Question

Gaseous propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right.\) ) at \(25^{\circ} \mathrm{C}, 1\) atm enters a reactor operating at steady state and burns with \(80 \%\) of theoretical air entering separately at \(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). An equilibrium mixture of \(\mathrm{CO}_{2}, \mathrm{CO}, \mathrm{H}_{2} \mathrm{O}(\mathrm{g}), \mathrm{H}_{2}\), and \(\mathrm{N}_{2}\) exits at \(1227^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). Determine the heat transfer between the reactor and its surroundings, in kJ per kmol of propane entering. Neglect kinetic and potential energy effects.

Short answer

The heat transfer between the reactor and surroundings is -2137.93 kJ per kmol of propane entering.

Step-by-step solution

- Write the balanced chemical equation

The balanced combustion reaction of propane (\text{C}_3\text{H}_8) with theoretical air is: \[ \text{C}_3\text{H}_8 + 5\text{O}_2 + 5 \times 3.76 \text{N}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} + 18.8 \text{N}_2 \] Note that each mole of oxygen is accompanied by 3.76 moles of nitrogen.

- Determine the amount of air used

Given that 80% of the theoretical air is used, calculate the actual moles of air: \[ \text{Actual air} = 0.80 \times \text{Theoretical air} = 0.80 \times (5 + 5 \times 3.76) = 18.8 \text{ moles}\]

- Identify reaction products

At equilibrium, not all products are completely burned, so we have \text{CO}_2, \text{CO}, \text{H}_2\text{O}(\text{g}), \text{H}_2, and \text{N}_2. We write the general form of the reaction: \[ \text{C}_3\text{H}_8 + 5O_2 + 18.8N_2 \rightarrow a\text{CO}_2 + b\text{CO} + c\text{H}_2\text{O} + d\text{H}_2 + 18.8\text{N}_2 \]

- Apply mass balance for each component

For Carbon: \[ 3 = a + b \] For Hydrogen: \[ 8 = 2c + 2d \] For Oxygen: \[ 10 = 2a + b + c \]

- Solve the system of equations

From step 4, solve the equations to find values for a, b, c, and d: \[ a = 3, b = 0, c = 4, d = 0 \] Hence, the products are 3 moles of \text{CO}_2, 4 moles of \text{H}_2\text{O}, and 18.8 moles of \text{N}_2.

- Calculate enthalpy changes

Use enthalpies at standard conditions (\text{kJ/mol}) for reactants and products. From standard enthalpy tables, we get: \( \text{C}_3\text{H}_8 : -103.85 \), O_2 : 0, \text{N}_2 : 0, \text{CO}_2 : -393.5, \text{H}_2\text{O (g)} : -241.82

- Calculate energy released (entry condition)

Calculate heat of formation reactants and products at 298.15K: \( \text{ΔH}_{298.15K}^{reactants} = (-103.85) + 0 + 0 = -103.85 \text{kJ/mol} \) \( \text{ΔH}_{298.15K}^{products} = 3(-393.5) + 4(-241.82) + 18.8(0) = -2241.78 \text{kJ/mol} \)

- Enthalpy change at standard state

\( \text{ΔH}_{298.15K} = (-2241.78) - (-103.85) = -2137.93 \text{kJ/mol} \)

- Heat transfer between reactor and surroundings

Given constant specific heat values, the heat transfer (Q) to the surrounding can be achieved from \( Q = ΔH_{reaction} - (\text{H products - H reactants}) \)

Key concepts

Balanced Chemical Equation

When dealing with chemical reactions, starting with a balanced equation is crucial. In the case of propane (\text{C}_3\text{H}_8) combustion, the balanced equation ensures that the reactants and products include all elements in the correct proportions. For propane burning with theoretical air, the equation is:
\[ \text{C}_3\text{H}_8 + 5\text{O}_2 + 5 \times 3.76 \text{N}_2 \rightarrow 3 \text{CO}_2 + 4 \text{H}_2\text{O} + 18.8 \text{N}_2 \] Here, each mole of oxygen (O_2) comes with 3.76 moles of nitrogen (N_2). This ensures the equation reflects real air composition and makes it simpler to analyze mass balance and enthalpy changes.

Mass Balance

In chemical processes, a mass balance ensures that the total mass of reactants equals the total mass of products. This balance provides insight into the quantities of different substances involved. For propane combustion at steady state:
\[ \text{C}_3\text{H}_8 + 5O_2 + 18.8N_2 \rightarrow a\text{CO}_2 + b\text{CO} + c\text{H}_2\text{O} + d\text{H}_2 + 18.8\text{N}_2 \] To solve this, we create equations based on the conservation of atoms for each element involved. For example:
- Carbon (C): \[ 3 = a + b \] - Hydrogen (H): \[ 8 = 2c + 2d \] - Oxygen (O): \[ 10 = 2a + b + c \] Solving these equations leads to values for a, b, c, and d. It helps us understand the amounts of different products formed during the reaction.

Enthalpy Change

Enthalpy change (\text{ΔH}) during a reaction refers to the heat absorbed or released. This is essential to determine whether the reaction is exothermic (releases heat) or endothermic (absorbs heat). For propane combustion, we use standard enthalpy values (\text{kJ/mol}) from tables:
- \text{C}_3\text{H}_8: -103.85
- \text{O}_2: 0
- \text{N}_2: 0
- \text{CO}_2: -393.5
- \text{H}_2\text{O (g)}: -241.82
By calculating the total enthalpy of reactants and products at standard conditions (298.15K), we find the enthalpy change of the reaction:
\[ \text{ΔH}_{298.15K} = (\text{ΔH products}) - (\text{ΔH reactants}) \] This gives us the energy change associated with the reaction, providing insight into the energy dynamics.

Heat Transfer Calculation

Heat transfer (\text{Q}) between the reactor and surroundings is a critical aspect of thermodynamic analysis. It represents the energy exchanged due to reaction enthalpy and temperature changes. Using the enthalpy change from the reaction and considering the reactor operates at steady state, we can calculate \text{Q}. This is done as follows:
\[ Q = ΔH_{reaction} - (\text{H products} - \text{H reactants}) \] By introducing the specific heat values and temperature differences, this equation quantifies how much heat is transferred to/from the reactor to its surroundings. Understanding \text{Q} helps in designing efficient reactors and ensuring energy optimization during the process.