Why can we use standard entropy when determining temperature at which a reaction becomes spontaneous? In this chemistry problem, $\Delta H$ and $\Delta S$ are given at $298\text{ K}$, and it asks for the temperature at which the reaction becomes spontaneous:


  For the reaction $2\text{SO}_3 \rightarrow 2\text{SO}_2 + \text{O}_2$, $\Delta H_r^\circ = +198 \text{ kJ/mol}$ and $\Delta S_r^\circ = 190 \text{ J / mol}\cdot\text{K}$. At what temperature will the forward reaction will become spontaneous?


I understand to solve this problem; the Gibbs free energy equation,
$$\Delta G = \Delta H - T \Delta S,$$
can be used to determine the temperature by assuming $\Delta H$ and $\Delta S$ are constant and solving for $T$ when $\Delta G = 0$. This gives $T = 1042\text{ K}$.

However, I don't understand why is this assumption valid. Isn't it true that $\Delta S_r$ would change at different temperatures? Why is it that the value for $\Delta S$ for this reaction under standard conditions can be used to calculate $\Delta G$ at $1042 \text{ K}$?
The question implicitly assumes that $\Delta H^\circ$ and $\Delta S^\circ$ are independent of temperature.

The temperature dependences depends on the change in heat capcity ($\Delta C_p$) are given by $\Delta H^\circ(T_2) = \Delta H^\circ(T_1) + \Delta C_p(T_2-T_1)$ and $\Delta S^\circ(T_2) = \Delta S^\circ(T_1) + \Delta C_p \ln(T_2/T_1)$.  (These equations assume $\Delta C$ is constant.) 

So if $T_2-T_1$ or $\Delta C_p$ is small then $\Delta H^\circ$ and $\Delta S^\circ$ will be fairly constant.