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Ethanol undergoes combustion in oxygen to produce carbon dioxide gas and liquid water. The standard heat of combustion of ethanol, c2h5oh(l), is -1366.8 JK/mol. Given that [co2(g)] = -393.5kg/mol and [h20(l) = -285.8 KJ/ mol, what is the standard enthalpy of formation of ethanol?

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boffeemadrid

Answer:

[tex]-277.6\ \text{kJ/mol}[/tex]

Explanation:

Heat of combustion of [tex]C_2H_5OH(l)[/tex] = -1366.8 kJ/mol

Heat of formation of [tex]CO_2(g)[/tex] = -393.5 kJ/mol

Heat of formation of [tex]H_2O(l)[/tex] = -285.8 kJ/mol

The reaction is

[tex]C_2H_5OH(l)\rightarrow 2CO_2(g)+3H_2O(l)[/tex]

Assuming [tex]x[/tex] is the standard enthalpy of formation of ethanol

[tex]-1366.8=2\times (-393.5)+3\times (-285.8)-x\\\Rightarrow x=2\times (-393.5)+3\times (-285.8)+1366.8\\\Rightarrow x=-277.6\ \text{kJ/mol}[/tex]

The standard enthalpy of formation of ethanol is [tex]-277.6\ \text{kJ/mol}[/tex].

pstnonsonjoku

The standard enthalpy of formation of ethanol is -277.6 kJ/mol.

The equation of the reaction is;

C2H5OH(l) + 3O2(g)⟶ 2CO2(g) + 3H2O(l)

The heat of combustion of each specie is listed below;

C2H5OH(l) = -1366.8 JK/mol

CO2(g) = -393.5 kJ/mol

H2O(l) = -285.8 KJ/ mol

Standard heat of formation of ethanol;

-1366.8 = 2( -393.5) + 3( -285.8 ) - x

where x = heat of formation

x = 2( -393.5) + 3( -285.8 ) + 1366.8

x = -277.6 kJ/mol

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