How can I calculate Kp given Gibb's free energy? [closed]








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I need to solve a problem but I can't seem to figure out how to move from the given values to Kp. Here's the question:


2SO2(g) + O2(g) ⇌ 2SO3(g)


ΔrH° = −200 kJ.mol-1 and ΔrS° = −187 J.mol-1.K-1 at 25 °C.


Calculate the temperature at which Kp = 1.


I suppose I'm supposed to calculate ΔG first, and then Kc, but then I'm pretty much stuck.




Using the following relations:


$\Delta{G}^⦵ = \Delta{H}^⦵ -T\Delta{S}^⦵$ 


$K\_P = e^{\frac{-\Delta{G}^⦵}{RT}}$


How would you find the temperature at which $K\_P = 1$?



> 
>  $K\_P = 1$ when $\frac{-\Delta{G}^⦵}{RT} = 0$, which is when $\Delta{G}^⦵ = 0$
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Given what $\Delta{G}^⦵$ must equal, how could you find T?



> 
>  $\Delta{G}^⦵ = 0 = \Delta{H}^⦵ -T\Delta{S}^⦵ \therefore T = \frac{\Delta{H}^⦵}{\Delta{S}^⦵}$
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