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# ΔS ΔH ΔG problems watch

1. Can someone confirm the following answers are right? As they seem wrong to me but I don't see how?

Also, can someone please explain part C.

Thanks!

ΔH = -822
ΔS = -1183.5
ΔG = 351861
2. (Original post by Josh-H)

Can someone confirm the following answers are right? As they seem wrong to me but I don't see how?

Also, can someone please explain part C.

Thanks!
err, you gonna have to provide the data that you were given...
3. (Original post by charco)
err, you gonna have to provide the data that you were given...
Oops sorry :|

4. (Original post by Josh-H)
Oops sorry :|

2Fe + 3/2O2 --> Fe2O3

you don't have to calculate ΔH as you are given the enthalpy of formation of Fe2O3 = -822 kJ

ΔS = entropy products - entropy reactants = 90 - (3/2 x 205) - (2 x 27) = -271.5 J K-1

ΔG298k = ΔH - TΔS = -822 - (298 x -0.2715)

ΔG298k = -741.1 kJ

The last part of your problem is to find the temperature change at which the reaction is just not possible i.e. when ΔG = 0

if ΔG = 0, then ΔH = TΔS

You can take it from there...
5. (Original post by charco)
2Fe + 3/2O2 --> Fe2O3

you don't have to calculate ΔH as you are given the enthalpy of formation of Fe2O3 = -822 kJ

ΔS = entropy products - entropy reactants = 90 - (3/2 x 205) - (2 x 27) = -271.5 J K-1

ΔG298k = ΔH - TΔS = -822 - (298 x -0.2715)

ΔG298k = -741.1 kJ

The last part of your problem is to find the temperature change at which the reaction is just not possible i.e. when ΔG = 0

if ΔG = 0, then ΔH = TΔS

You can take it from there...
Sorry, I don't understand how to work out the temperature range from that?
6. (Original post by Josh-H)
Sorry, I don't understand how to work out the temperature range from that?
Reaction is spontaneous when delta G is just greater than or equal to zero, so work out T with this inequality. Hence it is a range not just one particular temperature.
7. (Original post by shengoc)
Reaction is spontaneous when delta G is just greater than or equal to zero, so work out T with this inequality. Hence it is a range not just one particular temperature.
So the reaction is spontaneous when T is greater than or equal to 3028K ?

*fingers crossed*
8. To work out the total enthalphy change, you have to do the sum of the enthalpy change of the products, minus that of the reactants.
9. (Original post by Josh-H)
Oops sorry :|

so for 2Fe + 3/2O2 --> Fe2O3

-822 - ((2 x 27) + (3 x 205 /2))
10. (Original post by Josh-H)
So the reaction is spontaneous when T is greater than or equal to 3028K ?

*fingers crossed*
ΔH = TΔS

T = ΔH/ΔS = 822/0.2715 = 3028 K

So when the temperature reaches above 3028 K it becomes unfeasible ( i.e. not spontaneous)
11. (Original post by charco)
ΔH = TΔS

T = ΔH/ΔS = 822/0.2715 = 3028 K

So when the temperature reaches above 3028 K it becomes unfeasible ( i.e. not spontaneous)
Ah yeh, silly mistake from me! Thanks!
12. (Original post by charco)
ΔH = TΔS

T = ΔH/ΔS = 822/0.2715 = 3028 K

So when the temperature reaches above 3028 K it becomes unfeasible ( i.e. not spontaneous)
\
Would you mind helping me with question 7 please?

I've got an answer of -726 kj mol^-1 for ΔH but that doesn't seem correct as in part ii that means its spontaneous for both, and that's rare in this sort of question.

a) Calculate ΔH, ΔS and ΔG for the following reaction at 298K:

Fe2O3(s) + 3H2(g) -> 2Fe(s) + 3 H2O(g)

b) will this reaction be spontaneous at i) 20C ii) 500C
c) Calculate the temperature range at which it's spontaneous.
13. (Original post by Josh-H)
\
Would you mind helping me with question 7 please?

a) Calculate ΔH, ΔS and ΔG for the following reaction at 298K:

Fe2O3(s) + 3H2(g) -> 2Fe(s) + 3 H2O(g)

b) will this reaction be spontaneous at i) 20C ii) 500C
c) Calculate the temperature range at which it's spontaneous.

