Question

Part II-Short Answer 16.) a.) Find the enthalpy change of given reaction, using the enthalpies of formation provided. [unbalanced] C2H6 (e)+O2 (e) > H20 @)+CO2 (e) AH (kI/mol) Compound CH6(e) O2(e) H2O(e) CO2 (1) -83.7 -241.8 -393.5 Is this reaction exothermic or endothermic?

Answer 1

Consider a balanced reaction , 2 C ₂ H _{6 (g)} + 7 O _2(g)   4 CO _2(g) + 6 H₂O (g)

The standard enthalpy change of reaction is given as , r H ⁰ = H ⁰_f ( products ) -  H ⁰_f ( reactants )

H ⁰_f (products) = 4 H ⁰_f CO _2(g) + 6 H ⁰_f H₂O (g)

H ⁰_f (products) =4 (- 393.5 kJ mol ^-1 )  + 6 ( - 241.8 kJ mol ^-1 ) = - 3024.8 kJ mol ^-1

H ⁰_f (reactants) = 2 H ⁰_f C ₂ H _{6 (g)} +7 H ⁰_f O _2(g)

H ⁰_f (reactants) = [ 2 ( - 83.7 ) + 7 ( 0) ] kJ mol ^-1 = - 167.4 kJ mol ^-1

r H ⁰  = - 3024.8 kJ mol ^-1  - (- 167.4 kJ mol ^-1 ) = - 2857 kJ / mol

ANSWER : r H ⁰  = - 2857 kJ / mol

A reaction in which energy is released is called exothermic reaction. In above reaction, 2857 kJ energy is released , hence above reaction is exothermic reaction.