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When complete combustion of ethanol takes place at standard conditions it releases heat of -1370 KJ/mol. Find the standard heat of formation (in kJ/mol) of ethanol.

Given, \Delta H_f ^0 CO_2 = - 394 \ kJ/mol\Delta H_f ^0 H_2O = - 286 \ kJ/mol, :

Option: 1

-1754


Option: 2

1754


Option: 3

384


Option: 4

-276


Answers (1)

best_answer

As we have learnt,

\Delta H_f ^0 O_2 is equal to Zero because O2 is in standard state.

 First write the balanced equation of the reaction,

C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O

\Delta H_c = (2\times \Delta H_f^0CO_2 + 3\times \Delta H_f^0H_2O) - (\Delta H_f^0C_2H_5OH + 3\times \Delta H_f^0O_2)

We have given, \Delta H_c = -1370 \ kJ/mol

Let standard enthalpy of formation of ethanol is x.

-1370 = 2\times(-394) + 3\times (-286) - (x) -3\times(0)

x = -276\ kJ/mol

 

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Gunjita

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