In the chemical reaction $X\rightleftharpoons Y$ , the standard Gibb’s energy depends on temperature T(in kelvin) as 

$\Delta G_{\text{reaction }}\left(\text{ in }\mathrm{KJ}.{\mathrm{mole}}^{-1}\right)=120-\frac{3}{8}T$

The major component in the equilibrium mixture at given temperature is

$\Delta G_{\text{reaction }}=\Delta G^0+2.303RT\log \left(Q\right)$

At equilibrium $\Delta G_{\text{reaction }}=0$ , by comparing the given equation at equilibrium 

$120=\frac{3}{8}T$

T = 320 K

Is temperature is less than 320 K reaction quotient (Q) will be greater than equilibrium constant. Which drive the equilibrium backward  $\left(\Delta G_{\text{Bachward }}=-ve\right)$make ‘X’ as major product