What is the enthalpy change for the reaction: 2H2(g) + O2(g) → 2H2O(l)?

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Question: What is the enthalpy change for the reaction: 2H2(g) + O2(g) → 2H2O(l)?

Options:

Correct Answer: -571.6 kJ

Solution:

The enthalpy change for the formation of 2 moles of water is -571.6 kJ.

What is the enthalpy change for the reaction: 2H2(g) + O2(g) → 2H2O(l)?

What is the enthalpy change for the reaction: 2H2(g) + O2(g) → 2H2O(l)?

Step 1: Identify the reaction. We have 2 moles of hydrogen gas (H2) reacting with 1 mole of oxygen gas (O2) to form 2 moles of liquid water (H2O).
Step 2: Understand enthalpy change. Enthalpy change (ΔH) tells us how much energy is released or absorbed during a reaction.
Step 3: Look up the enthalpy of formation for water. The standard enthalpy change for forming 1 mole of water from its elements is -285.8 kJ.
Step 4: Calculate the total enthalpy change for 2 moles of water. Since we are forming 2 moles, we multiply -285.8 kJ by 2.
Step 5: Perform the calculation: -285.8 kJ * 2 = -571.6 kJ.
Step 6: Conclude that the enthalpy change for the reaction is -571.6 kJ, indicating that energy is released.

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