Question

PART 1. Which of the following reactions are spontaneous (favorable). Check all that apply.

A. 2Mg(s)+O2(g)--->2MgO(s) delta G=-1137kj/mol

B.NH3(g)+HCl(g)--->NH4Cl(s) delta G=-91.1 kj/mol

C.AgCl(s)--->Ag+(aq)+Cl-(aq) delta G=55.6 kj/mol

D.2H2(g)+O2(g)--->2H2O(g) delta G=456 kj/mol

E.C(s)+H2O(l)--->CO(g)+H2(g) delta G=90.8 kj/mol

F.CH4(g)+2O2(g)--->CO2(g)+2H2O(l) delta G=-820 kj/mol

PART 2. Calculate the standard entropy, ΔS°rxn, of the following reaction at 25.0 °C using the data in this table. The standard enthalpy of the reaction, ΔH°rxn, is –633.1 kJ·mol–1.

3C2H2(g)--->C6H6(l)   ΔS°rxn=____JxK-1xmol-1

Then calculate Gibbs free energy for ΔG°rxn in kjxmol-1

Finally, determine which direction the reaction is spontaneous as written at 25.0 °C and standard pressure.

Forward, reverse, both, or neither?

Answer 1

PART 1. The following reactions are spontaneous (favorable), since delta G has a negative value.

A. 2Mg(s)+O2(g)--->2MgO(s) delta G=-1137kj/mol

B.NH3(g)+HCl(g)--->NH4Cl(s) delta G=-91.1 kj/mol

F.CH4(g)+2O2(g)--->CO2(g)+2H2O(l) delta G=-820 kj/mol

PART 2.

3C₂H₂(g) C₆H₆(l)   

ΔH°_rxn = –633.1 kJ·mol^–1

T = 25 ^oC =298 K

ΔS°_rxn = (–633.1 kJ·mol^–1) / 298 K = 2.12  kJ·mol^–1 K^-1 = 2124 J·mol^–1 K^-1

ΔG°_rxn = ΔH°_rxn - T ΔS°_rxn = -633.1 kJ·mol^–1 - (298 K x 2.12 kJ·mol^–1 K^-1) =  -633.1 kJ·mol^–1 - 631.76  kJ·mol^–1 = -1264.86   kJ·mol^–1

Since ΔG°rxn has a negative value, the reaction is spontaneous in forward direction.