Question

CO(g) + O2(g) -> CO2(g) (unbalanced) Using the equation and thermodynamic data: AH(rxn) --566.0 kJ (for the balanced equation) Substance sºu/mol*K) O2(g) 205.0 CO(g) 197.7 CO2(g) 213.8 NOTE: Write all answers to four significant figures. a. What is the Assys(J/mol*K) - b. What is the Assurr(J/mol*K) = c. What is the Asuniv(J/mol*K) - d. At what temperature(°C) will the reaction go from spontaneous to non-spontaneous?

Answer 1

a) First, we will balance the reaction, by adding 1/2 to the coefficient of oxygen

CO +1/202 + CO2

Then we will find Entropy Change of the system

ASsys = Sproducts – Spactants = ASys = Sco: - ($%, +sco) ASsys = [213.8 - (0.5 x 205.5 * 197.7)]J/mol. K AS y = -86.4J/mol K

Thus, the answer will be -86.4 J/mol.K (Up to one decimal place)

b)

For the surrounding, we will assume ambient temperature of 298.15 K

Enthalpy of reaction = -566.0 kJ = -566.0 kJ x (1000 J/kJ) = -566 x 10³ J

The entropy change will be,

AH ASsurr = - -566.0 x 103J - = 1898.37J/K 298.15K

Thus, the entropy change for surrounding will be 1898.4 J/K

c) Entropy of the universe will be,

AS univ = ASsys + ASgurr = [(-86.4) + 1898.37]J/K = 1811.97J/K

Thus, the answer will be 1812.0 J/K

d)

We assume the temperature to be T. The reaction will be unfavourable when the total entropy of the universe becomes negative.

The total entropy will be given as

ASuniv = ASsys + Assurr = ASsys –

Thus, for reaction becoming non-spontaneous, we have,

Suniu < 0 → ASsys + ASsurr < 0 = ASsys - I AHO > I. AH - ASsys -566.0 x 10'J T<- -86.4J/K T< 6550.9K

Thus, the answer will be 6550.9 K