The standard enthalpy of formation (delta H degree_f) of CH_3CH_2CH_2CH_2OH (I)

Question

The standard enthalpy of formation (delta H degree_f) of CH_3CH_2CH_2CH_2OH (I) is -289.2 kJ/mol. The chemical equation associated with this reaction is: CH_3CH_2CH_2CH_2OH (I) + 6 O_2 (g) rightarrow 4 CO_2 (g) + 5 H_2O (g) CH_3CH_2CH_2CH_2OH (I) + 13/2 O_2 (g) rightarrow 4 CO_2 (g) + 5 H_2O (g) 4 C (s, graphite) + 1/2 O_2 (g) + 5 H_2 (g) rightarrow CH_3CH_2CH_2CH_2OH (I) 4 CO_2 (g) + 5 H_2O (g) rightarrow CH_3CH_2CH_2CH_2OH (I) + 6 O_2 (g) CH_3 (g) + CH_2 (g) + CH_2 (g) + CH_2 (g) + 1/2 O_2 (g) + 1/2 H_2 (g) rightarrow CH_3CH_2CH_2CH_2OH (I)

Explanation / Answer

when compound is formed from its elemnts in their standard state the heta of reaction is equal to enthalphy fo fofrmation of compound . Compound CH3CH2CH2CH2OH is formed from carbon which standradr state is C ( graphite) and Oxygen and Hydrogen whose standard state is O2 , H2

Hence the equation is

4 C ( graphite) + 1/2 O2 (g) + 5 H2 (g) --> CH3CH2CH2CH2OH (l)

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