthalpy of formation of reactant that is $O_2\left(g\right)$ is $0.0\text{kJ}\cdot {\text{mol}}^{-1}$.

The total enthalpy of formation of reactants is calculated by the expression shown below.

  ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(reac\tan ts\right)}={\text{n}}_{CO{\left(N{H}_2\right)}_2}\Delta {\text{H}}_f^{\circ }\left(CO{\left(N{H}_2\right)}_2,s\right)+{\text{n}}_{O_2}\Delta {\text{H}}_f^{\circ }\left(O_2,g\right)$        (2)

Where,

• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(reac\tan ts\right)}$ is total enthalpy of formation of reactants.
• ${\text{n}}_{CO{\left(N{H}_2\right)}_2}$ is the number of moles of $CO{\left(N{H}_2\right)}_2\left(s\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(CO{\left(N{H}_2\right)}_2,s\right)$ is the standard enthalpy of formation of $CO{\left(N{H}_2\right)}_2\left(s\right)$.
• ${\text{n}}_{O_2}$ is the number of moles of $O_2\left(g\right)$.
• $\Delta {\text{H}}_f^{\circ }\left(O_2,g\right)$ is the standard enthalpy of formation of $O_2\left(g\right)$.

The value of ${\text{n}}_{CO{\left(N{H}_2\right)}_2}$ is $1mol$.

The value of ${\text{n}}_{O_2}$ is $\frac{3}{2}mol$.

The value of $\Delta {\text{H}}_f^{\circ }\left(O_2,g\right)$ is $0.0\text{kJ}\cdot {\text{mol}}^{-1}$.

Substitute the value of ${\text{n}}_{CO{\left(N{H}_2\right)}_2}$, ${\text{n}}_{O_2}$ and $\Delta {\text{H}}_f^{\circ }\left(O_2,g\right)$ in equation (2).

  $\begin{array}{c}{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(reac\tan ts\right)}=\left(1mol\right)\times \Delta {\text{H}}_f^{\circ }\left(CO{\left(N{H}_2\right)}_2,s\right)+\left(\frac{3}{2}mol\right)\times \left(0.0\text{kJ}\cdot {\text{mol}}^{-1}\right)\\ =\Delta {\text{H}}_f^{\circ }\left(CO{\left(N{H}_2\right)}_2,s\right)\end{array}$

The standard enthalpy of formation of urea is calculated by the expression shown below.

  $\Delta \text{H}°={\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}-{\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }}\left(reac\tan ts\right)$        (3)

Where,

• $\Delta \text{H}°$ is standard enthalpy of reaction.
• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is total enthalpy of formation of products.
• ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(reac\tan ts\right)}$ is total enthalpy of formation of reactants.

The value of $\Delta \text{H}°$ is $-\text{632 kJ}$.

The value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ is $-965.17\text{kJ}$.

The value of ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(reac\tan ts\right)}$ is $\Delta {\text{H}}_f^{\circ }\left(CO{\left(N{H}_2\right)}_2,s\right)$.

Substitute the value of $\Delta \text{H}°$, ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(products\right)}$ and ${\displaystyle \sum \text{n}\Delta {\text{H}}_f^{\circ }\left(reac\tan ts\right)}$ in equation (3).

  $\begin{array}{c}-632\text{ kJ}=-965.17\text{kJ}-\Delta {\text{H}}_f^{\circ }\left(CO{\left(N{H}_2\right)}_2,s\right)\\ \Delta {\text{H}}_f^{\circ }\left(CO{\left(N{H}_2\right)}_2,s\right)=-965.17\text{kJ}+\text{632 kJ}\\ =-333.17kJ\end{array}$

Thus, after rounding to three significant figures, the standard enthalpy of formation of urea is $\underset{\_}{-333kJ\cdot mo{l}^{-1}}$.