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Tamil Nadu Board of Secondary EducationHSC Science Class 11

Calculate the standard heat of formation of propane, if its heat of combustion is −2220.2 kJ mol−1 the heats of formation of COX2X(g) and HX2OX(l) are −393.5 and −285.8 kJ mol−1 respectively. - Chemistry

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Question

Calculate the standard heat of formation of propane, if its heat of combustion is −2220.2 kJ mol−1 the heats of formation of \[\ce{CO2_{(g)}}\] and \[\ce{H2O_{(l)}}\] are −393.5 and −285.8 kJ mol−1 respectively.

Numerical

Solution

\[\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}\]

∆H°C = 2220.2 kJ mol−1 ……………..(1)

\[\ce{C + O2 -> CO2}\]

∆H°f = −393.5 kJ mol−1 ……….(2)

\[\ce{H2 + 1/2O2 -> H2O}\]

∆H°f = −285.8 kJ mol−1 ……………(3)

\[\ce{3C + 4H2 -> C3H8}\]

∆H°c = ?

(2) × (3)

\[\ce{3C + 3O2 -> 3CO2}\]

∆H°f = −1180.5 kJ ………..(4)

(3) × (4)

\[\ce{4H2 + 2O2 -> 4H2O}\]

∆H°f = 1143.2 kJ ……..(5)

(4) + (5) – (1)

\[\ce{3C + 3O2 + 4H2 + 2O2 + 3CO2 + 4H2O -> 3CO2 + 4H2O + C3H8 + 5O2}\]

∆H°f = –1180.5 – 1143.2 – (–2220.2) kJ

\[\ce{3C + 4H2 -> C3H8}\]

∆H°f = –103.5 kJ

Standard heat of formation of propane is ∆H°f (C3H8) = –103.5 kJ

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Chapter 7: Thermodynamics - Evaluation [Page 225]

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Samacheer Kalvi Chemistry - Volume 1 and 2 [English] Class 11 TN Board
Chapter 7 Thermodynamics
Evaluation | Q II. 33. | Page 225
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