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

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

Options:

  1. -483.6 kJ
  2. 241.8 kJ
  3. 0 kJ
  4. 483.6 kJ

Correct Answer: -483.6 kJ

Solution: 

The standard enthalpy change is 2 * (-241.8 kJ) = -483.6 kJ for the formation of 2 moles of H2O(g).

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

Practice Questions

Q1
What is the standard enthalpy change for the reaction: 2H2(g) + O2(g) → 2H2O(g) if the enthalpy of formation of H2O(g) is -241.8 kJ/mol?
  1. -483.6 kJ
  2. 241.8 kJ
  3. 0 kJ
  4. 483.6 kJ

Questions & Step-by-Step Solutions

What is the standard enthalpy change for the reaction: 2H2(g) + O2(g) → 2H2O(g) if the enthalpy of formation of H2O(g) is -241.8 kJ/mol?
  • Step 1: Identify the reaction: 2H2(g) + O2(g) → 2H2O(g).
  • Step 2: Understand that the enthalpy of formation of H2O(g) is given as -241.8 kJ/mol.
  • Step 3: Since the reaction produces 2 moles of H2O(g), we need to calculate the total enthalpy change for 2 moles.
  • Step 4: Multiply the enthalpy of formation of H2O(g) by the number of moles produced: 2 * (-241.8 kJ).
  • Step 5: Calculate the result: 2 * (-241.8 kJ) = -483.6 kJ.
  • Step 6: Conclude that the standard enthalpy change for the reaction is -483.6 kJ.
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