Standard entropies of X 2 , Y 2 and XY 3 are 60 , 40 and 50   J / K / mol respectively. At what…

So from this question we know that enthalpy for the reaction is $-30 \mathrm{KJ}$

$\frac{1}{2}{\mathrm{X}}_2 + \frac{3}{2}{\mathrm{Y}}_2 \to {\mathrm{XY}}_3$           $∆\mathrm{H} = -30 \mathrm{KJ}$

So we know from the concept of standard entropies is 

$$\begin{gathered} ∆\mathrm{S} = ∆\mathrm{S}\left(\mathrm{P}\right) - ∆\mathrm{S}\left(\mathrm{R}\right) \\ ∆\mathrm{S} = ∆\mathrm{S}\left({\mathrm{XY}}_3\right) -\frac{3}{2} ∆\mathrm{S}\left({\mathrm{Y}}_2\right)-\frac{1}{2}∆\mathrm{S}\left({\mathrm{X}}_2\right) \\ ∆\mathrm{S} = 50 - \frac{3}{2}\times 40-\frac{1}{2}\times 60 \\ ∆\mathrm{S} = -40 {\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1} \end{gathered}$$

So according to the second law of thermodynamics

$∆\mathrm{G} = ∆\mathrm{H}-\mathrm{T}∆\mathrm{S}$

For equilibrium the free Gibb's energy is $∆\mathrm{G} = 0$

$$\begin{gathered} ∆\mathrm{H} = \mathrm{T}∆\mathrm{S} \\ \mathrm{T} = \frac{∆\mathrm{H}}{∆\mathrm{S}} \\ \mathrm{T} = \frac{-30\times {10}^3\mathrm{J}}{-40{\mathrm{JK}}^{-1}} \end{gathered}$$

So the value of temperature is 

$\mathrm{T} = 750 \mathrm{K}$