Video: Identifying the Label Corresponding to the ΔH of the Reaction That Forms CO(g) and H₂O(l) from CH₄(g) and O₂(g) on an Enthalpy Diagram

Which label corresponds to the ΔH of the reaction that forms CO(g) and H₂O(l) from CH₄(g) and O₂(g)?

04:43

Video Transcript

Which label corresponds to the ΔH of the reaction that forms CO gas and H2O liquid from CH4 gas and O2 gas?

In this question, we’re being asked to use a diagram to identify the ΔH or enthalpy change of a given process where some reactants turn into some products. The reactants in question are methane gas and oxygen gas. The products being formed are carbon monoxide gas and liquid water. To make this symbol equation balance properly, we would need one and a half moles of oxygen molecules to react with one mole of methane molecules to form one mole of carbon monoxide molecules and two moles of water molecules.

ΔH represents the enthalpy change for a reaction. This is the change in enthalpy between the products and the reactants in a reaction taking place at constant pressure. Notice here that enthalpy is designated as a vertical scale on the left side of the diagram. Put simply, enthalpy is the heat content or internal energy of the chemicals involved in the reaction.

In the diagram here, we see reactants at the top and two possible reaction pathways leading to products at the bottom of the diagram. In all three reactions, the products have a lower enthalpy than the reactants. So, the reactions are all exothermic. The numbers quoted with kilojoules as units in the diagram are, in fact, the ΔH values for each reaction step. They’re all negative as they’re all exothermic reactions.

The diagram can be viewed as two alternative pathways for the complete combustion of one mole of methane. Reaction one represents the complete combustion of one mole of methane in a single step to form complete oxidation products. This reaction would release 890 kilojoules of heat energy to the surroundings. As carbon dioxide is formed as a product here, negative 890 kilojoules is not the label for the ΔH of the reaction that we are seeking. The complete combustion of methane does not form carbon monoxide, so its ΔH is not the correct answer. The label at the top of the diagram is simply a label for the reactants; it’s not a label for a ΔH for a reaction. And it’s also not a correct answer either.

In reaction two, we see the incomplete combustion of one mole of methane. Limited oxygen has reacted with the methane. And the carbon has not been fully oxidized to carbon dioxide. Instead, carbon monoxide or CO has been formed. If we subtract a half a mole of oxygen from both sides of this equation that didn’t react, we get the same equation that we were looking for in the first place with the correct reactants and products as specified in the question.

Negative 607 kilojoules is the label for the ΔH or enthalpy change for reaction that we are looking for as specified in the question. It could, therefore, be the correct answer. Before we commit to this answer, let us check what is happening in reaction three.

Reaction three represents the combustion of one mole of carbon monoxide. By canceling out the two moles of liquid water which are present on both sides of the equation, we can see that the only new product formed is one mole of carbon dioxide. Since the question specifies a reaction that forms carbon monoxide and liquid water and in reaction three no new liquid water is formed, then negative 283 kilojoules is not the correct answer either.

The final label in the diagram is just a label for the overall products of these two alternative reaction pathways. It’s not a label for a ΔH, and it’s not the correct answer either. The correct answer is, therefore, negative 607 kilojoules.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.