I've got an answer of -726 kj mol^-1 for ΔH but that doesn't seem correct as in part ii that means its spontaneous for both, and that's rare in this sort of question.

ΔH(rxn) = ΔH(products) - ΔH(reactants) = (3 x ΔHf(H2O)) - ΔHf(Fe2O3) = 3 x (-242) -(-822) = -726 + 822 = +96 kJ
14. (Original post by charco)
ΔH(rxn) = ΔH(products) - ΔH(reactants) = (3 x ΔHf(H2O)) - ΔHf(Fe2O3) = 3 x (-242) -(-822) = -726 + 822 = +96 kJ
Right ok, I'll give that a try now. Thanks for the help.
15. (Original post by charco)
ΔH(rxn) = ΔH(products) - ΔH(reactants) = (3 x ΔHf(H2O)) - ΔHf(Fe2O3) = 3 x (-242) -(-822) = -726 + 822 = +96 kJ
ΔG at 20C = 55.6 kjmol-1
ΔG at 500C = -10.7 kjmol-1

Correct?

Spontaneous when T < 695.7K

Just one confusion, I would have thought that a higher temperature would have meant it would have happened spontaneously, not the other way around Can someone explain why its T < 695.7k?
16. (Original post by Josh-H)
ΔG at 20C = 55.6 kjmol-1
ΔG at 500C = -10.7 kjmol-1

Correct?

Spontaneous when T < 695.7K

Just one confusion, I would have thought that a higher temperature would have meant it would have happened spontaneously, not the other way around Can someone explain why its T < 695.7k?
You are thinking of two different situations here, one is thermodynamic and the other one is kinetic. It is true that reactions can and will in most cases occur faster at higher temperature as activation energy is achieved faster. this is kinetic effect!

thermodynamic effect however depends on whether the reaction has positive or negative gibbs energy(ie spontaneous or not). as entropy is small relative to enthalpy in most cases,
negative enthalpy => negative gibbs => spontaneous, but doesn't necessarily means it will be fast, could be seconds, minutes, hours, months, years.

ie change from diamond to graphite at room temperature, the gibbs energy for this reaction is negative, suggesting it is is spontaneous, but actually takes a very long time to happen(thank god! those girls would be very upset if this isn't true).
17. (Original post by Josh-H)

Fe2O3(s) + 3H2(g) -> 2Fe(s) + 3 H2O(g)

ΔG at 20C = 55.6 kjmol-1
ΔG at 500C = -10.7 kjmol-1

Correct?

Spontaneous when T < 695.7K

Just one confusion, I would have thought that a higher temperature would have meant it would have happened spontaneously, not the other way around Can someone explain why its T < 695.7k?

ΔH = +96 kJ
ΔS = S(products) - S(reactants) = 621 - 483 = 138 JK<sup>-1</sup> = 0.138 kJ K<sup>-1</sup>

@20&#186;C: ΔG = ΔH -TΔS = 96 - 293 x (0.138) = 55.6 kJ

@500&#186;C: ΔG = ΔH -TΔS = 96 - 773 x (0.138) = -10.7 kJ

(Note that the units are not per mole as it refers to the stoichiometric quantities as expressed by the coefficients of the balanced equation)

The fact that it is an endothermic reaction means that it is LESS likely to occur as the temperature increases.

This is shown by Gibbs equation.

In terms of the entropy of the universe, this reaction (being endothermic) 'steals' entropy from the surroundings by absorbing energy. This is only compensated for by the increased entropy due to a greater number of particles being formed.

BUT as the temperature increases, the entropy 'stolen' from the surroundings is no longer compensated enough by the increased number of particles. The reaction becomes non-spontaneous.

As Shengoc says, increased heat increases RATE of reaction, but not necessarily the thermodynamic easibility. That depends on the overall (universal) entropy change, factored by the negative of Gibbs free energy.

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Updated: February 15, 2010
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