Answer: It can be made spontaneous by decreasing the pressure at 298 K.
For the reaction to be spontaneous, $\Delta G = -ve$
At equilibrium, $\Delta G =0$
$\Delta G = \Delta G^\circ - 2.303$ RT log $(p_{O_2})^{\frac{1}{2}}$
$\Rightarrow 0 = +11210 - 2.303\times (8.31) \times 298$ log $(p_{O_2})^{\frac{1}{2}}$
Solving $(p_{O_2})$ = 0.000116 atm = 0.089 torr
Therefore, at a pressure of less than 0.089 torr, the reaction is spontaneous.