- #1

zenterix

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- 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.

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.

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

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.

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