PHYSICAL CHEMISTRY IN BRIEF

CHAP. 5: THERMOCHEMISTRY [CONTENTS] 138

Note: In the enthalpy balance (5.18) it is assumed that the chemical reaction proceeds

quantitatively, and that no inert substances (i.e. substances which do not participate in

the reaction) are present in the system. If it is not so, members corresponding to the

heating of those reactants that do not react during the reaction and of inert substances

have to be added to the equation.

The theoretical flame temperature is the adiabatic temperature of reaction during
burning. The real mean flame temperature is lower than the theoretical temperature due to
losses.

Example

Calculate the theoretical flame temperature of methane burnt by air when the air is in a 50

percent surplus. For convenience, suppose that the air is composed of oxygen (xO 2 = 0. 2 ) and

nitrogen (xN 2 = 0. 8 ). Gases enter the reaction at temperature 298.15 K. The reaction proceeds at

constant pressure 101.325 kPa. Use data from the examples in5.3and5.2.2for the calculation.

Solution

We need 2 mol O 2 to burn 1 mol CH 4. Since the air is in a 50 percent surplus, it enters the

reaction with three moles of oxygen. Given that the amount of nitrogen in air is four-times

higher (xN 2 /xO 2 = 0. 8 / 0 .2 = 4), a total of 4×3 = 12 mol N 2 is present in the system during

combustion. The whole process can be formally described using an equation in the form

CH 4 + 3O 2 + 12N 2 =CO 2 + 2H 2 O+O 2 + 12N 2 ,

whose right side lists not only the products of the chemical reaction, water and carbon dioxide,

but also excess oxygen and (theoretically inert) nitrogen contained in the air. We use the result

from the example in5.2.2,∆rH(T = 298. 15 K) =−802 324J, and apply equation (5.18) to

obtain

0 = −802 324 +

∫Tprod

298. 15

[Cpm(CO 2 ) + 2Cpm(H 2 O) +Cpm(O 2 ) + 12Cpm(N 2 )] dT

= −802 324 + 581. 9 ×(Tprod− 298 .15) =⇒ Tprod= 1676. 9 K.