Question Video: Calculating Standard Enthalpy of Combustion of Methane Using Standard Enthalpies of Formation of Methane and Carbon Dioxide | Nagwa Question Video: Calculating Standard Enthalpy of Combustion of Methane Using Standard Enthalpies of Formation of Methane and Carbon Dioxide | Nagwa

# Question Video: Calculating Standard Enthalpy of Combustion of Methane Using Standard Enthalpies of Formation of Methane and Carbon Dioxide

Using the data in the table, calculate to the nearest kJ/mol the standard enthalpy of combustion of methane.

05:48

### Video Transcript

Using the data in the table, calculate to the nearest kilojoule per mole the standard enthalpy of combustion of methane.

A standard enthalpy of combustion can be written as ΔH c with a standard character. The Δ means a change, H means an enthalpy, c indicates combustion, and the ⦵ character indicates that everything is in its standard state. The definition of the standard enthalpy of combustion of a substance is the change in enthalpy of the system due to the complete combustion of the substance with oxygen per mole of substance with all substances in their standard states. The standard state of a substance is its reference state at one bar of pressure. So, to work out the standard enthalpy of combustion of methane, we need to construct the equation for its combustion with oxygen.

This is the balanced chemical equation for the reaction of oxygen with methane with all components in their standard states. Even though when methane reacts with oxygen water is likely to be produced as a gas, here it is a liquid because that is the reference state for water. The enthalpy change of this reaction will be equal to the standard enthalpy of combustion of methane. We’ll need to use the values in the table and construct a Hess cycle so that we can calculate the reaction enthalpy.

We’ve been given the standard enthalpies of formation of methane, carbon dioxide, and water. The definition of a standard enthalpy of formation of a substance is the change in enthalpy of the system due to the formation of the substance from its constituent elements per mole of substance with all substances in their standard states. For example, this is the reaction equation for the formation of methane from carbon and hydrogen. The standard state of carbon is solid graphite. And the standard state of hydrogen is diatomic hydrogen gas.

So we can see that one molecule of methane and two of oxygen can be formed from one atom of carbon, two hydrogen molecules, and two oxygen molecules. Since the standard state of oxygen is the oxygen molecule in its gas state, the enthalpy of formation for oxygen is zero by definition. Since the equation is balanced, we can use the atoms that form the reactants to also form the products, giving us a complete Hess cycle.

A Hess cycle is useful because of Hess’s law which states that the enthalpy change for a reaction is equal to the sum of the enthalpy changes for any set of reactions with the same overall reaction. Which means that the enthalpy change for going this way is the same as the enthalpy change for going this way. The enthalpy change for the formation of the reactants in this equation from their constituent elements is the enthalpy of formation of methane.

Remember that the enthalpy of formation of oxygen is zero. As for the formation of the products, the combined enthalpy of that reaction is equal to the enthalpy of formation of carbon dioxide plus twice the enthalpy of formation of water. Because of Hess’s law, using our Hess cycle, we’ve demonstrated that the standard enthalpy of combustion of methane is equal to the negative enthalpy of formation of methane plus the enthalpy of formation of carbon dioxide plus two times the enthalpy of formation of water.

Remember, when you reverse a reaction, you also reverse the sign of the enthalpy. So going from methane and oxygen to carbon, hydrogen, and oxygen requires a change in enthalpy of minus the enthalpy of formation of methane. Now, all we need to do is substitute in the values from the table to work out the enthalpy of the reaction. So the standard enthalpy of combustion of methane is equal to minus minus 74.6 plus minus 393.51 plus two times minus 285.83. Which is equal to 74.6 minus 393.51 minus 571.66, which is equal to minus 890.57 kilojoules per mole.

We’re asked to give the answer to the nearest kilojoule per mole. So our final answer is minus 891 kilojoules per mole of methane. The minus sign means that, for each mole of methane combusted with oxygen, 891 kilojoules of energy is transferred to the surroundings. So the reaction of methane with oxygen under standard conditions is exothermic.

A quicker way of answering this question would have been to use the rule that an enthalpy of combustion is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants. This arises because of Hess’s law and means that you don’t need to draw the Hess cycle. Either way, our final answer, to the nearest kilojoule per mole, for the standard enthalpy of combustion of methane is minus 891 kilojoules per mole of methane.

## Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions