Consider the following data 

Heat of combustion of H_2(g) = -241 kj .mol^-1. 

Heat of combustion of C_(s) = -393.5 kj .mol^-1. 

Heat of combustion of C₂H₅OH_(l) = -1234 kj .mol^-1.

Heat of formation of C₂H₅OH_(l) is - x kj .mol^-1. Then the value of 'x' is 

 

Calculation of Heat of Formation of Ethanol (C₂H₅OH)

To calculate the heat of formation (ΔH_f) of ethanol (C₂H₅OH), we can use the following approach based on the given heat of combustion values:

Write the combustion reactions for each substance:

• Combustion of H₂(g):

  2H₂(g) + O₂(g) → 2H₂O(l) ΔH = -241 kJ/mol

• Combustion of carbon (C(s)):

  C(s) + O₂(g) → CO₂(g) ΔH = -393.5 kJ/mol

• Combustion of ethanol (C₂H₅OH(l)):

  C₂H₅OH(l) + O₂(g) → 2CO₂(g) + 3H₂O(l) ΔH = -1234 kJ/mol

Write the formation reaction of ethanol:

The formation reaction of ethanol from its elements in their standard states is:

2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l) ΔH_f = -x kJ/mol

Apply Hess's Law:

We can derive the heat of formation of ethanol using the heats of combustion by combining these reactions. The idea is to reverse and add the combustion reactions to obtain the formation reaction of ethanol.

• The combustion of ethanol gives:

  C₂H₅OH(l) + O₂(g) → 2CO₂(g) + 3H₂O(l) ΔH = -1234 kJ/mol

• We need to reverse the combustion of CO₂ and H₂O and add them to get the formation of ethanol. The reactions would be:

  • Reverse the combustion of CO₂:

    2CO₂(g) → 2C(s) + 2O₂(g) ΔH = +787 kJ/mol

  • Reverse the combustion of H₂O:

    3H₂O(l) → 3H₂(g) + ½O₂(g) ΔH = +723 kJ/mol

Summing these gives:

ΔH_{formation of C2H5OH(l)} = -1234 + 787 + 723 = 276 kJ/mol

Thus, the heat of formation of C₂H₅OH(l) is -276 kJ/mol.

The value of x is 276.