To reverse the given reaction, we need to change the direction of the reaction, which in turn will change the sign of the enthalpy change (ΔH). The given reaction is:
[tex]\[
\ce{C2H4 + 3 O2 -> 2 CO2 + 2 H2O} \quad (\Delta H = -1410 \text{ kJ})
\][/tex]
When we reverse this reaction, the reactants become products and the products become reactants. The reversed reaction is:
[tex]\[
\ce{2 CO2 + 2 H2O -> C2H4 + 3 O2}
\][/tex]
For the enthalpy change associated with the reverse reaction, the sign of ΔH will change. Therefore, the enthalpy change for the reversed reaction becomes:
[tex]\[
\Delta H = +1410 \text{ kJ}
\][/tex]
Thus, the final value for the enthalpy of reaction you use for this reversed intermediate reaction is:
[tex]\[
\boxed{1410 \text{ kJ}}
\][/tex]
So, the correct answer is B. 1410 kJ
