Question

1.      Calculate the standard free energy change at 500 K for the following reaction.

Cu_(s) + H₂O_(g) à CuO_(s) + H_2(g)

	ΔH˚f (kJ/mol)	S˚ (J/mol·K)
Cu(s)	0	33.3
H2O(g)	-241.8	188.7
CuO(s)	-155.2	43.5
H2(g)	0	130.6

2.      When solid ammonium nitrate dissolves in water, the resulting solution becomes cold. Which is true and why?

a.      ΔH˚ is positive and ΔS˚ is positive

b.      ΔH˚ is positive and ΔS˚ is negative

c.       ΔH˚ is negative and ΔS˚ is positive

d.      ΔH˚ is negative and ΔS˚ is negative

e.      ΔG is negative independent of temperature

3.      For a reaction, the value of the equilibrium constant, K, at 298 K is 5.0 x 10⁸. The standard free energy change for this reaction is what?

4.      Given the reaction P_2(g) + 3Cl_2(g)   --> 2PCl_3(g) the standard free energy change is -642.9 kJ at 298 K. If the partial pressure of phosphorus (P₂) gas is 1.5 atm, chlorine gas is 1.6 atm, and phosphorus trichoride gas 0.65 atm, calculate the free energy change for the reaction.

Answer 1

deltaG= deltaH-TdeltaS

deltaG for Cu(s) = 0 -500k(0.0333kJ/molK) = -16.65kJ/mol (keeping the given values of delta H,T,deltaS)

deltaG for H2O(g) = -241.8kJ/mol - 500k*(0.1887J/molK) = -94.59kJ/mol

deltaG for CuO(s) = -155.2kJ/mol -500k*(0.0435kJ/molK) = -176.95kJ/mol

deltaG for H2(g) = 0 - 500K*(0.1306kJ/molK) = -65.3kJ/mol

DeltaGrxn = sum of deltaG for products - sum of deltaG for reactants

= (-176.5+(-65.3)) - (-16.65+(-94.59))

= -241.8 - (-111.24) = -130.56kJ/mol

deltaG rxn = -130.56kJ/mol