Does Combusting Different Amounts of Propane Change Final Temperature?

• Chemistry
• zenterix
zenterix
Homework Statement
Consider the combustion of propane gas by oxygen in the Earth's air. Assume air is a mixture of ##N_2## and ##O_2## in a 4 to 1 ratio.

Assume that the flame burns under adiabatic and constant pressure conditions.

In addition, we are given that

$$\mathrm{\Delta_f H^\circ (C_3H_8,g,298.15K)=-104\frac{kJ}{mol}}$$

$$C\mathrm{_P(C_3H_8,g,298.15K)=84\frac{J}{mol\cdot K}}$$

$$\mathrm{\Delta_f H^\circ (H_2O,g,298.15K)=-241.80\frac{kJ}{mol}}$$

$$\mathrm{\Delta_f H^\circ (CO_2,g,298.15K)=-393.51\frac{kJ}{mol}}$$
Relevant Equations
What is the final temperature ##T_f## for the adiabatic flame when 1.01 mol of propane are burnt with a stoichiometric amount of air?

Assume all ##C_P##'s are independent of ##T##.
The combustion reaction for 1 mol of propane is

$$\mathrm{C_3H_8(g)+5O_2(g)+20N_2(g)\rightarrow 3CO_2(g)+4H_20(g)+20N_2}\tag{1}$$

and by using the given enthalpies of formation we can easily compute

$$\Delta_{rxn} H^\circ=-2043.73\text{kJ}$$

This enthalpy of reaction is for an isothermal combustion. That is, in (1), both reactants and products are at the same temperature of 298.15K.

When the reaction actually happens, as the propane is combusting, the heat generated stays in the system and temperature of the products is raised.

Schematically, we have

The top path is the isothermal combustion (for which we calculated the reaction enthalpy).

The bottom path has two steps: first, the reaction occurs adiabatically and at constant pressure, generating products at a higher temperature ##T_f##; second, the products are cooled at constant pressure to 298.15K.

The enthalpy of reaction of the first step is zero since it is an adiabatic process.

The enthalpy of reaction of the second step is just ##\int_{T_f}^{298.5K} C_P dT## where ##C_P## is the constant pressure heat capacity of the products at temperature ##T_f##.

Since both the top and bottom paths start and end in the same states we can equate their enthalpies of reaction.

$$-2043.73\text{kJ}=\int_{T_f}^{298.5K} C_P dT\tag{2}$$

where

$$C_P=(3\cdot 37.25+4\cdot 33.577+20\cdot 29.12)\mathrm{\frac{J}{mol\cdot K}=828.458\frac{J}{K}}\tag{3}$$

We can solve (2) for ##T_f## and we find

$$T_f=2764.17\text{K}$$

Note, however, that these calculations seem to be for the combustion of 1 mol of propane. The problem asks for the combustion of 1.01 mol of propane.

This is where I have questions.
It seems the reaction becomes

$$\mathrm{1.01C_3H_8(g)+5.05O_2(g)+20.2N_2(g)\rightarrow 3.03CO_2(g)+4.04H_20(g)+20.2N_2}\tag{1a}$$

and the enthalpy of reaction is just ##1.01\cdot (-2043.73)\text{kJ}##.

The enthalpy of the bottom path in the picture above is

$$\int_{T_f}^{298.15} C_PdT$$

which we equation to ##1.01\cdot (-2043.73)\text{kJ}##.

Note, however, that we have more product at ##T_f## than we did before. Thus, it seems that the heat capacity of these products at ##T_f## is just 1.01 times what we had before

$$C^{new}_P=1.01\cdot 828.458\mathrm{\frac{J}{K}}$$

But then, when we write

$$\int_{T_f}^{298.15} C^{new}_PdT=\int_{T_f}^{298.15}1.01C_PdT=1.01\cdot (-2043.73)\text{kJ}$$

don't the factors of 1.01 cancel?

There seems to be something wrong since the cancelling would mean that ##T_f## doesn't depend on how much propane burns.

Last edited:
Stoichiometry means that the fuel to air ratio stays the same. Equation (1) and (1a) are the same and the flame temperature is indeed independent of the actual amount of propane. The flame temperature only changes when the ratio between propane and air changes.

You are using more propane, but you are using more air too, and the amounts of reaction products than get heated is also greater.

Here is the thing. I reached the result that the temperature of the flame is 2764.14K.

This problem is from an automated grading system on EdX. Their temperature for the flame is 2740K.

I can obtain this result if I do the following

$$C^{new}_P=1.01\cdot C_P=836.74$$

$$\int_{T_f}^{298.15} C^{new}_PdT=(249.47-0.836T_F)\text{kJ}$$

and I now equate this to the enthalpy of reaction for the initial reaction.

$$(249.47-0.836T_F)\text{kJ}=-2043.73$$

and solve

$$T_F=2740$$

In other words, to compute the temperature when we burn 1.01 mol of propane, they seem to have updated the heat capacity of the products to a account for the extra amount of products, but they did not do the same for the original reaction enthalpy.

Are they correct in this?

Cp is the heat capacity per mole To get the enthalpy change, you multiply the Cp by the number of moles, and by the temperature change.

Chestermiller said:
Cp is the heat capacity per mole To get the enthalpy change, you multiply the Cp by the number of moles, and by the temperature change.
This is clear. After all, this is exactly the calculation of ##C^{new}_P## in post #4: I multiplied the molar heat capacity ##C_P## by the number of moles, 1.01.

The change in enthalpy to cool the products down from ##T_f## to ##298.15\text{K}## is then ##\int_{T_f}^{298.15} C_P^{new}dT##.

This is all clear.

The question is, can we equate this to the enthalpy of reaction of (1) in the OP, or must we equate it to the enthalpy of reaction of (1a)?

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