Just going through my thermodynamics chapter I came across 2 contrasting statements:

  1. Change in enthalpy=enthalpy of (products-reactants) at constant pressure and temperature.

  2. Change in enthalpy=Energy required to break bonds in molecules of reactants -energy released in the formation of bonds in molecules of products.

So, if I consider these statements individually then these sound correct to me but they are different in the sense that in the first one it is (products-reactants) and in the second one (reactant -products)?

Is the book wrong or am I mistaken in my fundamentals ?


They are basically same statements. We can use Hess law to elaborate these two statements. For example, we can use the following reaction,
Here, Delta reaction = Enthalpy of CH4 - Enthalpy of C + 2(Enthalpy of H2).......(1)
We can break the reaction events as,
1) Breaking of 4 C-H bonds 2) Formation of 2 H-H bonds
So, delta H = Energy required to break 4 C-H bonds + Energy required to forming two H-H bonds.......(2)
Now we always take the reactor to be the system and that's why heat input to the reactor is positive and heat going out of the reactor is negative. In the formation of bonds, the energy are released, so the formation energy of H-H is negative with respect to our thermodynamic system.From 2
So, delta H = Energy required to break 4 C-H bonds - Energy released in the formation of two H-H bonds.......(3)
Now, energy required to break 4 C-H bond= Energy of C + Enrgy of 4 H - energy of CH4
=-Energy of reactants...(4)
Similarly, energy required to form 2 H-H bond= Energy of 2 H2 - energy of 4 H
=Energy of products (As C is in reference state).........(5)
Adding 4 and 5 yields, equation (1), which proves that both statements are same.


Bond breaking is endothermic, bond making is exothermic. To use bond strengths, sum the strengths of the broken bonds and sum the made bonds. Since the breaking is endothermic and the making is exothermic you can add the positive to the negative and get the enthalpy change for the reaction.

With heats of formation, it’s as if the above process has already been done with reference to the constituent elements of the substances involved. So, subtracting the heats of formation of the reactants from the heats of formation of the products is like doing the above process but with the actual values of the bonds in the substances, which can vary depending upon molecular structure.

Look at water as an example. The strength of a $\ce{O-H}$ bond is about 110 kcal/mol. $\ce{O=O}$ is about 120 and $\ce{H-H}$ is about 105. To make 2 mol of water you have to raise the potential energy of the system by 330 kcal by breaking the oxygen and hydrogen molecules into atoms. You then make 4 mol of $\ce{O-H}$ bonds in making 2 mol of water, lowering the potential energy by 440. This is a net change of −110 kcal for the process. −110 kcal is about 460 kJ. I rounded a lot and there’s always a difference in the bond strengths in a table and those in a molecule. A heat of formation table will tell you that a mol of water is about 240 kJ/mol lower in potential energy than the oxygen and hydrogen it was made from. You can take the 2(−240 kJ/mol) and subtract 1(0) for the oxygen and 2(0) for the hydrogen and end up with −480 kJ/mol, about what we got from the bond strength calculation.


1) change in enthalpy = enthalpy of (products-reactants) at constant pressure and temperature.

$$\Delta H_{p,T} = H_\text{products}-H_\text{reactants}$$

2) change in enthalpy = energy required to break bonds in molecules of reactants − energy released in the formation of bonds in molecules of products

$$\Delta H_{p,T} = (-H_\text{reactants})-(-H_\text{products})= H_\text{products}-H_\text{reactants}$$

because the energy required is negative of enthalphy of formation/ enthalpy of substance.

  • $\begingroup$ how can you say that the energy required is negative of enthalpy of substance ? I am sorry but I didn't understand the last part. $\endgroup$ – Karan Singh Nov 4 '14 at 5:38
  • $\begingroup$ @KaranSingh energy released is taken negative and absorbed positive; convention $\endgroup$ – RE60K Nov 4 '14 at 7:18
  • $\begingroup$ ok so according to that-how is Energy required to break bonds in molecules of reactants negative as you have shown? we need to provide energy to break bonds hence it should be positive?!?!: $\endgroup$ – Karan Singh Nov 4 '14 at 7:42

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