Question

EXPLAIN IN DETAIL , PLEASE( LIKE WHERE DO U GET THAT FROM? OR WHY DO WE HAVE TO DO THAT?)

15. Given the enthalpy changes for the following reactions... Equation 1 6MnO(s) + O2(g) → 2 MnO4(s) Equation 2 2Mn2O3(s) + O2(g) → 4MnO2(s) AH° = -464kJ AH° = -142 kJ AH° = -136 kJ Equations 3 Mno(s) + 1/2O2(g) → MnO2(s) It is possible to determine the enthalpy change for the reaction below: "Final Equation" 4Mn304(s) + O2(g) → 6Mn2O3(s) AH = ? Using this information, calculate AH of the "Final Equation." a. b. -278 kJ -46.3 kJ +142 kJ +1075 kJ -41.8 kJ d. e.

Answer 1

Lets number the reaction as 1, 2, 3, 4 from top to bottom

required reaction should be written in terms of other reaction

This is Hess Law

required reaction can be written as:

reaction 4 = -2 * (reaction 1) -3 * (reaction 2) +12 * (reaction 3)

So, ΔHo rxn for required reaction will be:

ΔHo rxn = -2 * ΔHo rxn(reaction 1) -3 * ΔHo rxn(reaction 2) +12 * ΔHo rxn(reaction 3)

= -2 * (-464.0) -3 * (-142.0) +12 * (-136.0)

= -278 KJ

Answer: -278 KJ