Enthalpy and temperature change

Suppose you don't let any of the heat (- 891 kJ mol^-1) escape from the flame (this is adiabatic conditions). You know that the mixture will be hot. To work out how hot, you need to know what you are heating up and what the heat capacities of the materials are.

In air, for every one oxygen there are four nitrogen. I could therefore write the reaction as

The heat produced by the reaction is heating the reaction products and the nitrogen. Note that it's going to be hot so the water produced will be a gas.

This is how you go about it.

Under adiabatic conditions no heat is entering or leaving the system so

D H(system) = 0

You know, however, that there are enthalpy changes taking place within the system

there is a reaction (here producing heat)

there is a temperature change.

D H(system) = D H (reaction process) + D H (temperature change) = 0

where

D H (temperature change) = C_p(everything being heated) x D T

In this equation the C_p is the total for everything being heated, not the molar heat capacity of any particular product.

If you want this more mathematically

where the C_p values now are the molar heat capacities and the m values are the number of moles of each component.

For one mole of reaction D H (reaction process) = - 891 000 J

and D H (temperature change) = [37 + (2 x 34) + (4 x 29)] DT

so 221 DT + (-891 000) = 0

DT = 891 000/221

DT = 4032 K

This is a temperature change, so the actual flame temperature, if the reactants were at 298 K, would be

T = 4330 K ( or 4057 °C)

Click to remove answer

Standard molar heat capacities
	C_p / J K^-1 mol^-1
CO₂(g)	37	Click for answer
N₂(g)	29
H₂O(g)	34	Click for help