Question

please solve this for me step by step on a piece of paper.

Concept Check 6: Consider the decomposition of yellow mercury(ll) oxide: Hg0(s, yellow) – Hg(1) +502(9) Calculate the standard free energy change, AG298 using: a) Standard free energies of formation b) Standard enthalpies of formation and standard entropies. c) Do the results indicate the rxn. to be spontaneous or nonspontaneous?

Answer 1

6. (a) We know that free energy change of a reaction is the free energy change of formation of products - free energy change of formation of reactants

So standard free energy change for given reaction

AG995
=
AG products – AG reactants

=

AG9196) +4GC, - AGP190

=0+(1/2)x0-(-58.5 kJ/mol)=58.5 kJ/mol

(b) Enthalpy change for a reaction is Enthalpy change for formation of products-Enthalpy change for formation of reactants

So Enthalpy change for given reaction=

AH,98 = AH products – AH eactants

=

AH;19(1) + 4H®, - AH100

=0+0-(-90.8 kJ/mol)=90.8 kJ/mol

Entropy change for a reaction is entropy of products-entropy of reactants

So entropy change for given reaction

AS998 = Sproducts – Seactants

=

$491) +-S., - $490

=75.9 J/molK + (1/2)205.2 J/molK-70.3 J/molK

=75.9 JmolK+102.6 J/molK-70.3 J/molK

=108.2 J/molK

Free energy change for given reaction= where T=Temperature=298 K

AG998 = AH - TAS reactants

=90.8 kJ/mol-298 K x 108.2 J/molK/1000 J/kJ

=90.8 kJ/mol-32.2 kJ/mol=58.6 kJ/mol

c) A reaction is spontaneous at a given temperature if its free energy change $\Delta G^$ is negative and it is non-spontaneous if $\Delta G^$ is positive. We can see in parts (a) and (b) that $\Delta G^$ is a positive value, so this reaction is non-spontaneous at 298 K.