# Chemistry question- enthalpy of formation

Hi, please could i have help on this question? I dont understand why i need the enthalpy of formation value of sodium?
Here is the question: https://i.ibb.co/G0c3nZZ/IMG-0217.jpg
Thanks!
Original post by anonymous294
Hi, please could i have help on this question? I dont understand why i need the enthalpy of formation value of sodium?
Here is the question: https://i.ibb.co/G0c3nZZ/IMG-0217.jpg
Thanks!

It shouldn’t matter either way. ΔfH^Θ [Na (s)] would be 0 anyway, since the enthalpy of formation of any element in it’s standard state will be 0. Essentially the equation that defines the enthalpy of formation of an element in it’s standard state is no reaction at all:

Element in standard state —> element in standard state

You should spot that you can split the direct conversion of Na (s) to Na^+ (g) into the two following steps:

The atomisation of Na (s), Na (s) —> Na (g)

The first ionisation energy of sodium, Na (g) —> Na^+ (g) + e^-

See that there isn’t actually any need for the enthalpy of formation of sodium? But even if you were to factor it in, because it has a value of 0, it wouldn’t change the final answer.

Like another one of your threads, the answer looks extremely dubious. Technically you can work out the enthalpy of formation of Na^+ (g) correctly regardless of whether you factor in the enthalpy of formation of sodium.
(edited 3 months ago)
Original post by TypicalNerd
It shouldn’t matter either way. ΔfH^Θ [Na (s)] would be 0 anyway, since the enthalpy of formation of any element in it’s standard state will be 0. Essentially the equation that defines the enthalpy of formation of an element in it’s standard state is no reaction at all:

Element in standard state —> element in standard state

You should spot that you can split the direct conversion of Na (s) to Na^+ (g) into the two following steps:

The atomisation of Na (s), Na (s) —> Na (g)

The first ionisation energy of sodium, Na (g) —> Na^+ (g) + e^-

See that there isn’t actually any need for the enthalpy of formation of sodium? But even if you were to factor it in, because it has a value of 0, it wouldn’t change the final answer.

Like another one of your threads, the answer looks extremely dubious. Technically you can work out the enthalpy of formation of Na^+ (g) correctly regardless of whether you factor in the enthalpy of formation of sodium.

I’m so sorry, the answer is that both 1 and 2 are needed to work out the enthalpy of formation of gaseous ions. Why would i need the enthalpy of atomisation of sodium when sodium already exists as atoms or is it to convert it form a solid to a gaseous state? Thanks
Original post by anonymous294
I’m so sorry, the answer is that both 1 and 2 are needed to work out the enthalpy of formation of gaseous ions. Why would i need the enthalpy of atomisation of sodium when sodium already exists as atoms or is it to convert it form a solid to a gaseous state? Thanks

its to convert it from solid to gas
Original post by j3ssica jon3s
its to convert it from solid to gas

This is correct.

In the question, the product is shown to be in the gaseous state and so the enthalpy change associated with changing the state from solid to gas is needed. Another big hint is that the sodium is ionised and that the first ionisation energy of sodium requires it to be in the gas state.

As the enthalpy change of atomisation is the enthalpy change defined as the enthalpy change associated with the formation of 1.00 mol of gaseous atoms from the element in it’s standard state.
Original post by TypicalNerd
This is correct.

In the question, the product is shown to be in the gaseous state and so the enthalpy change associated with changing the state from solid to gas is needed. Another big hint is that the sodium is ionised and that the first ionisation energy of sodium requires it to be in the gas state.

As the enthalpy change of atomisation is the enthalpy change defined as the enthalpy change associated with the formation of 1.00 mol of gaseous atoms from the element in it’s standard state.

Thank you!