# Role of heat in combustion

What role does heat play in the combustion process.Specifically, does the amount of heat determine whether complete or incomplete combustion occurs, and is the amount of heat impacted by the amount of oxygen

• Welcome to chemistry S.E?Why did you tag your question with inorganic-chemistry? Which combustion do you refer? – G M Mar 27 '14 at 22:23

Yes, temperature and heat do affect the spontaneity and the equilibrium of reactions. The term heat can be referred to enthalpy, which can be quite confusing in context, as I believe you are asking about raising the temperature.

Let's first take a look at the equation for the Gibb's Free Energy and raising temperature:

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$

The Gibb's Free energy does not only relate to the spontaneity of a reaction, the magnitude of Gibb's Free energy describes how much "free energy" is available, or how much work is possible.

Since, entropy, enthalpy are state functions, we can calculate them by using Hess's Law and using tabulated values for standard thermodynamic data:

$$\Delta H^\circ_{rxn} = \Sigma \Delta H^\circ_{products} - \Sigma \Delta H^\circ_{reactants}$$

$$\Delta S^\circ_{rxn} = \Sigma \Delta S^\circ_{products} - \Sigma \Delta S^\circ_{reactants}$$

For simplicity's sake, I will use the combustion of methane that mannaia has so kindly provided and run the reaction at STP using heats of formation:

$\Delta H^\circ_{\text{comb of} \ \ce{CH_4}} = -890 \text{kJ} \cdot \text{mol}^{-1}$

$\Delta H^\circ_{\text{f of} \ \ce{H_2O} \ \text{(l)}} = -285.83 \text{kJ} \cdot \text{mol}^{-1}$

$\Delta H^\circ_{\text{f of} \ \ce{CO_2} \ \text{(g)}} = -285.83 \text{kJ} \cdot \text{mol}^{-1}$

$$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$$

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$ $$\Delta G^\circ = -604.53 \text{kJ} \cdot \text{mol}^{-1} - (298.15 \text{K}) *97.39 \text{J} \cdot \text{mol}^{-1} = -633.7 \text{kJ} \cdot \text{mol}^{-1}$$

If we use the enthalpy of combustion for methane:

$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$ $$\Delta G^\circ = -1569.34 \text{kJ} \cdot \text{mol}^{-1} - (298.15 \text{K}) *97.39 \text{J} \cdot \text{mol}^{-1} = -1598.38 \text{kJ} \cdot \text{mol}^{-1}$$

What would occur if we raise the T term? Obviously, the Gibbs Free Energy would become larger in magnitude and more negative, thus being more spontaneous. Inversely, if we lower the temperature to such a degree that $T\Delta S^\circ = \Delta H^\circ$, the reaction would not occur at all, and if $T\Delta S^\circ > \Delta H^\circ$, we would need to input energy in order for the reaction to proceed forward.

In general:

If $\Delta H<0$ and (-) and $\Delta S>0$ (+), the reaction is spontaneous at any temperature.

If $\Delta H>0$ and (+) and $\Delta S<0$ (-), the reaction is NOT spontaneous at any temperature.

If $\Delta H<0$ and (-) and $\Delta S<0$ (-), the reaction is spontaneous at low enough temperatures.

If $\Delta H>0$ and (+) and $\Delta S>0$ (+), the reaction is spontaneous at high enough temperatures.

It is possible to think of heat as a product or reactant. Combustion reactions are highly exothermic. Le Chatlier's principle includes raising the temperature as a stress. Alternatively, we can think about the equilibrium of this reaction to see how much product will be formed by looking at the magnitude of the equilibrium constant, $K$.

Accordingly the van't Hoff isochore equation is written as:

$$ln \frac{K_2}{K_1} = \frac{\Delta H}{R}(\frac{1}{T_1} - \frac{1}{T_2})$$

In summary, an increase in temperature favors the formation of products if the reaction is endothermic. Inversely, a decrease in temperature favors the formation of products if the reaction is exothermic.

