What is the enthalpy change (ΔH) for the reaction: 2H2(g) + O2(g) → 2H2O(g) if Δ

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Question: What is the enthalpy change (ΔH) for the reaction: 2H2(g) + O2(g) → 2H2O(g) if ΔHf for H2O(g) is -241.8 kJ/mol?

Options:

Correct Answer: -483.6 kJ

Solution:

The enthalpy change for the reaction is 2 times the standard enthalpy of formation of H2O(g), which is -241.8 kJ/mol, giving -483.6 kJ.

What is the enthalpy change (ΔH) for the reaction: 2H2(g) + O2(g) → 2H2O(g) if ΔHf for H2O(g) is -241.8 kJ/mol?

What is the enthalpy change (ΔH) for the reaction: 2H2(g) + O2(g) → 2H2O(g) if ΔHf for H2O(g) is -241.8 kJ/mol?

Step 1: Identify the reaction: 2H2(g) + O2(g) → 2H2O(g).
Step 2: Understand that ΔHf for H2O(g) is given as -241.8 kJ/mol.
Step 3: Note that the reaction produces 2 moles of H2O(g).
Step 4: Calculate the total enthalpy change (ΔH) by multiplying the ΔHf of H2O(g) by the number of moles produced: 2 moles × -241.8 kJ/mol.
Step 5: Perform the multiplication: 2 × -241.8 = -483.6 kJ.
Step 6: Conclude that the enthalpy change (ΔH) for the reaction is -483.6 kJ.

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