# Enthalpy of reaction with equal amounts of gaseous products and reactants

At a constant pressure and temperature the enthalpy change for a reaction (assuming ideal gas law) is given as

$$\Delta_\mathrm{rxn}H = \Delta n_\mathrm{g}C_pT,$$

where $$\Delta n_\mathrm{g}$$ is the amount difference of gaseous products and reactants. Now, consider the reaction of combustion of carbon:

$$\ce{C(s) + O2(g) -> CO2(g)}$$

We know that the combustion reactions are exothermic, but here since $$\Delta n_\mathrm{g} = 0,$$ does it mean that the change in the reaction enthalpy turns out to be zero?

• Your initial claim is rather unreasonable. Commented May 16 at 11:05
• ΔH =∑Bond energies of bonds broken −∑Bond energies of bonds formed Commented May 16 at 11:09
• @UnlogicalChymist The first equation refers to this quantity $\Delta_\mathrm{r}H - \Delta_\mathrm{r}U$, under the assumption of ideal gas behaviour and disregarding the molar volumes of species which are not gases. If it was true, all reactions that show no net change of gaseous species will have a zero value of standard enthalpy of reaction... Commented May 16 at 11:12
• Metal Storm you corrected his equation. Please correct your statement. H=U + PV, if PV does not change then DeltaH = DeltaU not zero. Commented May 16 at 20:52
• (To OP) There is no universal value of $C_p$ which will be applicable for the reactants and the products side of the reaction. This shows the formula you have shared can't be true.
– Dodo
Commented May 17 at 1:14

For an irreversible, complete reaction of the form:

$$\ce{aA +bB->cC +dD}$$

The standard molar change in enthalpy is given by:

$$\Delta\overline{H}^o=c\;\Delta\overline{H}_{f_C}^o\;+\;d\;\Delta\overline{H}_{f_D}^o\;-\;\left(a\;\Delta\overline{H}_{f_A}^o\;+\;b\;\Delta\overline{H}_{f_B}^o\right)$$

And the standard change in enthalpy by:

$$\Delta H^o=\frac{n_{Ao}\;\Delta \overline{H}^o}{a}$$

The total amount of products once reaction finalizes is:

$$n=n_{Ao}\left(1+\frac{n_{Bo}}{n_{Ao}}+\frac{\Delta n}{a}\right)\implies n_{Ao}=\frac{n}{1+\frac{n_{Bo}}{n_{Ao}}+\frac{\Delta n}{a}}$$

Substituting above:

$$\Delta H^o=\frac{n\;\Delta \overline{H}^o}{a\left(1+\frac{n_{Bo}}{n_{Ao}}\right)+\Delta n}$$

If $$\;\Delta n=0\;$$, then:

$$\Delta H^o=\frac{n\;\Delta \overline{H}^o}{a\left(1+\frac{n_{Bo}}{n_{Ao}}\right)}$$

Which means the standard change in enthalpy for the reaction can only be zero if the standard molar change is itself zero, but that is rare to find in practice, although theoretically possible.

However, having equal amounts of reactants and products does not result in an enthalpy change of zero:

$$\Delta \overline{H}^o=0\;\;\rightarrow\;\;\Delta H^o=0$$

$$\Delta n=0\;\;\nrightarrow\;\; \Delta H^o=0$$