Question
The bond enthalpy of $\mathrm{F}_{2}(g)$ is $156.9 \mathrm{kJ} / \mathrm{mol}$. Calculate $\Delta H_{\mathrm{f}}^{\circ}$ for $\mathrm{F}(g)$.
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Step 1: We start by writing the equation for the formation of fluorine atom from fluorine molecule: \[\frac{1}{2}F_{2}(g) \rightarrow F(g)\] Show more…
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(a) From the following data calculate the bond enthalpy of the $\mathrm{F}_{2}^{-}$ ion. $$\begin{array}{ll} \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{F}(g) & \Delta H_{\mathrm{rxn}}^{\circ}=156.9 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{F}^{-}(g) \longrightarrow \mathrm{F}(g)+e^{-} & \Delta H_{\mathrm{rxn}}^{\circ}=333 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{F}_{2}(g) \longrightarrow \mathrm{F}_{2}(g)+e^{-} & \Delta H_{\mathrm{rxn}}^{\circ}=290 \mathrm{kJ} / \mathrm{mol} \end{array}$$ (b) Explain the difference between the bond enthalpies of $\mathrm{F}_{2}$ and $\mathrm{F}_{2}^{-}$
Given the following data: $$ \begin{array}{ll}{\mathrm{NO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}(g)} & {\Delta H=233 \mathrm{kJ}} \\ {2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g)} & {\Delta H=-427 \mathrm{kJ}}\end{array} $$ $$ \mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad \Delta H=-199 \mathrm{kJ} $$ Calculate the bond energy for the $\mathrm{O}_{2}$ bond, that is, calculate $\Delta H$ for: $$ \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{O}(g) \qquad \Delta H=? $$
The enthalpy change for the following reaction is $368 \mathrm{~kJ}$. Calculate the average $\mathrm{O}-\mathrm{F}$ bond energy. $$ \mathrm{OF}_{2}(g) \longrightarrow \mathrm{O}(g)+2 \mathrm{~F}(g) $$ (a) $184 \mathrm{~kJ} / \mathrm{mol}$ (b) $368 \mathrm{~kJ} / \mathrm{mol}$ (c) $536 \mathrm{~kJ} / \mathrm{mol}$ (d) $736 \mathrm{~kJ} / \mathrm{mol}$
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