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Entropy

The temperature of 2.00 mol Ne(g) is increased from 25◦ C to 200◦ C at constant pressure. Assume the heat capacity of Ne is 20.8 J/K·mol. Calculate the change in the entropy of neon. Assume ideal behavior.

\Delta S=nCln(\frac {T_2}{T_1})

\Delta S=(2 mol)(20.8 J/K*mol)ln(\frac {473}{298})

\Delta S=19.2 J/K

Just want to make sure I'm doing this correctly.

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Reply 1

shengoc

D-Day

\Delta S=nCln(\frac {T_2}{T_1})

\Delta S=(2 mol)(20.8 J/K*mol)ln(\frac {473}{298})

\Delta S=19.2 J/K

Just want to make sure I'm doing this correctly.

The formula is right.

Reply 2

D-Day

Ok, another question.

Estimate the minimum temperature at which magnetite can be reduced to iron by graphite.

Fe_3O_4 (s) + 2 C(s) \to 2 CO_2 (g) + 3 Fe(s)

I have values for delta H, delta G and S. I just don't know what to use them in.

Reply 3

EierVonSatan

D-Day

Ok, another question.

Estimate the minimum temperature at which magnetite can be reduced to iron by graphite.

Fe_3O_4 (s) + 2 C(s) \to 2 CO_2 (g) + 3 Fe(s)

I have values for delta H, delta G and S. I just don't know what to use them in.

dG = dH - TdS, for a reaction to go the change in dG must be negative :smile:

Reply 4

D-Day

Ah. I knew that, I just had a brain fart. So solve for zero :p:

Reply 5

D-Day

Chemistry is certainly not my strong subject. Physics :love:

Reply 6

EierVonSatan

D-Day

Chemistry is certainly not my strong subject. Physics :love:

Pssh, this is practically physics :p:

Reply 7

D-Day

This is not working. :headdesk:

Reply 8

D-Day

Ok now I got an answer of 34,736.8K. I think that may be slightly off :rolleyes:

Reply 9

EierVonSatan

D-Day

Ok now I got an answer of 34,736.8K. I think that may be slightly off :rolleyes:

lol what are your values?

Reply 10

D-Day

Magnetite:
dH=-1,117 kJ/mol
S=146 J/mol K

Graphite:
dH=0
S=6 J/mol K

CO₂:
dH=-393.5 kJ/mol
S=214 J/mol K

Iron:
dH=0
S=27 J/mol K

\Delta H_{reaction}=\sum {\Delta H_{products}}-\sum {\Delta H_{reactants}}

S_{reaction}=\sum {S_{products}}-\sum {S_{reactants}}

So:

\Delta H_{reaction}=2(-393.5)-(-1117)=330 kJ/mol

S_{reaction}=\frac {((214+27)-146)}{1000}=.0095 kJ/mol K

The 1,000 is to convert J to kJ.

Then into

\Delta G=\Delta H - T\Delta S

0=330-T(.0095)

T=34739.84211 K

Tell me what I did wrong :frown:

Reply 11

EierVonSatan

Actually you need 3 irons and 2 carbon dioxides in your entropy calculation...

Reply 12

D-Day

oh **** me. I figured it was something silly.

Reply 13

D-Day

And that's still roughly twice as much as my smallest possible choice.
edit: biggest

Reply 14

EierVonSatan

D-Day

And that's still roughly twice as much as my smallest possible choice.

I get 940K?

Reply 15

D-Day

Answer choices are:

1. 787 K
2. Not possible at any temperature
3. 984 K
4. Possible at any temperature
5. 535 K
6. 670 K
7. 1790 K

Answers should be exact when done correctly, so I don't think 909 is right :s-smilie:

Reply 16

EierVonSatan

well heres my calculation:

enthalpy = 2(-393.5) - (1117) = 330
entropy = 2(0.214) + 3(0.027) - 0.146 - 2(0.006) = 0.351
so dG = 330 - T(0.351) = 0

T = 940K :dontknow: