# Problem: (i) Consider this reactionC2H5OH + 3O2 → 2CO2 + 3H2O Change in Heat = -1370 kJ/molWhat is the enthalpy for this reaction reversed?Reaction reversed: 2CO2 + 3H2O → C2H5OH + 3O2Change in Heat = ? kJ/mol(ii) Calculate the enthalpy of the reaction2NO(g) + O2(g) → 2NO2(g)Given the following reactions and enthalpies of formation:½ N2(g) + O2(g) → NO2(g) Change in Heat a = 33.2 KJ½ N2(g) + ½ O2 (g) → NO(g) Change in Heat b = 90.2 KJExpress your answer numerically in kilojoules.Change in Heat = ? kJ/molPlease show all your steps and explain why you did what.

###### FREE Expert Solution

(i) Consider this reaction

C2H5OH + 3O2 → 2CO2 + 3H2O  Change in Heat = -1370 kJ/mol

When we reverse a reaction, we get the same numerical value for the enthalpy change except that we change the sign.

We initially have a negative charge so we make it positive for the reverse reaction.

Reaction reversed: 2CO2 + 3H2O → C2H5OH + 3O2
Change in Heat = + 1370 kJ/mol ###### Problem Details

(i) Consider this reaction
C2H5OH + 3O2 → 2CO2 + 3H2O Change in Heat = -1370 kJ/mol

What is the enthalpy for this reaction reversed?

Reaction reversed: 2CO2 + 3H2O → C2H5OH + 3O2
Change in Heat = ? kJ/mol

(ii) Calculate the enthalpy of the reaction
2NO(g) + O2(g) → 2NO2(g)
Given the following reactions and enthalpies of formation:
½ N2(g) + O2(g) → NO2(g) Change in Heat a = 33.2 KJ
½ N2(g) + ½ O2 (g) → NO(g) Change in Heat b = 90.2 KJ