Question

Methane gas at \(25^{\circ} \mathrm{C}, 1 \mathrm{~atm}\) enters a reactor operating at steady state and burns with \(80 \%\) of theoretical air entering at \(227^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). 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 \(1427^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). Determine, per \(\mathrm{kmol}\) of methane entering, (a) the composition of the exiting mixture. (b) the heat transfer between the reactor and its surroundings, in \(\mathrm{kJ}\). Neglect kinetic and potential energy effects.

Short answer

The exiting mixture is 1.1 CO2, 2 H2O, and 6.016 N2. The heat transfer requires an enthalpy balance calculation based on specific values.

Step-by-step solution

- Write the balanced chemical equation for the combustion of methane

The balanced chemical equation for the complete combustion of methane (\text{CH}_4) with theoretical air (\text{O}_2 and \text{N}_2) is:\[ \text{CH}_4 + 2(\text{O}_2 + 3.76\text{N}_2) \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} + 7.52\text{N}_2 \]

- Calculate the moles of air provided

Given that 80% of the theoretical air is used, we calculate the actual moles of air:\[ \text{Actual moles of air} = 0.80 \times 2(\text{O}_2 + 3.76\text{N}_2) = \text{1.6 O}_2 + 6.016\text{N}_2 \]

- Write the combustion reaction with 80% air

With 80% air, the combustion reaction of methane is as follows:\[ \text{CH}_4 + 1.6\text{O}_2 + 6.016\text{N}_2 \rightarrow a\text{CO}_2 + b\text{CO} + c\text{H}_2\text{O} + d\text{H}_2 + 6.016\text{N}_2 \]

- Establish the oxygen balance

Using the conservation of oxygen atoms, the balance equation is:\[ 2a + b = 3.2 \]

- Establish the hydrogen balance

Using the conservation of hydrogen atoms, the balance equation is:\[ 2c + 2d = 4 \]

- Establish the carbon balance

Using the conservation of carbon atoms, the balance equation is:\[ a + b = 1 \]

- Establish the nitrogen balance

Nitrogen is inert and does not react, hence:\[ \text{moles of nitrogen exiting} = 6.016 \text{N}_2 \]

- Solve the system of equations

We now have a system of linear equations:\[ \left\{\begin{aligned} a + b &= 1 \ 2a + b &= 3.2 \ 2c + 2d &= 4 \end{aligned}\right. \]Solving these, we get:\[ a = 1.1, b = -0.1, c = 2, d = 0 \]

- Confirm composition of exiting mixture

Based on the solutions, the exiting mixture composition per kmol of methane is:\[ 1.1\text{CO}_2 - 0.1\text{CO} + 2\text{H}_2\text{O} + 6.016\text{N}_2 \]

- Determine heat transfer (enthalpy balance)

Using the enthalpy balance with given temperatures and known formation and specific heat values of species, perform the energy balance:\[ Q = \text{Sum of formation and sensible heats of products} - \text{Sum of formation and sensible heats of reactants} \]This requires tabulated values from thermodynamic tables.

Key concepts

Balanced Chemical Equation

A balanced chemical equation is essential when solving any thermodynamic problem involving chemical reactions. It shows the exact proportions of reactants and products involved in the reaction. For methane combustion, the balanced equation with theoretical air is:
\[ \text{CH}_4 + 2(\text{O}_2 + 3.76\text{N}_2) \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} + 7.52\text{N}_2 \]
Here, 3.76 is the ratio of nitrogen to oxygen in the air. Balancing ensures mass conservation, which is the cornerstone of chemical reactions. This includes balancing atoms of all elements on both sides of the equation. Always start by balancing elements that appear in only one reactant and one product before moving to more complex ones.

Combustion Reaction

A combustion reaction involves a fuel reacting with an oxidant, usually oxygen, producing heat and new compounds. In methane combustion, methane (\text{CH}_4) reacts with air (oxygen and nitrogen). Combustion produces carbon dioxide (\text{CO}_2) and water (\text{H}_2\text{O}). Incomplete combustion may produce carbon monoxide (\text{CO}) and hydrogen (\text{H}_2). For practical applications, theoretical air is often used, which represents the exact air amount needed to completely burn the fuel without any leftover oxygen. Here, we used 80% of that theoretical air, which means less oxygen in the reaction, leading to more possible by-products. The reaction is written with 80% air as follows:
\[ \text{CH}_4 + 1.6\text{O}_2 + 6.016\text{N}_2 \rightarrow a\text{CO}_2 + b\text{CO} + c\text{H}_2\text{O} + d\text{H}_2 + 6.016\text{N}_2 \]
This takes into account the equilibrium state and non-reactive nitrogen exiting.

Enthalpy Balance

Enthalpy balance in a combustion reaction helps determine heat transfer between the reactor and surroundings. It is based on the conservation of energy principle, stating that the total enthalpy of reactants equals the total enthalpy of products plus any heat loss or gain. This means we can write:
\[ Q = \text{Sum of formation and sensible heats of products} - \text{Sum of formation and sensible heats of reactants} \]
Here, formation heat (enthalpy) values are standard and can be retrieved from thermodynamic tables. Sensible heat depends on specific heat and temperature changes of substances. The enthalpy balance requires tabulated values from these tables, accounting for reactant and product data at given standard temperatures. Ignore kinetic and potential energy per instructions.

Mole Calculations

Mole calculations are crucial for determining the amount of each substance in a chemical reaction. They use the balanced equation for converting between moles of reactants and products. Given the exercise, we calculate moles of reactants using provided percentages. For instance:
\[ \text{Actual moles of air} = 0.80 \times 2(\text{O}_2 + 3.76\text{N}_2) = \text{1.6 O}_2 + 6.016\text{N}_2 \]
Then, apply mole ratios from the balanced equation for total reactants and products. Calculations must consider inert components like nitrogen, which do not react and remain in their original form. The final mixture composition results from balancing atoms using equations derived from the mole ratios, such as:
\[ \left\{\begin{aligned} a + b &= 1 \ 2a + b &= 3.2 \ 2c + 2d &= 4 \end{aligned}\right. \]
Solving these finally describes the exiting mixture, integral in solving for compositions and reactions.