The heat of formation of ethene, \( \mathrm{C_2H_4} \) is 50 \( \mathrm{kJmol^{-1}} \), and that of ethane, \( \mathrm{C_2H_6} \) is -82\( \mathrm{kJmol^{-1...

Assessment: JAMB UTME - Chemistry - 2019 Subject: Chemistry

Question 1 Report

The heat of formation of ethene, \( \mathrm{C_2H_4} \) is 50 \( \mathrm{kJmol^{-1}} \), and that of ethane, \( \mathrm{C_2H_6} \) is -82\( \mathrm{kJmol^{-1}} \). Calculate the heat evolved in the process:

\[ \mathrm{C_2H_4 + H_2 \rightarrow C_2H_6} \]

Answer Details
The heat evolved in a chemical reaction can be calculated by subtracting the heat of formation of the reactants from the heat of formation of the products. In this case, the reactants are ethene (C2H4) and hydrogen (H2), and the product is ethane (C2H6). The heat of formation of ethene is 50 kJ/mol and that of hydrogen is 0 kJ/mol (because hydrogen is a reference element). The heat of formation of ethane is -82 kJ/mol. So, the heat evolved in the reaction is given by: Heat evolved = (Heat of formation of products) - (Heat of formation of reactants) = (-82 kJ/mol) - (50 kJ/mol + 0 kJ/mol) = -82 kJ/mol - 50 kJ/mol = -132 kJ/mol. Therefore, the heat evolved in the process is -132 kJ.

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