Use the data given below to construct a Born-Haber cycle to determine the second ionization energy of Ca.
$$\begin{array} { l c }
& \Delta H ^ { \circ } \left( \mathrm { kJ } \mathrm { mol } ^ { - 1 } \right) \
\mathrm { Ca } ( s ) \rightarrow \mathrm { Ca } ( g ) & 193 \
\mathrm { Ca } ( g ) \rightarrow \mathrm { Ca } ^ { + } ( g ) + \mathrm { e } ^ { - } & 590 \
2 \mathrm { O } ( g ) \rightarrow \mathrm { O } _ { 2 } ( g ) & - 498 \
\mathrm { O } ( g ) + \mathrm { e } ^ { - } \rightarrow \mathrm { O } ^ { - } ( g ) & - 141 \
\mathrm { O } ^ { - } ( g ) + \mathrm { e } ^ { - } \rightarrow \mathrm { O } ^ { 2 - } ( g ) & 878 \
\mathrm { Ca } ( s ) + \frac { 1 } { 2 } \mathrm { O } _ { 2 } ( g ) \rightarrow \mathrm { CaO } ( s ) & - 635 \
\mathrm { Ca } ^ { 2 + } ( g ) + \mathrm { O } ^ { 2 - } ( g ) \rightarrow \mathrm { CaO } ( s ) & - 3414
\end{array}$$

Question

Use the data given below to construct a Born-Haber cycle to determine the second ionization energy of Ca. 
$$\begin{array} { l c } 
& \Delta H ^ { \circ } \left( \mathrm { kJ } \mathrm { mol } ^ { - 1 } \right) \
\mathrm { Ca } ( s ) \rightarrow \mathrm { Ca } ( g ) & 193 \
\mathrm { Ca } ( g ) \rightarrow \mathrm { Ca } ^ { + } ( g ) + \mathrm { e } ^ { - } & 590 \
2 \mathrm { O } ( g ) \rightarrow \mathrm { O } _ { 2 } ( g ) & - 498 \
\mathrm { O } ( g ) + \mathrm { e } ^ { - } \rightarrow \mathrm { O } ^ { - } ( g ) & - 141 \
\mathrm { O } ^ { - } ( g ) + \mathrm { e } ^ { - } \rightarrow \mathrm { O } ^ { 2 - } ( g ) & 878 \
\mathrm { Ca } ( s ) + \frac { 1 } { 2 } \mathrm { O } _ { 2 } ( g ) \rightarrow \mathrm { CaO } ( s ) & - 635 \
\mathrm { Ca } ^ { 2 + } ( g ) + \mathrm { O } ^ { 2 - } ( g ) \rightarrow \mathrm { CaO } ( s ) & - 3414
\end{array}$$

Quizplus · Accepted Answer

The second ionization energy of Ca is the energy required to remove an electron from Ca⁺(g). This can be calculated using the Born-Haber cycle:Ca(s) → Ca(g) ΔH = 193 kJ mol^-1Ca(g) → Ca⁺(g) + e^- ΔH = 590 kJ mol^-1Ca⁺(g) → Ca²⁺(g) + e^- ΔH = ?Ca²⁺(g) + O^2-(g) → CaO(s) ΔH = -3414 kJ mol^-1O(g) + e^- → O^-(g) ΔH = -141 kJ mol^-1O^-(g) + e^- → O^2-(g) ΔH = 878 kJ mol^-12 O(g) → O₂(g) ΔH = -498 kJ mol^-1Ca(s) + ½ O₂(g) → CaO(s) ΔH = -635 kJ mol^-1The overall reaction is:Ca(s) → Ca²⁺(g) + 2 e^- ΔH = ?The enthalpy change for this reaction is the sum of the enthalpy changes for the individual steps:ΔH = 193 kJ mol^-1 + 590 kJ mol^-1 + ? - 3414 kJ mol^-1 - 141 kJ mol^-1 - 878 kJ mol^-1 - 498 kJ mol^-1 - 635 kJ mol^-1ΔH = -1010 kJ mol^-1Therefore, the second ionization energy of Ca is 1010 kJ mol^-1.

Use the Data Given Below to Construct a Born-Haber Cycle