Calculate the carbon carbon double bond energy in ethane from the following reaction, H2C=CH2(g) + H2(g) –> H3C−CH3(g) ΔH = −138 kJ/mol. If Bond enthalpies are: C−C = 348; H−H = 436; C−H = 412 in KJ/mol.

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Class 11 Chemistry Practice Test: Thermodynamics - Enthalpies for Different Types of Reactions — practice the complete quiz, review flashcards, or try a random question.

Enthalpies for Different Types of Reactions are: 1.Heat of Formation : It is “The amount of heat evolved (or) absorbed during the formation of one mole of the compound at constant temperature from the constituent elements in the standard state. Eg : Heat or formation of CO2 C(s) + O2(g) →CO2(g) : ∆H = -393.5KJ   2. Enthalpy of fusion : The enthalpy change that accompanies melting of one mole of a solid substance in standard state is called enthalpy of fusion. H2O(s) →H2O(l) ; ∆H = 6.0KJ /mole    3, Enthalpy of Vapourisation : Amount of heat required to vapourise one mole of a liquid... Show more

Calculate the carbon carbon double bond energy in ethane from the following reaction, H<sub>2</sub>C=CH<sub>2</sub>(g) + H<sub>2</sub>(g) –> H<sub>3</sub>C−CH<sub>3</sub>(g) ΔH = −138 kJ/mol. If Bond enthalpies are: C−C = 348; H−H = 436; C−H = 412 in KJ/mol.






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