Chemistry Question, Reactions?

Question

Can you calculate the energy change associated with each of the reactions. Identify the reaction as being endothermic or exothermic.

Reactant-------Product

H2O2 ---------> H20

Answer 1

The enthalpy ΔH of hydrogen peroxide at STP is -136.1 kj/mol. That of water is -241.8. This is assuming all your species are gases.

Subtract product enthalpies from reactant enthalpies:

(-241.8 kj/mol) - (-136.1 kj/mol) = -105.7 kj/mol

The negative sign in front of 105.7 indicates that it is an exothermic reaction. This means that the heat of product exceeds that of reactant. This is energetically favorable in chemistry, or spontaneous. A positive sign would have indicated an endothermic reaction, with energy being required to make the reaction spontaneous.

Answer 2

First, get the equation correct for the decomposition of hydrogen peroxide (not that it will affect the DH).  Then consider the physical states of these.   This is why IT IS SO VERY IMPORTANT TO ADD THE STATE SYMBOLS!

It is also possible that these substances can be considered to be in the liquid state.  It would be even more likely that H2O2 would be in aqueous solution. Only the O2 is a gas, and it's elemental (meaning that it's heat of formation is zero).

H2O2(l) -->  H2O(l) + 1/2 O2(g)

DHrx = SUM(DHf products) - SUM(DHf reactants)

DHrx = -285.9 kJ/mol - (-187.8 kJ/mol) = -98.1 kJ

The reaction is exothermic.

======== Follow up =========

The only way to know if a reaction is spontaneous, or not, is to look at the difference in Gibbs free energy, DG.

DG = DH - TDS

Gibbs free energy also contains a term for the change in entropy of the system.  We would predict that a 1 molecule of a gas or liquid going to 2 molecules of gas, or a molecule of a liquid and a gas would be an increase in entropy.  An increase in entropy, coupled with an exothermic reaction ensures that DG will be negative, indicating a spontaneous reaction.