Question

Calculate heat of formation of ethane from following.

& {{C}_{2}}{{H}_{6}}(g)\to 2C(s)+3{{H}_{2}} \\\ & (\therefore \Delta H=31KJ/mol) \\\ \end{aligned}$$

Answer 1

Heat of formation is also called enthalpy of formation or standard heat of formation. It is nothing but the amount of heat absorbed or liberated during the reaction. If some amount of heat absorbed during the reaction then enthalpy of the reaction is positive. If some amount of heat is liberated during the reaction then enthalpy of the reaction will be negative.

Complete step by step answer:
- In the question it is given that ethane gas is undergoing dissociation and forming carbon and hydrogen as the products.
- The enthalpy of the reaction is 31 KJ/mol.
- We can see that the change in enthalpy is positive means some amount of heat is absorbed during the reaction to convert ethane to carbon and hydrogen.
- Means we have to apply some amount of energy to convert ethane to carbon and hydrogen.
- In the question it is mentioned that we have to calculate the heat of formation of ethane, which means two moles of carbon combines with 3 moles of hydrogen and forms ethane as the product.
$2C(s)+3{{H}_{2}}\to {{C}_{2}}{{H}_{6}}(g)$
- The above reaction is the reverse reaction of dissociation of ethane.
- Means in the formation of ethane from carbon and hydrogen the heat should be released.
Therefore the heat of formation of ethane will be ‘$- \ 31 KJ/mol$’.

Note: If we know the heat of formation of a dissociation reaction then we can easily find the heat formation of association reaction. Just we have to change the sign of the enthalpy to get the heat of formation.