Enthalpy of a reaction changes with temperature, does this mean bond strengths change with temperature?

Question

During chemical reactions, the bonds between atoms break or form to either absorb or release energy. The result is a change to the potential energy of the system. The heat absorbed or released from a system under constant pressure is known as enthalpy.

Mathematically, we can think of the enthalpy of reaction as the difference between the potential energy from the product bonds and the potential energy of the reactant bonds.

But we know that enthalpy of a reaction depends on the temperature according to Kirchoff's equations. So does this mean that bond energies also change with temperature.

Also can someone explain to me at the molecular level why does enthalpy of the reaction change with temperature?

And since the enthalpy of a reaction depend on the constant pressure at which reaction is carried out, does this mean bond strengths also depend on pressure?

Answer 1

No: bond strengths as such do not change with temperature. The potential energy surface holding the atoms together, i.e. all the bonds, is constant for a given molecule.

However, all molecules have vibrational and rotational energy levels and molecules also have bulk kinetic energy due to thermal motion in a gas or liquid. As the temperature increases the bulk thermal energy increases, i.e. molecules move faster, but also more vibrational and rotational energy levels become excited, which means that the vibrational and rotational quantum numbers increase. This means that less extra energy is needed to break a bond if the temperature is already high, compared to that needed at low temperature, simply because the bond is already has a lot of internal (vibrational/rotational) energy.

Not all molecules have exactly the same energy but instead there is an exponential distribution of energies (the Boltzmann distribution) where the higher the energy is the fewer molecules are there that have this energy.

The subject of Statistical Mechanics describes these molecular effects and also forms the foundation for Classical Thermodynamics, i.e. what we usually call 'Thermodynamics', and which was developed before atomic/molecular theory.

Answer 2

I think the answer is already in the question. You're talking about enthalpy, so we're measuring change in energy under constant pressure. This means that the system is able to do work as a result of volume changes. Volumes do change as a function of temperature, and different substances have different coefficients of thermal expansion, so there is a enthalpy differential when this is taken into account.

Answer 3

I'd like to address the one part of this question not treated in the answers:

Also can someone explain to me at the molecular level why does enthalpy of the reaction change with temperature?

This is a really interesting question I'd not considered before.

As Zhe mentioned in his answer, $\Delta H_{rxn}$ includes a $\Delta (pV)_{rxn}$ term. At constant $p$, this becomes $p \Delta V_{rxn}$, and, with gas-phase reactions, $\Delta V_{rxn} \propto T$.

But this is an added macroscopic effect that exists in addition to the direct temperature effect on enthalpy at the molecular level. Thus, to simplify things, and get more directly at your question, I'd suggest reframing it to look at the T-dependence not of enthalpy, but of the internal energy, at constant volume (which removes the added pV term).

Let's start phenomenologically, with the thermodynamics:

$$\left( \frac{\partial \Delta U_{rxn}}{\partial T}\right)_V = \Delta C_{V\, rxn}$$

Thus what's causing $\Delta U_{V\,rxn}$ to change with temperature is the difference between the heat capacity of the reactants and products.

How can we understand this at a molecular scale? As we increase the temperature, thermal energy will flow into the reactants and products. This includes energy flowing into their bonds, thus putting their bonds at a higher-energy state, making them easier to break.

If both reactants and products have the same constant-V heat capacity, this effect will be the same for both, and there would be no net change in $\Delta U_{V\,rxn}$. But suppose the reactants have a higher constant_V heat capacity than the products $(\Delta C_{V\, rxn} < 0)$. In this case the reactants are raised in energy more than the products (their bonds become easier to break), causing $\Delta U_{V\,rxn}$ to become more favorable (more negative):

$$\left( \frac{\partial \Delta U_{rxn}}{\partial T}\right)_V = \Delta C_{V\, rxn} < 0$$

Answer 4

The connection between quantum mechanics and classical thermodynamics is not straightforward. Or perhaps I should say it isn't simple. Quantum Mechanics, as you may know, isn't an easy subject. Richard Feynman famously said:"... no one understands quantum mechanics." So, looking too hard for intuitive understanding of the way the world works isn't likely to be productive. As has been widely discussed since it was invented/discovered in 1925, you can't use your intuition to understand it, you have to use the math. Does temperature affect bond strength? Temperature is a measure of molecular motion. Will motion affect bond energy? It depends, if the two atoms involved in the bond have sufficiently different momentum, the bond will break. So as you raise temperature of a material, you get closer to the point of disintegration. How about pressure? Well, you may know that water ice has various crystal structures depending on T and P? You can't explain that if you assume there's no change in energy and entropy of the system at different P or T. But chemicals are generally assumed to be at their LOWEST energy state, and compressing them (ignoring the PV work) will result in more repulsion between electrons, right? So you'll have to overcome that repulsive energy hence compression adds to potential energy.