SELECT * FROM question_mgmt as q WHERE id=2800 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=47 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 11th chemistry Thermodynamics calculate the standard enthalpy of formation of ch

Question:

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.

Answer:

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

\therefore ΔfHθ[CH3OH(l)]

= -239 kJ mol-1


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