Hess's Law, enthalpy,entropy,Gibbs free energy change Problem.

Question

For the following unbalanaced reaction at 25°C

PbCO3(s)-->PbO(s)+CO2(g)

Calculate the standard enthalpy change, change in H° for the reaction

Ccalculate the standard entropy change, change in S° for the reaction.

Using the answers from above calculate the standard Gibbs free energy change, change in G° for the reaction.

Is the reaction spontaneous or not at standard conditions

Answer 1

PbCO₃(s)-- --> PbO(s)+CO₂(g) ... balanced equation (metal carbonate => metal oxide + CO₂)

You need to look up the standard ∆H for formation of CO2(g) and for PbO(s), add them together and subtract that from the ∆H formation for PbCO3(s). That will give you ∆Hº for the reaction.

Do the same for standard entropy (Sº) for products and reactants to get Sº reaction

∆G = ∆H - T∆S plug in the values you obtained above and use 298K as the temperature.

If ∆G is negative, the reaction is spontaneous. If positive, it is not spontaneous. If zero, it is at equilibrium.