Heat of combustion of hydrogen and methane

Question

8. The heat of combustion of hydrogen gas per gram is about 2.5 times that of natural gas. The main component of natural gas is methane gas.

Also, the combustion of hydrogen is “cleaner” than that of natural gas, since the only product is water. Thus, it seems a likely choice for a major fuel source in the future.

Prove that the heat of combustion of hydrogen per gram is about 2.5 times that of natural gas. Begin by calculating the heat associated with the combustion of each fuel in kJ/g of fuel.

Answer 1

We can calculate heat of combustion (H_C) of a reaction just using the balanced equilibrium equation of the combustion process. To calculate methane's H_C, we will use the balanced methane formation equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

We will use this equation and each element's heat of formation (H_F) to figure out the total heat of combustion. Luckily for us, any elements in their standard state (diatomics, single elements, etc) have an H_F of 0 kJ/mol, which means we can basically ignore that 2O₂ in the equation and focus on the other parts. To find our H_C, we simply need the H_F values of each molecule (which can be found on a standardized table) and follow a simple formula:

H_C = H_F(products) - H_F(reactants)

After searching H_F tables, we can see that the H_F of each component are as follows:

CH₄: -74.6 kJ/mol

CO₂: -393.51 kJ/mol

H₂O: -285.83 kJ/mol x2 = -571.66 kJ/mol

I multiplied the water by 2 because it takes two molar equivalents to complete the combustion based on the balanced equation above.

We can apply these values to our (products - reactants) equation above and we get approx. -890.6 kJ/mol. we can convert moles to grams using standard stoichiometry. We know that 1 mole of methane is 16.043g, so to convert, we simply divide by this number. We then get a heat of combustion of about -55.51 kJ/g.

Now that you can do the heat of combustion, try your hand at calculating the heat of combustion of hydrogen gas using this equation:

H₂ + 1/2O₂ → H₂O

Remember, elemental reactants/products will have an H_F of 0, so the only H_F you'll need is for 1 water molecule (which we know - look above!). Convert from kJ/mol to kJ/g for Hydrogen and you're good to go. You should get a value of about -141.6 kJ/g.

Comparing your H_C values, you'll notice that -141.6 kJ/g is about 2.5x that of -55.51 kJ/g.

NOTE:

Many reactions have very specific, known H_C values in kJ/mol. They are a chemical property of that compound and can be found in H_C tables! Tables are the "fast track" and more practical way of finding these values, but it IS important to know how to calculate them by hand. I just want to make you aware that tables exist.