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Chemistry · Ch 6 — Thermodynamics

A Useful New State Function

6.2.2(a)

A Useful New State Function

A Useful New State Function

When a chemical reaction is carried out in a beaker or flask open to the atmosphere, the pressure remains constant (atmospheric pressure) but the volume may change. Under these conditions, the heat absorbed by the system, qpq_p, is not equal to the change in internal energy ΔU\Delta U. We already know that at constant volume, ΔU=qV\Delta U = q_V because no work is done. But at constant pressure, the system can do expansion work (or have work done on it) as its volume changes.

From the first law of thermodynamics, ΔU=q+w\Delta U = q + w. If the only work is pressure-volume work, then at constant pressure pp, the work done by the system is w=−pΔVw = -p\Delta V. So we write:

ΔU=qp−pΔV\Delta U = q_p - p\Delta V

Let the initial state be denoted by subscript 1 and the final state by subscript 2. Then:

U2−U1=qp−p(V2−V1)U_2 - U_1 = q_p - p(V_2 - V_1)

Rearranging this equation gives:

qp=(U2+pV2)−(U1+pV1)q_p = (U_2 + pV_2) - (U_1 + pV_1)

The quantity U+pVU + pV appears so naturally that it is given its own name and symbol. We define a new thermodynamic function called enthalpy, HH (from the Greek enthalpien, meaning "to warm" or "heat content"):

H=U+pVH = U + pV

Using this definition, the equation above becomes:

qp=H2−H1=ΔHq_p = H_2 - H_1 = \Delta H

This is a crucial result. Although qq (heat) is a path-dependent quantity, HH is a state function because it depends only on UU, pp, and VV — all of which are state functions. Therefore, ΔH\Delta H is independent of the path taken, and consequently qpq_p is also independent of path. This makes enthalpy an enormously useful quantity for studying reactions carried out under constant pressure — which is how most chemical reactions are performed.

Important

When heat is absorbed or released by a system at constant pressure, we are actually measuring the change in enthalpy, ΔH\Delta H.

ΔH\Delta H is negative for exothermic reactions, which evolve heat during the reaction, and ΔH\Delta H is positive for endothermic reactions, which absorb heat from the surroundings.

At constant volume (ΔV=0\Delta V = 0), the pΔVp\Delta V term vanishes, so ΔH=ΔU=qV\Delta H = \Delta U = q_V.

For finite changes at constant pressure, we can write:

ΔH=ΔU+Δ(pV)\Delta H = \Delta U + \Delta(pV)

Since pressure is constant, Δ(pV)=pΔV\Delta(pV) = p\Delta V, so:

ΔH=ΔU+pΔV\Delta H = \Delta U + p\Delta V

This relationship connects the enthalpy change to the internal energy change and the pressure-volume work done by the system.


Relationship Between ΔH\Delta H and ΔU\Delta U for Reactions Involving Gases

For reactions involving gases, the pΔVp\Delta V term can be expressed in terms of the change in the number of moles of gas. If we treat the gases as ideal, then pV=nRTpV = nRT. At constant temperature and pressure:

pΔV=ΔngRTp\Delta V = \Delta n_g RT

where Δng\Delta n_g is the change in the number of moles of gaseous species in the reaction (moles of gaseous products minus moles of gaseous reactants). Therefore:

ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT

This equation allows us to convert between ΔH\Delta H and ΔU\Delta U for any reaction, provided we know the temperature and the change in the number of moles of gas.

Tip

If Δng=0\Delta n_g = 0 (same number of moles of gas on both sides), then ΔH=ΔU\Delta H = \Delta U. If Δng>0\Delta n_g > 0 (more gas molecules produced), then ΔH>ΔU\Delta H > \Delta U. If Δng<0\Delta n_g < 0 (gas consumed), then ΔH<ΔU\Delta H < \Delta U. …