As, Uncle Al has stated:

Ambient oxygen makes no difference

The atmosphere is composed of about 70% nitrogen and 20% oxygen. If this reaction was to be observed in nature, there would be almost a limitless amount of oxygen for the reaction to rpoceed. Temperature does play a role in the stoichiometry of the reaction and thus whether or not the combustion is complete or incomplete. However, it is possible to limit one of the reactants as mannaia has done, specifically oxygen in a laboratory setting.

As related to the kinetic molecular theory of matter, increased temperature will increase the frequency of collisions of particles with sufficient energy to cleave bonds and reform them, driving the reaction further to the right. This is commonly done with the commercial production of ammonia using Van Haber process. Although this is thermodynamically unfavorable, when run at high temperatures, the rate of formation of ammonia is much faster.

In regards to say, an open environment, the temperature would increase the kinetic energy of the oxygen molecules, and so the particles would be spaced further apart. Describing this quantitatively... well, an experiment should be carried out. I would assume if one was so inclined to find if that would increase the equilibrium, rate, or completeness of combustion I'm sure they would find differences.

• could you please elaborate your sentence: "While you can limit one of the reactants ..., relatively speaking, it does not have an effect." Relatively to what ? Effect on what ? Thanks ! – mannaia Mar 28 '14 at 8:25
• check also the value of $\Delta H$ for methane combustion. thanks again ! – mannaia Mar 28 '14 at 11:11
• hopefully one last point ... Uncle All's statement ("Ambient oxygen makes no difference") referred to a specific case, but it can't be generalized. What happens if oxygen is not enough, for methane combustion?(chemistry.stackexchange.com/questions/9552/…). – mannaia Mar 28 '14 at 17:23
• @mannaia Thanks! I used the heat of formation for methane instead of combustion. I corrected that as well as making the last two paragraphs more clear and concise. And as you stated in the comment above and in your answer below, limiting one of the reactions will also limit completeness of combustion. – Jun-Goo Kwak Mar 28 '14 at 20:39
• could you still have a look at the numbers. I think there's something wrong in $\Delta G$ calculation. Then, why do you calculate $\Delta G$ twice and get two different values for the same reaction at STP ? – mannaia Mar 30 '14 at 7:50

Indirectly yes, the amount of oxygen will have an impact on the amount of heat. This because, if there is enough oxygen, combustion will tend to be complete.

Consider methane combustion, complete and then incomplete:

$$\ce{CH4(g) + 2O2(g) ->CO2(g) + 2H2O(l)} \qquad \Delta H^{o}_{comb}=-891\,kJ/mol$$

$$\ce{CH4(g) + 3/2O2(g) ->CO(g) + 2H2O(l)} \qquad \Delta H^{o}_{comb}=-608\,kJ/mol$$

The value of enthalpy, $\Delta H^{o}_{comb}$ ($25^{o}$) is larger for a complete combustion, as reported above.

That means that more heat (= enthalpy in conditions of constant pressure) is released if the combustion is complete.

• I think the question was a bit different: how the heat itself affect on combustion process. – pmod Mar 27 '14 at 23:14
• @pmod Well, I've tried to address the point "amount of heat impacted by the amount of oxygen" and differentiate between incomplete and complete combustion. But, anyway, I haven't probably interpreted the question in the good way (the title and body of the question look different to me). – mannaia Mar 28 '14 at 7:23

Short answer is, heat is a byproduct of the chemical reaction. It is released by the reaction. It is such that we now use combustion for (most of the time) the sole purpose of getting the heat out of it. We don't want the CO2, but it comes with it (if you burn hydrocarbons).

• Welcome to Chemistry.SE. Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$. – Martin - マーチン Apr 8 '15 at 7:14

In contrast, the explosion of triacetone triperoxide (83% the pop of TNT) has hard by zero exotherm. Ambient oxygen makes no difference. Much of the acetone can be recovered and reused - a near-zero carbon footprint, highly recyclable terrorist explosive. Why do people complain? J. Am. Chem. Soc. 127, 1146 (2005)

TATP, $\mathrm{Δ_{f}H°}$ = +151.4 kJ/mol
DOI: 10.1002/prep.201100100