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Calculate the standard enthalpy of formation of CH3OH(l) from the following data: 
CH3​OH(I)+3/2​O2​(g)→CO2​(g)+2H2​O(l);  △r​Hθ=−726kJmol−1
C(graphite)​+O2​(g)→CO2​(g);    △c​Hθ=−393kJmol−1
H2​(g)+1/2​O2​(g)→H2​O(l);    △f​Hθ=−286kJmol−1
  • a)
    -339 kJ mol−1
  • b)
    -239 kJ mol−1
  • c)
    -269 kJ mol−1
  • d)
    -209 kJ mol−1
Correct answer is option 'B'. Can you explain this answer?
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To calculate the standard enthalpy of formation of CH3OH(l) from the given data, we need to use the enthalpy of formation of CH3OH(I) and the enthalpy of formation of O2(g).

The balanced chemical equation for the formation of CH3OH(l) is:

CH3OH(I) + 3/2O2(g) → CH3OH(l)

From the equation, we can see that the stoichiometric coefficient of CH3OH(I) is 1, and the stoichiometric coefficient of O2(g) is 3/2.

Using the enthalpy of formation data, we have:

ΔHf(CH3OH(I)) = -238.7 kJ/mol (Given)
ΔHf(O2(g)) = 0 kJ/mol (Standard enthalpy of formation of O2(g) is defined as 0 kJ/mol)

The standard enthalpy of formation of CH3OH(l) can be calculated using the formula:

ΔHf(CH3OH(l)) = Σ(nΔHf(products)) - Σ(nΔHf(reactants))

where n is the stoichiometric coefficient of each component.

In this case, the equation becomes:

ΔHf(CH3OH(l)) = (1)(ΔHf(CH3OH(I))) + (3/2)(ΔHf(O2(g))) - 0

ΔHf(CH3OH(l)) = (1)(-238.7 kJ/mol) + (3/2)(0 kJ/mol) - 0

ΔHf(CH3OH(l)) = -238.7 kJ/mol

Therefore, the standard enthalpy of formation of CH3OH(l) is -238.7 kJ/mol.
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Calculate the standard enthalpy of formation of CH3OH(l) from the following data:CH3OH(I)+3/2O2(g)→CO2(g)+2H2O(l); rHθ=−726kJmol−1C(graphite)+O2(g)→CO2(g); cHθ=−393kJmol−1H2(g)+1/2O2(g)→H2O(l); fHθ=−286kJmol−1a)-339 kJmol−1b)-239 kJmol−1c)-269 kJmol−1d)-209 kJmol−1Correct answer is option 'B'. Can you explain this answer?
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Calculate the standard enthalpy of formation of CH3OH(l) from the following data:CH3OH(I)+3/2O2(g)→CO2(g)+2H2O(l); rHθ=−726kJmol−1C(graphite)+O2(g)→CO2(g); cHθ=−393kJmol−1H2(g)+1/2O2(g)→H2O(l); fHθ=−286kJmol−1a)-339 kJmol−1b)-239 kJmol−1c)-269 kJmol−1d)-209 kJmol−1Correct answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Calculate the standard enthalpy of formation of CH3OH(l) from the following data:CH3OH(I)+3/2O2(g)→CO2(g)+2H2O(l); rHθ=−726kJmol−1C(graphite)+O2(g)→CO2(g); cHθ=−393kJmol−1H2(g)+1/2O2(g)→H2O(l); fHθ=−286kJmol−1a)-339 kJmol−1b)-239 kJmol−1c)-269 kJmol−1d)-209 kJmol−1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the standard enthalpy of formation of CH3OH(l) from the following data:CH3OH(I)+3/2O2(g)→CO2(g)+2H2O(l); rHθ=−726kJmol−1C(graphite)+O2(g)→CO2(g); cHθ=−393kJmol−1H2(g)+1/2O2(g)→H2O(l); fHθ=−286kJmol−1a)-339 kJmol−1b)-239 kJmol−1c)-269 kJmol−1d)-209 kJmol−1Correct answer is option 'B'. Can you explain this answer?.
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