What is the enthalpy change for the reaction: N2(g) + 3H2(g) → 2NH3(g) if ΔHf° f

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Question: What is the enthalpy change for the reaction: N2(g) + 3H2(g) → 2NH3(g) if ΔHf° for NH3 is -45.9 kJ/mol?

Options:

Correct Answer: -137.7 kJ

Solution:

The enthalpy change for the reaction is calculated as ΔH = 2 * ΔHf°(NH3) = 2 * (-45.9 kJ) = -91.8 kJ.

What is the enthalpy change for the reaction: N2(g) + 3H2(g) → 2NH3(g) if ΔHf° for NH3 is -45.9 kJ/mol?

What is the enthalpy change for the reaction: N2(g) + 3H2(g) → 2NH3(g) if ΔHf° for NH3 is -45.9 kJ/mol?

Step 1: Identify the reaction: N2(g) + 3H2(g) → 2NH3(g).
Step 2: Understand that ΔHf° (enthalpy of formation) for NH3 is given as -45.9 kJ/mol.
Step 3: Note that the reaction produces 2 moles of NH3.
Step 4: Calculate the total enthalpy change for the formation of 2 moles of NH3 by multiplying the ΔHf° by the number of moles: ΔH = 2 * ΔHf°(NH3).
Step 5: Substitute the value: ΔH = 2 * (-45.9 kJ).
Step 6: Perform the multiplication: ΔH = -91.8 kJ.
Step 7: Conclude that the enthalpy change for the reaction is -91.8 kJ.

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