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What is the standard enthalpy change (ΔH°) for the formation of 1 mole of CO2(g)

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Question: What is the standard enthalpy change (ΔH°) for the formation of 1 mole of CO2(g) from C(s) and O2(g) given the following reactions: C(s) + O2(g) → CO2(g) ΔH° = -393.5 kJ?

Options:

  1. -393.5 kJ
  2. 393.5 kJ
  3. 0 kJ
  4. -273.15 kJ

Correct Answer: -393.5 kJ

Solution: 

The standard enthalpy change for the formation of CO2(g) is directly given as ΔH° = -393.5 kJ.

What is the standard enthalpy change (ΔH°) for the formation of 1 mole of CO2(g)

Practice Questions

Q1
What is the standard enthalpy change (ΔH°) for the formation of 1 mole of CO2(g) from C(s) and O2(g) given the following reactions: C(s) + O2(g) → CO2(g) ΔH° = -393.5 kJ?
  1. -393.5 kJ
  2. 393.5 kJ
  3. 0 kJ
  4. -273.15 kJ

Questions & Step-by-Step Solutions

What is the standard enthalpy change (ΔH°) for the formation of 1 mole of CO2(g) from C(s) and O2(g) given the following reactions: C(s) + O2(g) → CO2(g) ΔH° = -393.5 kJ?
  • Step 1: Identify the reaction given: C(s) + O2(g) → CO2(g).
  • Step 2: Note the enthalpy change (ΔH°) provided for this reaction, which is -393.5 kJ.
  • Step 3: Understand that this value represents the standard enthalpy change for the formation of 1 mole of CO2(g) from its elements in their standard states.
  • Step 4: Conclude that the standard enthalpy change (ΔH°) for the formation of 1 mole of CO2(g) is -393.5 kJ.
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