Question

At 1 atm and 25 ˚C the molar entropy change for the reaction CO(g) + 1 2 O2(g)  CO2(g) (1) is ΔSo = - 86.46 J K-1. The molar heat capacities at constant pressure at 1 atm and 25 ˚C for CO, O2, and CO2 are, respectively, 29.123, 29.362, and 37.12 J K-1 mol-1. Assume that the heat capacities are independent of temperature and calculate the entropy change for reaction (1) at 100 ˚C and 1 atm.

Answer 1

Equation relating ΔS and T is

ΔS = (q/T)

Where    ΔS = change in entropy   ; T=temperature in K    ; q = Heat absorbed /released in J/mol .

So we need to calculate q

q = Cp Δ T

Where Cp is the molar heat capacities at constant pressure.

So we need to calculate q when temparature changes from 25^oC to 100^oC i.e 298 K to 373 K.

Δ T = 75 K

Cp for CO = 29.123 J/mol K

Cp for O2 = 29.362 J/mol K

Cp for CO2 = 37.12 J/mol K

for CO

q = 29.123 J/mol K * 75 K = 2184.225 J/mol ; ΔS = (2184.225 / 373) = 5.86 J/K

for O2

q = 29.362 J/mol K * 75 K = 2202.150 J/mol ; ΔS = (2202.150 / 373) = 5.90 J/ K

for CO2

q = 37.12 J/mol K * 75 K = 2784.000 J/mol ; ΔS = (2202.150 / 373) = 7.46 J/K

So ΔS at 373 K for the reaction CO(g) + 1/2 O2(g) ------------> CO2(g)

ΔS_rxn = ΔS_products - ΔS_reactants = 7.46 - (5.86 + (5.90/2) ) = -1.35 J/ K