# Why does an exothermic/endothermic reaction have a negative/positive enthalpy

So regarding change in enthalpy and exothermic/endothermic reactions, I think I have some conceptual understanding but I'm missing some things.

My current understanding is that:

• Given a reaction, if the energy added to break the reactant bonds is greater than the energy released when forming the product bonds, the reaction is endothermic / +Δ𝐻.
• Opposite of above is true; energy added < energy released, then exothermic / -Δ𝐻
• Enthalpy of formation is synonymous with bond energy

For example, if I have an exothermic reaction:

• Energy added < energy released
• Δ𝐻 is negative because more energy was released than added
• Δ𝐻 is the negative of the difference between energy added and energy released

But where I'm falling short is with the equation Δ𝐻 = Δ𝐻F(products) − Δ𝐻F(reactants)

In the example above:

• Energy released is Δ𝐻F(products)
• Energy added < energy released

So how does Δ𝐻 end up negative? I know it should be negative, but I'm not sure why it doesn't turn out that way in this equation.

Are some of the values supposed to be negative, or is enthalpy of formation not the exact same as bond energy?

I feel as if I oversimplified / dumbed down things too much when I tried to conceptualize the material and now I'm missing some fine print / technical knowledge.

Why does an exothermic/endothermic reaction have a negative/positive enthalpy?

it is because of the definition of terms enthalpy, exothermic and endothermic.

• Exothermic or $$\Delta H \lt 0$$ means released heat at constant pressure.
• Endothermic or $$\Delta H \gt 0$$ means absorbed heat at constant pressure.

Be aware that enthalpy change reflects the reaction net heat outcome at constant pressure, where part of the net energy change may be consumed or released by volume work. There may be a reaction that is slightly exothermic ( $$\Delta U \lt 0$$) at constant volume, but is slightly endothermic at constant pressure ($$\Delta H = \Delta U + p \Delta V \gt 0$$) if there is volume expansion and work done against external pressure. But that would be rather a rare case, as usually $$|\Delta U| \gg |\Delta (pV)|$$.

Enthalpy of formation is NOT synonymous with bond energy.

Enthalpy of substance formation is heat released at constant pressure if ( often just formally) prepared from chemical elements at their standard state. This enthalpy is negative if formed bonds are overall stronger than bonds within elements, so energy is released.

Imagine reaction $$\ce{Cl2(g) + H2(g) -> 2 HCl(g)}$$.

Imagine the formation of the bond $$\ce{H-Cl}$$ had released the energy equal the average of the respective energies of bonds $$\ce{H-H}$$ and $$\ce{Cl-Cl}$$ and then the enthalpy of $$\ce{HCl}$$ formation would have been zero. That would not have meant the bond formation energy was zero too.

$$\Delta H_\mathrm{r} = \Delta H_\mathrm{f,products} − \Delta H_\mathrm{f,reactants}$$

This equation means that the enthalpy of reaction

$$\ce{"reactants" -> "products"}$$

is the difference of heat released at constant pressure between

"formation of products from standard-state elements"
and
"formation of reactants from standard-state elements"

If formation of products releases more heat than formation of reactants, it is clear the reaction is exothermic.