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Academic session 2026-27, Chapter 5 in the current curriculum.
Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter Thermodynamics. This page offers a step-by-step solution to the specific question from Exercise 1, Question 14:
Calculate the standard enthalpy of formation of CH3OH(l) from the following data:
CH3OH (l) + 3/2 O2(g) → CO2(g) + 2H2O(l) ; ΔrH0 = –726 kJ mol–1
C(g) + O2(g) → CO2(g) ; ΔcH0 = –393 kJ mol–1
H2(g) + 1/2 O2(g) → H2O(l) ; ΔfH0 = –286 kJ mol–1.
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.Calculate the standard enthalpy of formation of CH3OH(l) from the following data:
CH3OH (l) + 3/2 O2(g) → CO2(g) + 2H2O(l) ; ΔrH0 = –726 kJ mol–1
C(g) + O2(g) → CO2(g) ; ΔcH0 = –393 kJ mol–1
H2(g) + 1/2 O2(g) → H2O(l) ; ΔfH0 = –286 kJ mol–1.
The reaction that takes place during the formation of CH3OH(l) can be written as:
C(s) + 2H2O(g) + ½O2(g) → CH3OH(l) (1)
The reaction (1) can be obtained from the given reactions by following the algebraic calculations as:
Equation (ii) + 2 × equation (iii) - equation (i)
ΔfHθ [CH3OH(l)] = ΔcHθ + 2ΔfHθ [H2O(l)] - ΔrHθ
= (-393 kJ mol-1) + 2 (-286 kJ mol-1) - (-726 kJ mol-1)
= (-393 - 572 + 726) kJ mol-1
ΔfHθ[CH3OH(l)]
= -239 kJ mol-1
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 14: Calculate the standard enthalpy of formation of CH3OH(l) from the following data: CH3OH (l) + 3/2....
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