[hellayeah]

Hello. I need some help with the following problems. If anyone would be so kind as to help me solve them. Thank you!

Mercury has an entropy of vaporization of 92.92 J/K and an enthalpy of vaporization of 58.51 kJ/mol. What is the normal temperature of vaporization of Mercury?
What I intended to do here was to divide the enthalpy over the entropy and get the temperature. Not sure if that's correct though.

For the reaction 2CO₂(g) + O₂(g) --> CO₂(g) at 298K we know that ∆S° = -173 J/Kmol and that ∆H°= .566 kJ/mol. Calculate the ∆G°  at 450K. (Suppose that ∆S° and ∆H° are independent from temperature.
I used Gibb's equation for this.

To form SO₂(g), the value of ∆G°=-71.8kJ. Determine de equilibrium constant for this reaction at 298K.
I'm completely stuck on this one.


Thank you so much to the ones that decide to help me out

[tamim83]

 :delta:G = -RT ln K, K is your equilibrium constant. 

[savy2020]

Mercury has an entropy of vaporization of 92.92 J/K and an enthalpy of vaporization of 58.51 kJ/mol. What is the normal temperature of vaporization of Mercury?
What I intended to do here was to divide the enthalpy over the entropy and get the temperature. Not sure if that's correct though.

That's correct. But do you know why you did that?

For the reaction 2CO₂(g) + O₂(g) --> CO₂(g) at 298K we know that ∆S° = -173 J/Kmol and that ∆H°= .566 kJ/mol. Calculate the ∆G°  at 450K. (Suppose that ∆S° and ∆H° are independent from temperature.
I used Gibb's equation for this.

I don't think "∆G° at 450K" makes sense, because ∆G° is defined as the fee energy at 25°C and 1bar(10⁵Pa)

To form SO₂(g), the value of ∆G°=-71.8kJ. Determine de equilibrium constant for this reaction at 298K.
I'm completely stuck on this one.

Well as above poster says you could use ∆G° = -RT ln K,

[tamim83]

Quote from: hellayeah on May 22, 2010, 09:20:46 AM
For the reaction 2CO2(g) + O2(g) --> CO2(g) at 298K we know that ∆S° = -173 J/Kmol and that ∆H°= .566 kJ/mol. Calculate the ∆G°  at 450K. (Suppose that ∆S° and ∆H° are independent from temperature.
I used Gibb's equation for this.
I don't think "∆G° at 450K" makes sense, because ∆G° is defined as the fee energy at 25°C and 1bar(105Pa)

You can calculate delta G at 298K and then use the Gibbs Helmholtz equation to get delta G at 450K since delta H is independent of temperature.  Of course, you also have to assume that the pressure is constant as well. 

[savy2020]

^Well yes of course we can. But the question must then ask for '∆G at 450K' .
'∆G° at 450K' doesn't mean anything.
It may be just a minor difference, a 'o' at the top 

[tamim83]

Very true .   A typo perhaps?