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:



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> 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?
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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. 


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