Enthalpies of Formation from Methane & Oxygen

In summary, we discussed the production of three different carbon-containing products (soot, CO2, and CO2) through the reaction of methane gas with oxygen. We wrote three balanced equations and calculated the standard enthalpies for each reaction. We then discussed why, in the presence of adequate oxygen, CO2 is the predominant product of methane combustion, considering the amount of energy produced per mole of methane.
  • #1
chemistry4all
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Homework Statement



Enthalpies of Formation. burning methane in oxygen can produce three different carbon-containing products: soot (very fine particles of graphite), CO2 (g), and CO2 (g). A) Write three balanced equations for the reaction of methane gas with oxygen to produce these three products. In each case assume that H2O (l) is the only other product. B) Determine the standard enthalpies for the reactions in part A). C) Why, when the oxygen supply is adequate, is CO2 (g) the predominant carbon-containing product of the combustion of methane?


Homework Equations



Standard formations of enthalpies table was used for this question

The Attempt at a Solution



My attemp for A and B but have no idea how to figure out C I need help.

A) and B) toegther is as follows

1. 2CH4 (g) + 3(O2) (g) -----> 2CO (g) + 4H2O (l)
((2*-110.5)+(4*-285.8))-((3*0)+(2*-74.80)) = DH is -1214.6 kJ/mol

2. CH4 (g) + 2(O2) (g) -------> CO2 (g) + 2H2O (l)
((1*-393.5)+(2*-285.8)) - ((1*-74.8)+(2*0)) = DH is -890 kJ/mol

3. CH4 (g) + O2 (g) ----------> C (s) + 2H2O (l)
((1*1.88)+(2*-285.8)) - ((1*-74.80)+(1*0)) = DH is -494.92 kJ/mol

I used the enthalpies of formation data tables and the formula
DH= (sum of products) - (sum of reactants). Now how do I figure out C?

I notice that B with CO2 as the product is -890.3 kJ/mol which is between A and C. or is it because when there are more number of moles of O2 and less only 1 mole of methane then CO2 forms as a product. I am still confused on how to answer C. Any help please?:confused::confused:
 
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  • #2


Looks fine. Just consider: Since you have all the oxygen you need, which reaction produces the most energy per mole of methane?
 

What is the enthalpy of formation for methane?

The enthalpy of formation for methane (CH4) is -74.6 kilojoules per mole. This is the amount of energy released when one mole of methane is formed from its constituent elements, carbon and hydrogen, in their standard states.

How is the enthalpy of formation calculated?

The enthalpy of formation is calculated by taking the difference between the enthalpies of the products and the reactants in a chemical reaction. In the case of methane, the enthalpy of formation is calculated by subtracting the enthalpy of formation of carbon (graphite) and hydrogen gas from the enthalpy of formation of methane.

What is the significance of the enthalpy of formation?

The enthalpy of formation is a measure of the stability of a compound. A negative enthalpy of formation, as in the case of methane, indicates that the compound is stable and will release energy when formed. It is also used to calculate the enthalpy of a reaction and can be used to predict the feasibility of a reaction.

How does the enthalpy of formation of methane compare to other hydrocarbons?

The enthalpy of formation for methane is lower than that of other hydrocarbons with longer carbon chains. This is due to the fact that longer hydrocarbons have more bonds and therefore require more energy to break and form. Methane is the simplest hydrocarbon and has the lowest amount of energy required for its formation.

Can the enthalpy of formation be negative?

Yes, the enthalpy of formation can be negative as in the case of methane. This indicates that the formation of the compound is thermodynamically favorable and will release energy. However, not all compounds have a negative enthalpy of formation and some may have positive values, indicating that they require energy to form.

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