Question

If enthalpies of formation for $C _{2} H _{4}(g), CO _{2( g )}$ and $H _{2} O _{(1)}$ at $25^{\circ} C$ and $1\, atm$ pressure are $52,-394$ and - $286\, kJ / mol$ respectively, then enthalpy of combustion of $C _{2} H _{4}( g )$ will be

• A. - 141.2 kJ/mol
• B. #ERROR!
• C. #ERROR!
• D. - 1412 kJ/mol

Answer 1

The Correct Option is D

To determine the enthalpy of combustion of ethene (C_2H_4(g)), we will make use of the enthalpies of formation given in the question:

• Enthalpy of formation of C_2H_4(g): \Delta H_f^\circ = 52 \, \text{kJ/mol}
• Enthalpy of formation of CO_2(g): \Delta H_f^\circ = -394 \, \text{kJ/mol}
• Enthalpy of formation of H_2O(l): \Delta H_f^\circ = -286 \, \text{kJ/mol}

The balanced chemical equation for the combustion of ethene is:

C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(l)

The enthalpy change for the reaction (combustion) can be found using the standard enthalpies of formation:

\Delta H_{combustion} = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})

Substitute the values:

\Delta H_{combustion} = [2 \times (-394) + 2 \times (-286)] - [52]

Calculate the contributions of the products:

• 2 \times (-394) = -788 \, \text{kJ/mol} (from CO_2)
• 2 \times (-286) = -572 \, \text{kJ/mol} (from H_2O)

Add these to get the total enthalpy of the products:

\sum \Delta H_f^\circ(\text{products}) = -788 - 572 = -1360 \, \text{kJ/mol}

Calculate the enthalpy change for the combustion reaction:

\Delta H_{combustion} = -1360 - 52 = -1412 \, \text{kJ/mol}

Therefore, the enthalpy of combustion of ethene (C_2H_4(g)) is -1412 \, \text{kJ/mol}.

Hence, the correct answer is -1412 kJ/mol.