For a reaction to be spontaneous, the Gibbs free energy change ($\Delta G$) must be negative ($\Delta G < 0$). We know that, $\Delta G = \Delta H - T\Delta S$ So, $\Delta H - T\Delta S < 0 \implies T > \frac{\Delta H}{\Delta S}$ Given: $\Delta H = 170 \text{ kJ} = 170 \times 10^3 \text{ J}$ $\Delta S = 170 \text{ J K}^{-1}$ Substituting the values: $T > \frac{170 \times 10^3 \text{ J}}{170 \text{ J K}^{-1}} = 1000 \text{ K}$ Thus, the reaction will be spontaneous at any temperature strictly greater than $1000 \text{ K}$. Among the given options, $1110 \text{ K}$ is the only temperature greater than $1000 \text{ K}$.