Find ∆H° in kilojoules for the reaciton of nitric oxide with oxygen, 2 NO(g) + O2(g) → N2O4(g), given the following data: N2O4(g) → 2 NO2(g) ∆H° = 55.3 kJ NO(g) + 1/2 O2(g) → NO2(g) ∆H° = -58.1 kJ
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Identify the given reactions and their enthalpy changes: (1) N2O4(g) → 2 NO2(g), ∆H° = 55.3 kJ and (2) NO(g) + 1/2 O2(g) → NO2(g), ∆H° = -58.1 kJ.
Write the target reaction: 2 NO(g) + O2(g) → N2O4(g).
Reverse reaction (1) to align with the target reaction: 2 NO2(g) → N2O4(g), which changes the sign of ∆H° to -55.3 kJ.
Multiply reaction (2) by 2 to balance the number of NO2(g) molecules in the reversed reaction (1): 2 NO(g) + O2(g) → 2 NO2(g), and multiply ∆H° by 2, giving -116.2 kJ.
Add the modified reactions to get the target reaction: (2 NO(g) + O2(g) → 2 NO2(g)) + (2 NO2(g) → N2O4(g)) results in 2 NO(g) + O2(g) → N2O4(g). Sum the ∆H° values: -116.2 kJ + (-55.3 kJ).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Hess's Law
Hess's Law states that the total enthalpy change for a chemical reaction is the sum of the enthalpy changes for the individual steps of the reaction, regardless of the pathway taken. This principle allows us to calculate the enthalpy change for a reaction by using known enthalpy changes of related reactions, making it essential for solving problems involving thermochemical equations.
The standard enthalpy of formation (∆H°f) is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. This concept is crucial for calculating the enthalpy change of a reaction, as it provides a reference point for the energy associated with the formation of reactants and products.
Enthalpy change (∆H) is a measure of the heat absorbed or released during a chemical reaction at constant pressure. It is a key concept in thermodynamics, allowing chemists to understand energy changes in reactions, predict reaction spontaneity, and calculate the energy required or released in various chemical processes.
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