# What is the total enthalpy for "C"(g) -> "C"^(4+)(g) + 4e^(-)?

Jun 6, 2017

$\text{148.0262 eV}$, or $\text{14280.78 kJ/mol}$.

Well, it's very unusual, but I guess we can do it... Break it into four steps.

The ionization energy is the energy put into the gaseous atom to remove an electron.

${\text{C"(g) -> "C}}^{+} \left(g\right) + {e}^{-}$, $\Delta {H}_{I {E}_{1}} = \text{11.260300 eV}$

${\text{C"^(+)(g) -> "C}}^{2 +} \left(g\right) + {e}^{-}$, $\Delta {H}_{I {E}_{2}} = \text{24.3846 eV}$

${\text{C"^(2+)(g) -> "C}}^{3 +} \left(g\right) + {e}^{-}$, $\Delta {H}_{I {E}_{3}} = \text{47.88778 eV}$

${\text{C"^(3+)(g) -> "C}}^{4 +} \left(g\right) + {e}^{-}$, $\Delta {H}_{I {E}_{4}} = \text{64.49351 eV}$

where the ionization energies were taken from NIST (most reliable for light elements).

(The electron volt, $\text{eV}$ is a unit of energy needed to accelerate the electron through a $\text{1 V}$ potential difference.)

The change in enthalpy, $\Delta H$, is additive, and a state function (a function with an initial and final state independent of the path).

So, we can build a series of steps that each have their own $\Delta H$, without really caring what particular path we take, as long as the initial and final states are what we want.

$\Delta {H}_{C \left(g\right) \to {C}^{4 +} \left(g\right)} = \Delta {H}_{I {E}_{1}} + \Delta {H}_{I {E}_{2}} + \Delta {H}_{I {E}_{3}} + \Delta {H}_{I {E}_{4}}$

$= \text{11.260300 eV" + "24.3846 eV" + "47.88778 eV" + "64.49351 eV}$

$=$ $\text{148.0262 eV}$

which is actually quite large, but that makes sense, right? You're ionizing carbon FOUR times, which is probably nearly impossible to accomplish in real life... (Even if it was, it would be short-lived.)

If you don't see how large this is, let's try it in $\text{kJ/mol}$.

color(blue)(DeltaH_(C(g) -> C^(4+)(g))) = 148.0262 cancel"eV" xx (1.602 xx 10^(-19) cancel"J")/cancel"eV" xx "1 kJ"/(1000 cancel"J") xx 6.0221413 xx 10^(23) ("mol C")^(-1)

$=$ $\textcolor{b l u e}{\text{14280.78 kJ/mol}}$

Compare that to the following ionization energies:

color(white)([(" ", color(black)("kJ/mol")),(color(black)("H"), color(black)(1312.0)),(color(black)("Na(2nd)"),color(black)(4562.4)),(color(black)("Mg(3rd)"),color(black)(7732.6)),(color(black)("Al(4th)"),color(black)(11577))])

It exceeds the fourth ionization energy of aluminum, which is the removal of its $2 s$ core electron, by $\text{2704 kJ/mol}$, which is also about $27$ times the enthalpy of the average chemical bond.