Question

please show neatly with steps thank you.

12) Given the enthalpy changes: 0 Inn A+B → A+DE+F AH=-35 kJ•mol- AH = +20 kJ•mol' All= +15 kJ.mol- F → C+E What is AH for the reaction 2A+B+D 2 F? A) -40 kJ•mol B) -30 kJ•mol' C) -70 kJ•mol D) 0 kJomol"

Answer 1

Reversing the last equation: E + C $\rightarrow$ F The enthalpy is the negative of the enthalpy change of the forward reaction: $\Delta$ H = -15 KJ/mol

Now adding the first two reactions and the equation we formed above: 2A + B + D $\rightarrow$ 2F

This means we need to add the individual enthalpies of the three separate equations: -35 + 20 - 15 = -30 KJ/mol