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Number 1

For the gas phase reaction, cyclopentene + I2 = cyclopentadiene + 2HI, we find that

ln Kp = 7.55 - (4817K / T)

(where the symbol K on the right hand side refers to the units Kelvin). What are the changes in entropy and enthalpy?

Possible answers are

ΔSo = 40.05 kJK-1mol-1 and ΔHo = 62.78 kJmol-1

ΔSo = 40.05 JK-1mol-1 and ΔHo = -62.68 kJmol-1

ΔSo = 62.78 JK-1mol-1 and ΔHo = -40.05 kJmol-1

ΔSo = 62.78 JK-1mol-1 and ΔHo = 40.05 kJmol-1

Why are they writing lnKp=7.55-(T/T)? I thought inK=-deltaG/RT.

Nu,ber 2 At 298 K the equilibrium constant for the gas phase dissociation N2O4 ⇋ 2NO2 is 0.14. What are the equilibrium partial pressures of N2O4 at total pressures of 1 atm and 10 atm, where the total pressure is the sum of the partial pressures of NO2 and N2O4?

Answer

a. at total pressures of 1 atm and 10 atm, we have partial pressures of 0.85 atm and 9.3 atm, respectively.

b. at total pressures of 1 atm and 10 atm, we have partial pressures of 0.69 atm and 8.9 atm, respectively.

c. at total pressures of 1 atm and 10 atm, we have partial pressures of 0.1 atm and 1 atm, respectively.

d. at total pressures of 1 atm and 10 atm, we have partial pressures of 0.57 atm and 6.7 atm, respectively.

My attempts Kp= 0.14. PN2O4=0.69 then PNO2= 1- 0.69 = 0.31. Kp= PNO^2/N2O4= (0.31)^2/0.69= 0.14, not sure if that is correct.

Number 3

Consider the isomerization of a homonuclear diatomic molecule. The two isomers, A and B, differ only in the bond length which, for A and B, are 1.5 Å and 1.1 Å, respectively. What is the value of the equilibrium ratio of NB/NA?

Answer

a. 0.54

b. 1.36

c. 0.73

d. 1.00

e. 1.86

My attempt: was leaning towards the answer being one, however now I am not so sure because the partition function of rotational motion is given by the characterisitc temperature of rotation and this value is effected by bond length. Please see attached scan for the working out, but I know that this should be multiplied by e^(eob-eoa)/KbT

For the gas phase reaction, cyclopentene + I2 = cyclopentadiene + 2HI, we find that

ln Kp = 7.55 - (4817K / T)

(where the symbol K on the right hand side refers to the units Kelvin). What are the changes in entropy and enthalpy?

Possible answers are

ΔSo = 40.05 kJK-1mol-1 and ΔHo = 62.78 kJmol-1

ΔSo = 40.05 JK-1mol-1 and ΔHo = -62.68 kJmol-1

ΔSo = 62.78 JK-1mol-1 and ΔHo = -40.05 kJmol-1

ΔSo = 62.78 JK-1mol-1 and ΔHo = 40.05 kJmol-1

Why are they writing lnKp=7.55-(T/T)? I thought inK=-deltaG/RT.

Nu,ber 2 At 298 K the equilibrium constant for the gas phase dissociation N2O4 ⇋ 2NO2 is 0.14. What are the equilibrium partial pressures of N2O4 at total pressures of 1 atm and 10 atm, where the total pressure is the sum of the partial pressures of NO2 and N2O4?

Answer

a. at total pressures of 1 atm and 10 atm, we have partial pressures of 0.85 atm and 9.3 atm, respectively.

b. at total pressures of 1 atm and 10 atm, we have partial pressures of 0.69 atm and 8.9 atm, respectively.

c. at total pressures of 1 atm and 10 atm, we have partial pressures of 0.1 atm and 1 atm, respectively.

d. at total pressures of 1 atm and 10 atm, we have partial pressures of 0.57 atm and 6.7 atm, respectively.

My attempts Kp= 0.14. PN2O4=0.69 then PNO2= 1- 0.69 = 0.31. Kp= PNO^2/N2O4= (0.31)^2/0.69= 0.14, not sure if that is correct.

Number 3

Consider the isomerization of a homonuclear diatomic molecule. The two isomers, A and B, differ only in the bond length which, for A and B, are 1.5 Å and 1.1 Å, respectively. What is the value of the equilibrium ratio of NB/NA?

Answer

a. 0.54

b. 1.36

c. 0.73

d. 1.00

e. 1.86

My attempt: was leaning towards the answer being one, however now I am not so sure because the partition function of rotational motion is given by the characterisitc temperature of rotation and this value is effected by bond length. Please see attached scan for the working out, but I know that this should be multiplied by e^(eob-eoa)/KbT