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Which reaction has an enthalpy change equal to the standard enthalpy change of formation of propane? A 3C(g) + 4H2(g) → C3H8(g) B 3C(g) + 8H(g) → C3H8(g) C 3C(s) + 4H2(g) → C3H8(g) D 3C(s) + 4H2(g) → C3H8(l) - CIE - A-Level Chemistry - Question 11 - 2014 - Paper 1
Question 11
Which reaction has an enthalpy change equal to the standard enthalpy change of formation of propane? A 3C(g) + 4H2(g) → C3H8(g) B 3C(g) + 8H(g) → C3H8(g) C 3C(s) + ... show full transcript
Worked Solution & Example Answer:Which reaction has an enthalpy change equal to the standard enthalpy change of formation of propane? A 3C(g) + 4H2(g) → C3H8(g) B 3C(g) + 8H(g) → C3H8(g) C 3C(s) + 4H2(g) → C3H8(g) D 3C(s) + 4H2(g) → C3H8(l) - CIE - A-Level Chemistry - Question 11 - 2014 - Paper 1
Step 1
Identify the standard enthalpy change of formation of propane
Answer
The standard enthalpy change of formation of a compound is defined as the enthalpy change when one mole of the compound is formed from its elements in their standard states. For propane (C3H8), the standard state for carbon is solid (C(s)) and for hydrogen is gas (H2(g)). Therefore, the formation of propane from its elements is represented as:
Step 2
Analyze the given reactions
Answer
Now, we need to analyze the options given in the question:
- A: 3C(g) + 4H2(g) → C3H8(g) (uses gaseous carbon, not standard state)
- B: 3C(g) + 8H(g) → C3H8(g) (uses gaseous carbon and hydrogen which is not standard state)
- C: 3C(s) + 4H2(g) → C3H8(g) (correct as it uses solid carbon in standard state)
- D: 3C(s) + 4H2(g) → C3H8(l) (correct but represents propane as a liquid, not its standard state)
From the analysis above, option C represents the reaction with the correct enthalpy change as it correctly uses the standard states for both elements.
Step 3