# Methane Combustion and Chemical Equilibrium

• HinduHammer
In summary, the problem involves calculating the concentration of different products in a natural gas fired water heater operating with 15% excess air. The given equations and equilibrium constant are used to balance the reaction and find the concentrations, but the presence of CO creates an imbalance in the equations. The use of K can help in solving for the concentrations using the ICE table method.
HinduHammer
This may belong in the Chemistry section but this is a problem from my Thermodynamics class, so I'll post it here and then move to the Chem section if need be.

## Homework Statement

Methane (CH4) is the principal constituent of natural gas and is often used to represent natural gas. Consider the operation of a natural gas fired water heater operating with 15% excess air. Assuming the fuel to be CH4 and the products of the reaction to be CO2, H2O, N2, O2 , and CO, calculate the concentration (percent by volume) of each product if the temperature in the products is 1000K.

## Homework Equations

CH4 + (a)(m)(O2 + 3.76N2 → bCO2 + cH2O + dN2 + eO2 + fCO

2CO2 $\Leftrightarrow$ 2CO + O2

where:
a=moles for stoichiometric combustion
m=excess air ratio (1.15 in this case)
b,c,d,e,f = moles of products

## The Attempt at a Solution

To balance equation:

From combustion reaction:
C: 1 = b+f
H: 4 = 2; c=2
O: 2*a*1.15 = c+2e+f
N: 3.76*1.15*a*2 = 2d

From dissociation reaction:
2b = 2f + e (2 CO2 makes 2CO and 1O2)

I still only have 5 equations and 6 unknowns.

There is also the equilibrium constant K that we are given a table of:

I know how to solve the problem when CO is not created. It becomes a much simpler problem. However, when CO enters the problem, I end up with not enough equations.

Thanks for your help!

Last edited by a moderator:
Are you given K value? Look to me like it can be just a stoichiometry problem, where you are to assume reaction proceeded to the end.

Alternatively, the only equilibrium to be taken into account is 2CO + O2 <-> 2CO2

Yes, we are given a K value. However, we haven't been taught this type of Chemistry yet, so I'm having difficulty solving the K equation.

I know that the equation for 2CO + O2 <-> 2CO2 is used to find concentrations of the substances, but I'm not exactly sure how...

Edit: K = 3.68 x 10^-21

After I set up K = ((Y_CO)^2 x (Y_O2))/ (Y_CO2)^2 I get stuck on solving for the terms. I assume Y_CO2 = 1 because it is the only term on the right side. However, I am unsure on how to find Y_CO and Y_O2.

Last edited:

Hello, thank you for your question. I would approach this problem by first determining the equilibrium constant for the given reaction, which is the ratio of the products to the reactants at equilibrium. In this case, the reaction is CH4 + (a)(m)(O2 + 3.76N2 → bCO2 + cH2O + dN2 + eO2 + fCO. The equilibrium constant can be calculated using the table provided, which gives us the values for the coefficients of each component in the reaction.

Next, I would use the given excess air ratio (m) and the stoichiometric combustion equation to determine the moles of each component in the reaction. This can be done by setting up a system of equations using the coefficients of each component and the known values for excess air and stoichiometric combustion. This will give us the values for a, b, c, d, e, and f.

After obtaining these values, I would use the equilibrium constant to calculate the concentrations of each product at equilibrium. The equilibrium constant equation is K = ([CO]^b * [H2O]^c * [N2]^d * [O2]^e * [CO2]^f)/([CH4]^a * [O2]^am * [N2]^(3.76am)). Using the values obtained in the previous step, we can solve for the concentrations of each product.

Finally, we can convert the concentrations to percent by volume by dividing each concentration by the total volume of the products (which can be calculated using the ideal gas law and the given temperature). This will give us the final answer for the concentration of each product in the natural gas combustion reaction.

I hope this helps guide you in solving the problem. It is important to note that the equilibrium constant may change at different temperatures, so this solution is only valid for the given temperature of 1000K. If you have any further questions, please let me know. Good luck!

## What is methane combustion?

Methane combustion is the chemical reaction where methane (CH4) combines with oxygen (O2) to produce carbon dioxide (CO2), water (H2O), and energy in the form of heat and light.

## What is the chemical equation for methane combustion?

The chemical equation for methane combustion is: CH4 + 2O2 → CO2 + 2H2O + heat.

## What is the role of chemical equilibrium in methane combustion?

Chemical equilibrium is the state in which the rate of the forward reaction (reactants turning into products) is equal to the rate of the reverse reaction (products turning into reactants). In methane combustion, chemical equilibrium is important because it determines the amount of methane and oxygen consumed, as well as the amount of carbon dioxide and water produced.

## What factors affect the chemical equilibrium in methane combustion?

The factors that affect chemical equilibrium in methane combustion include temperature, pressure, and the initial concentrations of methane and oxygen. Changes in any of these factors can shift the equilibrium in favor of the reactants or products.

## Can methane combustion be reversed to produce methane and oxygen again?

Technically, yes. However, the process of reversing methane combustion is not feasible or practical in most situations. It requires a significant amount of energy, and the resulting methane and oxygen would immediately react again to form carbon dioxide and water.

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