The bond enthalpy of N2(g) is 418.0 kJ/mol. Calculate ÄH°f for N(g).
I figured that if it gave the enthalpy for N2, then the enthalpy for N would be half of the given amount. But that wouldn't be bonded to anything??
Is there a standard calculation?
To calculate the standard enthalpy change of formation (ΔH°f) for N(g), you can use Hess's law and the given bond enthalpy for N2(g).
The standard enthalpy change of formation is defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states, with all reactants and products also in their standard states.
In this case, you want to calculate the enthalpy change of formation for nitrogen gas (N2). However, you are given the bond enthalpy for N2 and want to find the bond enthalpy for N.
To start, let's write the balanced chemical equation for the formation of N2(g):
N2(g) → 2N(g)
According to Hess's law, the enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual steps that make up the reaction. In this case, we will calculate the enthalpy change for the reverse of the bond-breaking process (since you have the bond enthalpy for N2) and the formation of N.
Step 1: Bond Breaking (endothermic):
N2(g) → 2N(g) (reverse of the formation reaction)
Since breaking bonds requires energy, the enthalpy change for this step is equal to the bond enthalpy of N2, which is 418.0 kJ/mol.
Step 2: Formation of N (exothermic):
2N(g) → N(g)
Since 2N(g) is forming from N(g), the enthalpy change for this step is equal to -ΔH°f for N(g).
Now, adding these two steps together gives us the overall reaction:
N2(g) → 2N(g) + ΔH°f for N(g)
Therefore, we have:
418.0 kJ/mol = -ΔH°f for N(g)
Rearranging the equation to solve for ΔH°f for N(g), we get:
ΔH°f for N(g) = -418.0 kJ/mol
So, the standard enthalpy change of formation for nitrogen gas (N(g)) is -418.0 kJ/mol.
To summarize, to calculate the standard enthalpy change of formation for N(g) from the given bond enthalpy for N2(g), you apply Hess's law and consider the bond-breaking process of N2(g) and the formation of N(g) separately. The enthalpy change for the reverse of the bond-breaking process is equal to the bond enthalpy of N2, and the enthalpy change for the formation of N(g) is equal to -ΔH°f for N(g). By adding these two steps together, you can solve for ΔH°f for N(g), which is found to be -418.0 kJ/mol.