Question

Use data from Appendix D in the textbook to determine whether the forward reaction is favored by high temperatures or low temperatures. (general chemistry principles and modern applications) PCl3(g)+Cl2(g)⇌PCl5(g)PCl3(g)+Cl2(g)⇌PCl5(g) SO2(g)+2H2S(g)⇌2H2O(g)+3S(s) 2N2(g)+3O2(g)+4HCl(g)⇌4NOCl(g)+2H2O(g)

Answer 1

A)

PCl3(g)+Cl2(g)⇌PCl5(g)

dG < dH - T*dS

dH - T*dS < 0

dH < T*dS

T > dH/dS

PCl3(g)+Cl2(g)⇌PCl5(g)

dH = (-374.9) - (-287.0 + 0) = -87.9 kJ/mol

dS = 364.6 - (311.8 + 223.1) = -170.3 J/K = -0.1703 kJ/K

T > dH/dS

T > 87.9 /0.1703

T > 516.14K , 243.3 °C

B)

SO2(g)+2H2S(g)⇌2H2O(g)+3S(s)

dG < dH - T*dS

dH - T*dS < 0

dH < T*dS

T > dH/dS

SO2(g)+2H2S(g)⇌2H2O(g)+3S(s) .

dH = (2*-241.8 + 3*0) - (-296.8 + 2*-20.6) = -145.6 kJ/mol

dS = (2*188.8 + 3*31.80) - (248.2 + 2*205.8) = -186.8 J/K = -0.1868 kJ/mol

T > dH/dS

t > 145.6 /0.1868 = 779 K

T > 779K, 506 °C

C)

2N2(g)+3O2(g)+4HCl(g)⇌4NOCl(g)+2H2O(g)

dG < dH - T*dS

dH - T*dS < 0

dH < T*dS

T > dH/dS

2N2(g)+3O2(g)+4HCl(g)⇌4NOCl(g)+2H2O(g)

dH = 4*51.71424 + 2*-241.818464 - ( 2*0 + 3*0 + 4*-92.29904) = 92.416

dS = 4*261.58368 + 2*188.715136 - ( 2*191.6 + 3*205.5 + 4 *186.9) = -323.53

T > 92.416*1000/(-323.53) = -285.6489

That is, this will never be spontaneous since it is endothermic and entropy decreases for the system