Question

match the reactions shown with the process that they are depicting.
prompts
1 f₂(g)→2f(g)
2 na(s)+½f₂(g)→naf(s)
3 na(g)→na⁺(g)+e⁻
4 f(g)+e⁻→f⁻(g)
answers
select match
bond enthalpy
electron affinity
ionization energy
standard enthalpy of formation

Explanation:

Step1: Analyze Reaction 1

Reaction 1: $\ce{F_{2(g)} -> 2F_{(g)}}$
This reaction involves breaking the $\ce{F-F}$ bond in $\ce{F_2}$ gas to form two $\ce{F}$ atoms. Bond enthalpy is the energy required to break one mole of bonds in the gas phase. So this reaction corresponds to bond enthalpy.

Step2: Analyze Reaction 2

Reaction 2: $\ce{Na_{(s)} + \frac{1}{2}F_{2(g)} -> NaF_{(s)}}$
This reaction is the formation of one mole of $\ce{NaF}$ from its constituent elements in their standard states (Na is solid, $\ce{F_2}$ is gas). The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states. So this reaction corresponds to standard enthalpy of formation.

Step3: Analyze Reaction 3

Reaction 3: $\ce{Na_{(g)} -> Na^+_{(g)} + e^-}$
This reaction shows a gaseous sodium atom losing an electron to form a gaseous sodium ion. Ionization energy is the energy required to remove an electron from a gaseous atom. So this reaction corresponds to ionization energy.

Step4: Analyze Reaction 4

Reaction 4: $\ce{F_{(g)} + e^- -> F^-_{(g)}}$
This reaction shows a gaseous fluorine atom gaining an electron to form a gaseous fluoride ion. Electron affinity is the energy change when an electron is added to a gaseous atom. So this reaction corresponds to electron affinity.

Answer:

1. $\ce{F_{2(g)} -> 2F_{(g)}}$ - Bond enthalpy
2. $\ce{Na_{(s)} + \frac{1}{2}F_{2(g)} -> NaF_{(s)}}$ - Standard enthalpy of formation
3. $\ce{Na_{(g)} -> Na^+_{(g)} + e^-}$ - Ionization energy
4. $\ce{F_{(g)} + e^- -> F^-_{(g)}}$ - Electron affinity