Question

If enthalpies of formation of \( {C}_2{H}_4(g) \), \( {CO}_2(g) \) and \( {H}_2{O}(l) \) at 25°C and 1 atm pressure are 52, 394 and -286 kJ/mol respectively, the change in enthalpy for combustion of \( {C}_2{H}_4 \) is equal to:

• A. -141.2 kJ/mol
• B. -1412 kJ/mol
• C. +14.2 kJ/mol
• D. +1412 kJ/mol

Hint:

To calculate the change in enthalpy for a reaction, use the enthalpy of formation values for products and reactants, applying Hess's Law. Remember, \( \Delta H_f^\circ \) for elements in their standard state (like \( {O}_2 \)) is zero.

Answer 1

The Correct Option is B

The change in enthalpy (\( \Delta H \)) for the combustion of a substance is given by the following equation based on Hess's Law: \[ \Delta H = \sum \Delta H_f^\circ ({products}) - \sum \Delta H_f^\circ ({reactants}) \] where: - \( \Delta H_f^\circ \) is the enthalpy of formation of each substance, - Products are \( {CO}_2(g) \) and \( {H}_2{O}(l) \), - Reactants are \( {C}_2{H}_4(g) \). The combustion reaction for \( {C}_2{H}_4(g) \) is: \[ {C}_2{H}_4(g) + 3 {O}_2(g) \rightarrow 2 {CO}_2(g) + 2 {H}_2{O}(l) \] Step 1: Writing the enthalpy change equation. \[ \Delta H = \left[ 2 \times \Delta H_f^\circ ({CO}_2) + 2 \times \Delta H_f^\circ ({H}_2{O}) \right] - \left[ \Delta H_f^\circ ({C}_2{H}_4) + 3 \times \Delta H_f^\circ ({O}_2) \right] \] Step 2: Substituting the values. From the given data: - \( \Delta H_f^\circ ({C}_2{H}_4) = 52 \, {kJ/mol} \), - \( \Delta H_f^\circ ({CO}_2) = -394 \, {kJ/mol} \), - \( \Delta H_f^\circ ({H}_2{O}) = -286 \, {kJ/mol} \), - \( \Delta H_f^\circ ({O}_2) = 0 \, {kJ/mol} \) (since it's an element in its standard state). So, we have: \[ \Delta H = \left[ 2 \times (-394) + 2 \times (-286) \right] - \left[ 52 + 3 \times 0 \right] \] \[ \Delta H = \left[ -788 + (-572) \right] - 52 \] \[ \Delta H = -1360 - 52 = -1412 \, {kJ/mol} \] Thus, the change in enthalpy for the combustion of \( {C}_2{H}_4 \) is \( -1412 \, {kJ/mol} \).