The bond enthalpy of F2(g) is 156.9 kJ/mol. Calculate ÄH°f for F(g).
1 kJ/mol
Wouldn't it just be 1/2 of 156.9?
F2 (g) ------> 2F (g)
so deltaHrxn = +156.9/2 = 78.45 kJ/mol
you need to divide by 2 because 156.9 kJ/mol you need to from 2 F-atoms, and you need to prorate per one atom.
Well, the bond enthalpy of F2 is indeed 156.9 kJ/mol, but I'm afraid that's not enough to calculate the standard enthalpy of formation (ΔH°f) for F.
The standard enthalpy of formation is defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states. In other words, it tells us how much heat is released or absorbed when a compound is formed.
To calculate ΔH°f for F, we need to know its standard enthalpy of formation. However, since F is an element in its standard state, its standard enthalpy of formation is defined as zero. So, ΔH°f for F(g) is actually zero.
But hey, don't be too hard on F. It may not have an exciting enthalpy of formation, but it sure makes a great addition to chemical reactions. Just remember, F may be quiet when it comes to enthalpy, but it can definitely make a lot of noise when reacting with other elements.
To calculate the standard enthalpy change (ΔH°f) for F(g), you need to use the bond enthalpy of the F-F bond and the definition of enthalpy of formation.
Enthalpy of formation (ΔH°f) refers to the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states.
The equation for the formation of F(g) from F2(g) is:
F2(g) → 2F(g)
According to the definition of bond enthalpy, it represents the energy required to break one mole of a specific bond in a compound. In this case, the bond enthalpy of the F-F bond is 156.9 kJ/mol. Breaking the F2 bond requires this amount of energy.
However, since we want to calculate the enthalpy change for the formation of F(g), we need to consider the reaction in the opposite direction.
In the reverse reaction, we have:
2F(g) → F2(g)
Since we're going in the opposite direction, the sign of the bond enthalpy should be flipped. Therefore, it becomes -156.9 kJ/mol.
Since the reverse reaction involves the formation of 1 mole of F(g) and the bond enthalpy is for 1 mole of F2(g), we need to divide the bond enthalpy by 2 to apply it to the formation of 1 mole of F(g):
-156.9 kJ/mol ÷ 2 = -78.45 kJ/mol
Therefore, the enthalpy of formation (ΔH°f) for F(g) is approximately -78.45 kJ/mol.
(Note: The negative sign indicates that the reaction is exothermic, meaning the formation of F(g) releases energy.)