At a constant temperature, the reaction X_ 2 (g) + 3Y_ 2 (g)...

For the reaction $X_{2}(g) + 3Y_{2}(g) \rightleftharpoons 2XY_{3}(g)$, the standard Gibbs free energy change is $\Delta G^{\circ}_{1} = -z$.

The target reaction is $XY_{3}(g) \rightleftharpoons \dfrac{1}{2}X_{2}(g) + \dfrac{3}{2}Y_{2}(g)$.

This reaction is obtained by reversing the first reaction and multiplying it by $\dfrac{1}{2}$.

When a reaction is reversed and multiplied by a factor $n$, the new standard Gibbs free energy change is given by $\Delta G^{\circ}_{2} = -n \Delta G^{\circ}_{1}$.

Here, $n = \dfrac{1}{2}$.

$\Delta G^{\circ}_{2} = -\dfrac{1}{2} \times (-z) = +\dfrac{z}{2}$.

Answer: $+z/2$