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

Question

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

Mateo Aguilar · Accepted Answer

#"148.0262 eV"#, or #"14280.78 kJ/mol"#.

It's quite unusual, but I think we can manage it if we divide it up into four steps.

The energy required to extract an electron from a gaseous atom is known as the ionization energy.

#"C"(g) -> "C"^(+)(g) + e^(-)#, #DeltaH_(IE_1) = "11.260300 eV"#

#"C"^(+)(g) -> "C"^(2+)(g) + e^(-)#, #DeltaH_(IE_2) = "24.3846 eV"#

#"C"^(2+)(g) -> "C"^(3+)(g) + e^(-)#, #DeltaH_(IE_3) = "47.88778 eV"#

#"C"^(3+)(g) -> "C"^(4+)(g) + e^(-)#, #DeltaH_(IE_4) = "64.49351 eV"#

where NIST, which is most trustworthy for light elements, provided the ionization energies.

(The electron volt, #"eV"# is a unit of energy needed to accelerate the electron through a #"1 V"# potential difference.)

The change in enthalpy, #DeltaH#, 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 #DeltaH#, without really caring what particular path we take, as long as the initial and final states are what we want.

#DeltaH_(C(g)->C^(4+)(g)) = DeltaH_(IE_1) + DeltaH_(IE_2) + DeltaH_(IE_3) + DeltaH_(IE_4)#

#= "11.260300 eV" + "24.3846 eV" + "47.88778 eV" + "64.49351 eV"#

#=# #"148.0262 eV"#

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

If you don't see how large this is, let's try it in #"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)#

#=# #color(blue)("14280.78 kJ/mol")#

Contrast that with the ionization energies listed below:

#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 #2s# core electron, by #"2704 kJ/mol"#, which is also about #27# times the enthalpy of the average chemical bond.

Xavier Avery · Answer

undefined The total enthalpy change for the reaction "C"(g) -> "C"^(4+)(g) + 4e^(-) is determined by the enthalpies of formation of the reactants and products involved in the reaction. Without specific enthalpy values provided, it's not possible to calculate the exact total enthalpy change for the given reaction.

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

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