Step-by-step Solution

1. Analyze the equilibrium plots to determine the equilibrium constants.

   • At temperature T₁: The first plot shows that at equilibrium, the concentration of the product P is greater than the concentration of the reactant A, i.e., $[P]_{eq, T_1} > [A]_{eq, T_1}$. The equilibrium constant is given by $K_1 = \frac{[P]_{eq, T_1}}{[A]_{eq, T_1}}$. Therefore, $K_1 > 1$.
   • At temperature T₂: The second plot shows that at equilibrium, the concentration of the reactant A is greater than the concentration of the product P, i.e., $[A]_{eq, T_2} > [P]_{eq, T_2}$. The equilibrium constant is given by $K_2 = \frac{[P]_{eq, T_2}}{[A]_{eq, T_2}}$. Therefore, $K_2 < 1$.

2. Determine the sign of the standard enthalpy change, ΔH°. We are given that $T_2 > T_1$. We observe that the equilibrium constant $K$ decreases as the temperature increases (since $K_1 > 1$ and $K_2 < 1$, we have $K_1 > K_2$). According to Le Chatelier's principle, if an increase in temperature shifts the equilibrium to the left (favoring reactants), the forward reaction must be exothermic. Therefore, $\Delta H^\circ < 0$. This is also described by the van't Hoff equation: $\ln \left( \frac{K_2}{K_1} \right) = \frac{\Delta H^\circ}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$ Since $T_2 > T_1$, the term $\left( \frac{1}{T_1} - \frac{1}{T_2} \right)$ is positive. Since $K_2 < K_1$, the term $\ln \left( \frac{K_2}{K_1} \right)$ is negative. For the equation to hold, $\Delta H^\circ$ must be negative.

3. Determine the sign of the standard entropy change, ΔS°. The standard Gibbs free energy change, $\Delta G^\circ$, is related to the equilibrium constant by $\Delta G^\circ = -RT \ln K$. It is also given by $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$.

   • At $T_1$, since $K_1 > 1$, $\ln K_1 > 0$. Thus, $\Delta G^\circ_{T_1} = -RT_1 \ln K_1 < 0$. The reaction is spontaneous.
   • At $T_2$, since $K_2 < 1$, $\ln K_2 < 0$. Thus, $\Delta G^\circ_{T_2} = -RT_2 \ln K_2 > 0$. The reaction is non-spontaneous.

   The reaction changes from spontaneous at a lower temperature ($T_1$) to non-spontaneous at a higher temperature ($T_2$). This happens when both $\Delta H^\circ$ and $\Delta S^\circ$ are negative. The reaction is enthalpy-driven, but opposed by entropy. At low temperatures, the favorable enthalpy term dominates, making $\Delta G^\circ$ negative. At high temperatures, the unfavorable entropy term ($-T\Delta S^\circ$, which is positive if $\Delta S^\circ < 0$) becomes larger and makes $\Delta G^\circ$ positive. Mathematically:

   • $\Delta G^\circ_{T_1} = \Delta H^\circ - T_1\Delta S^\circ < 0 \implies \Delta H^\circ < T_1\Delta S^\circ$
   • $\Delta G^\circ_{T_2} = \Delta H^\circ - T_2\Delta S^\circ > 0 \implies \Delta H^\circ > T_2\Delta S^\circ$ Combining these gives $T_2\Delta S^\circ < \Delta H^\circ < T_1\Delta S^\circ$. This implies $T_2\Delta S^\circ < T_1\Delta S^\circ$. Since $T_2 > T_1$, this inequality can only be true if $\Delta S^\circ < 0$.

4. Analyze the textual condition. The problem states that the "ratio of $lnK$ at ${T_1}$ to $lnK$ at ${T_2}$ is greater than ${{{T_2}} \over {{T_1}}}$", which means $\frac{\ln K_1}{\ln K_2} > \frac{T_2}{T_1}$. From our analysis of the plots, we found $\ln K_1 > 0$ and $\ln K_2 < 0$. This makes the left side of the inequality, $\frac{\ln K_1}{\ln K_2}$, negative. The right side, $\frac{T_2}{T_1}$, is positive since temperatures are positive. A negative number cannot be greater than a positive number. Therefore, this condition is mathematically impossible and likely a typo in the question. The problem should be solved based on the unambiguous information from the plots.

5. Evaluate the options based on the findings from the plots. Our analysis based on the plots concluded that $\Delta H^\circ < 0$ and $\Delta S^\circ < 0$. Also, the reaction is spontaneous at $T_1$, so $\Delta G^\circ < 0$ is a valid statement for the conditions at $T_1$.

   • A: $\Delta {H^ \circ } < 0, \Delta {S^ \circ } < O$ - This is consistent with our findings. Correct.
   • B: $\Delta {G^ \circ } < 0, \Delta {H^ \circ } > 0$ - This is incorrect because we found $\Delta H^\circ < 0$.
   • C: $\Delta {G^ \circ } < 0, \Delta {S^ \circ } < 0$ - This is consistent. We found $\Delta S^\circ < 0$, and $\Delta G^\circ < 0$ is true at temperature $T_1$. Correct.
   • D: $\Delta {G^ \circ } < 0, \Delta {S^ \circ } > 0$ - This is incorrect because we found $\Delta S^\circ < 0$.

Final Answer

Based on the analysis of the provided plots, both statements A and C are correct. The additional textual condition is inconsistent and must be disregarded.