Chapter 10, Problem 10.101PAE

10.101 Fluorine reacts with liquid water to form gaseous hydrogen fluoride and oxygen. (a) Write a balanced chemical equation for this reaction. (b) Use tabulated data to determine the free energy change for the reaction and comment on its spontaneity. (c) Use tabulated data to calculate the enthalpy change of the reaction. (d) Determine how much heat flows and in what direction when 34.5 g of fluorine gas is bubbled through excess water.

Interpretation Introduction

Interpretation:

Fluorine reacts with liquid water to form HF and Oxygen gas. The feasibility of this reaction can be identified by using free energy change. The free energy change, enthalpy change and entropy change can be calculated by using standard data.

Concept introduction: The free energy change of the reaction $\Delta G^0$ is calculated by using $\Delta H^0$ and $\Delta S^0$ at the given temperature. The formula that can be used is as follows.

$\Delta G^0=\Delta H^0-T\Delta S^0$

$\Delta H^0,\Delta S^0$ is calculated by using the following mathematical expressions.

$\Delta H^0={\displaystyle \sum \Delta H^0\left(\text{products}\right)-}{\displaystyle \sum \Delta H^0\left(\text{reactants}\right)}$

$\begin{array}{l}\Delta S^0={\displaystyle \sum \Delta S^0\left(\text{products}\right)-}{\displaystyle \sum \Delta S^0\left(\text{reactants}\right)}\\ \Delta G^0={\displaystyle \sum \Delta G^0\left(\text{products}\right)-}{\displaystyle \sum \Delta G^0\left(\text{reactants}\right)}\end{array}$

If the sign of the free energy change is negative, it indicates that the reaction is spontaneous.

Answer to Problem 10.101PAE

Solution: a) The balanced equation is $2{\text{F}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\stackrel{}{\to }4\text{HF}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)$

b) $\Delta G_f^0=\boxed{-617.6\text{kJ}}$

c) $\Delta H_f^0=\boxed{-513.6\text{kJ}}$

d) The amount of heat produced by 34.5 g of F₂ = 233 kJ

Given: From the standard data −

$\begin{array}{l}\Delta H_f^0\left[{\text{F}}_{\text{2}}\left(g\right)\right]=0\text{kJ/mol}\text{,}\Delta {\text{G}}_f^0\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]=-285.8\text{kJ/mol}\\ \Delta H_f^0\left[\text{HF}\left(\text{g}\right)\right]=-271.0\text{kJ/mol,}\Delta {\text{H}}_f^0\left[{\text{O}}_{\text{2}}\left(\text{g}\right)\right]=0\text{kJ/mol}\end{array}$

$\begin{array}{l}\Delta G_f^0\left[{\text{F}}_{\text{2}}\left(g\right)\right]=0\text{kJ/mol}\text{,}\Delta {\text{G}}_f^0\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]=-237.2\text{kJ/mol}\\ \Delta G_f^0\left[\text{HF}\left(\text{g}\right)\right]=-273.0\text{kJ/mol,}\Delta {\text{G}}_f^0\left[{\text{O}}_{\text{2}}\left(\text{g}\right)\right]=0\text{kJ/mol}\end{array}$

Explanation of Solution

a)

The balanced chemical equation of the given reaction is as follows.

$2{\text{F}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\stackrel{}{\to }4\text{HF}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)$

b) From the standard data −

$\begin{array}{l}\Delta G_f^0\left[{\text{F}}_{\text{2}}\left(g\right)\right]=0\text{kJ/mol}\text{,}\Delta {\text{G}}_f^0\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]=-237.2\text{kJ/mol}\\ \Delta G_f^0\left[\text{HF}\left(\text{g}\right)\right]=-273.0\text{kJ/mol,}\Delta {\text{G}}_f^0\left[{\text{O}}_{\text{2}}\left(\text{g}\right)\right]=0\text{kJ/mol}\end{array}$

$\Delta G_{rxn}^0={\displaystyle \sum \Delta G_f^0\left(\text{products}\right)-}{\displaystyle \sum \Delta G_f^0\left(\text{reactants}\right)}$

$\begin{array}{l}\Delta G_{rxn}^0=\left[4\times \Delta G_f^0\left[\text{HF}\left(\text{g}\right)\right]+1\times \Delta {\text{G}}_f^0\left[{\text{O}}_{\text{2}}\left(\text{g}\right)\right]\right]-\left[2\times \Delta G_f^0\left[{\text{F}}_{\text{2}}\left(g\right)\right]+2\times \Delta {\text{G}}_f^0\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]\right]\\ =4\left(-273\right)+0-2\left(0\right)-2\left(-237.2\right)\\ =\boxed{-617.6\text{kJ}}\end{array}$

c)

From the standard data −

$\begin{array}{l}\Delta H_f^0\left[{\text{F}}_{\text{2}}\left(g\right)\right]=0\text{kJ/mol}\text{,}\Delta {\text{G}}_f^0\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]=-285.8\text{kJ/mol}\\ \Delta H_f^0\left[\text{HF}\left(\text{g}\right)\right]=-271.0\text{kJ/mol,}\Delta {\text{H}}_f^0\left[{\text{O}}_{\text{2}}\left(\text{g}\right)\right]=0\text{kJ/mol}\end{array}$

$\Delta H_{rxn}^0={\displaystyle \sum \Delta H_f^0\left(\text{products}\right)-}{\displaystyle \sum \Delta H_f^0\left(\text{reactants}\right)}$

$\begin{array}{l}\Delta H_{rxn}^0=\left[4\times \Delta H_f^0\left[\text{HF}\left(\text{g}\right)\right]+1\times \Delta {\text{H}}_f^0\left[{\text{O}}_{\text{2}}\left(\text{g}\right)\right]\right]-\left[2\times \Delta H_f^0\left[{\text{F}}_{\text{2}}\left(g\right)\right]+2\times \Delta {\text{H}}_f^0\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]\right]\\ =4\left(-271\right)+0-2\left(0\right)-2\left(-285.2\right)\\ =\boxed{-513.6\text{kJ}}\end{array}$

d)

From the balanced equation, 2 mols $\left(\text{2}\text{mol×38}\text{g/mol}\text{=}\text{76}\text{g}\right)$ Fluorine $\left({\text{F}}_{\text{2}}\right)$ produce 513.6 kJ of heat.

Therefore, the amount of heat produced by 34.5 g of Fluorine $\text{=}\left(\frac{\text{513}\text{.6×34}\text{.5}}{\text{76}}\right)\text{kJ}$

$=\boxed{233\text{kJ}}$

Conclusion

The reaction is non-spontaneous as the sign of the free energy change is negative. The enthalpy change is negative and indicates that the reaction is an exothermic reaction.