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A few physical chem questions.

  • Thread starter Twickel
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  • #1
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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
 

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Answers and Replies

  • #2
148
2
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.
What this rather strange notation means is that if you consider the equilibrium at 481.7 K, then ln Kp will be equal to 7.55 –*10.00 = –2.45;
at 963.4 K it would be 7.55 – 5.00 = 2.55; and so on.

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.
Hang in there! Looks fairly OK to me (cept for typo in PNO2^2)

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
I have not examined nor analyzed your attachment. You are right in assuming that the rotational partition function is the key to this problem.
 
  • #3
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With regards to the rotational partition function, I narrowed it down to (1.5)^2/(1.1)^2 multiplied by e^(eob-eoa)/KbT, but I do not know the values of T so how do I complete this?

Or is that actaully part of the ratio since its the ratio multiplied by e^(eob-eoa)/KbT



You help is really really appreciated.
 
Last edited:
  • #4
148
2
OK then, Twickel, for the first question, write down the expression that connects delta G° with Kp, and the equation that connects delta G° with delta H° and delta S°. Equate the two different expressions for delta G° with one another, rearrange, and compare with the original question.

For the third question, I am coming around to thinking that the answer must be 1.0.
Why? because no temperature or atomic mass is given, and it is not even stated that we are dealing with a gaseous substance.
If no temperature is stated, then you certainly ought to be able to legitimately take the temperature as 0 K, and at that temperature the answer will certainly be 1.0!

But teachers often do not like that sort of argument.

In my teaching days, I would have rejected a question like this. Why? Because it is more likely to confuse and take up the time of a good student who is on top of the subject than a bad one who does not really understand it, and that makes it an inappropriate exercise and an inappropriate assessment or self-assessment.
 
  • #5
10
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Thank you I figured out one and I just maked a low T approx for the second, so it is one.

Q)The bond vibrational frequencies of F2, Li2, KCl and K2 are 274 x 1011 s-1, 105 x 1011 s-1, 84 x 1011 s-1 and 27 x 1011 s-1. Considering gases of the same low density of each species, which gas would have the highest vibrational molar heat capacity at 50K?

I know vib heat capacity always the same and differs only by the point ( char temp) of which this kicks in. I calculated the vib char tempe for all the species and they are F2= 1315K Li2=504K KCL= 403K and K2= 129K therefore at 50K all the heat capacities are the same?
 
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