Kaplan free energy question

[joonkimdds]

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The balanced equation below is a nonspontaneous reaction(delta H = 131kJ/mol and delta S = 134J/(mol*K)). Assuming that delta H and delta S don't vary with temperature, at what temperature will the reaction become spontaneous?

C(s) + H2O (l) => CO(g) +H2(g)

A. 1250 celcius
B. 1022 celcius
C. 978 celcius
D. 900 celcius
E. 749 celcius

answer = E.

First of all
G = delta H - T(delta S)
delta H = 131000J/Mol
delta S = 134J/(mol*K)
If G = negative = spontaneous.

so basically if T(delta S) is bigger than delta H, it will make G into negative
and thus it will be spontaneous.
I changed all the answer choices into Kelvin and substituted them to T
and all of them were negative, so shouldn't answer be
all the above?

If i put 749 celcius => 1022.15Kelvin into the equation
G = 131000 - (1022.15*134) = 131000 - 136968.1 = -5968.1

 

[PD3]

The quickest way to do it is to estimate. YOu know that the reaction will change from being nonspontaneous to spontaneous when delta G is equal to zero. So, G equal to zero and you get this:

T=(H\S)

This gives you almost 1000K, then take off 273 and it brings it to about 700. THus E is the answer.

 

[jdmsamurai]

The quickest way to do it is to estimate. YOu know that the reaction will change from being nonspontaneous to spontaneous when delta G is equal to zero. So, G equal to zero and you get this:

T=(H\S)

This gives you almost 1000K, then take off 273 and it brings it to about 700. THus E is the answer.

QFT!!!

 

[joonkimdds]

The quickest way to do it is to estimate. YOu know that the reaction will change from being nonspontaneous to spontaneous when delta G is equal to zero. So, G equal to zero and you get this:

T=(H\S)

This gives you almost 1000K, then take off 273 and it brings it to about 700. THus E is the answer.

but even if it is the closest among the answer choices,
749 celcius makes G = -5968.1 that is far from 0.

 

[PD3]

I guess you could plug in the next highest one and see what it tells you.