For the reaction $X \rightleftharpoons Y$ the sign of standard Gibbs energy change determines which side is thermodynamically favored.

• If $\Delta_r G^\circ < 0$, products are favored, so $Y$ is the major component.
• If $\Delta_r G^\circ > 0$, reactants are favored, so $X$ is the major component.

Given: $\Delta_r G^\circ = 120 - \frac{3}{8}T \quad (\text{kJ mol}^{-1})$

Find the temperature at which $\Delta_r G^\circ = 0$: $120 - \frac{3}{8}T = 0$ $\frac{3}{8}T = 120$ $T = 120 \times \frac{8}{3} = 320\,\text{K}$

So:

• For $T < 320\,\text{K}$, $\Delta_r G^\circ > 0$, hence $X$ is favored.
• For $T > 320\,\text{K}$, $\Delta_r G^\circ < 0$, hence $Y$ is favored.

Check each option:

A: $Y$ if $T=300$ K Since $300 < 320$, $\Delta_r G^\circ > 0$ so $X$ is favored, not $Y$. Hence, A is incorrect.

B: $Y$ if $T=280$ K Since $280 < 320$, $\Delta_r G^\circ > 0$ so $X$ is favored, not $Y$. Hence, B is incorrect.

C: $X$ if $T=350$ K Since $350 > 320$, $\Delta_r G^\circ < 0$ so $Y$ is favored, not $X$. Hence, C is incorrect.

D: $X$ if $T=315$ K Since $315 < 320$, $\Delta_r G^\circ > 0$ so $X$ is favored. Hence, D is correct.

Therefore, the correct option is: $\boxed{D}$