A reaction has \(K=1.9 \times 10^{-14}\) at \(25^{\circ} \mathrm{C}\) and $K=9.1 \times 10^{3}$ at \(227^{\circ} \mathrm{C}\) . Predict the signs for $\Delta G^{\circ}, \Delta H^{\circ},\( and \)\Delta S^{\circ}\( for this reaction at \)25^{\circ} \mathrm{C}\( . Assume \)\Delta H^{\circ}\( and \)\Delta S^{\circ}$ do not depend on temperature.

The Van't Hoff equation is given by: \[\frac{d(\ln K)}{dT} = \frac{\Delta H^{\circ}}{RT^2}\] We will first use this equation to solve for ΔH°.

In order to find ΔH°, we will take the derivative of ln(K) with respect to T. To do this, we can use the information of the two equilibrium constants K1 at T1 = 25°C and K2 at T2 = 227°C. First, convert the temperatures from degree Celsius to Kelvin: \(T_1 = 25 + 273.15 = 298.15 K\) \(T_2 = 227 + 273.15 = 500.15 K\) Now, we can write: \[\frac{d(\ln K)}{dT} = \frac{\ln(K_2) - \ln(K_1)}{T_2 - T_1}\] \[\frac{d(\ln K)}{dT} = \frac{\ln(\frac{K_2}{K_1})}{T_2 - T_1}\] Substitute the given values of K1, K2, T1 and T2: \[\frac{d(\ln K)}{dT} = \frac{\ln(\frac{9.1\times 10^{3}}{1.9 \times 10^{-14}})}{500.15 - 298.15} = 0.08379 K^{-1}\] Using the Van't Hoff equation, we can now find ΔH°: \[\Delta H^{\circ} = \frac{d(\ln K)}{dT} \cdot RT^2\] \[\Delta H^{\circ} = 0.08379 \times 8.314 \times (298.15)^2 = 65686 J/mol\]

Now that we have the value of ΔH°, we can determine its sign. Since ΔH° is positive (65686 J/mol), this indicates that the reaction is endothermic. This means that energy is absorbed by the system. Now we can determine the sign of ΔS° by checking which direction the value of K moves as temperature increases. In this case, at 25°C, K is very small (1.9 × 10⁻¹⁴) practically zero. However, at 227°C, K has increased significantly (9.1 × 10³). Therefore, we can deduce that ΔS° is positive, meaning the entropy of the system increases.

Since we know the signs of ΔH° and ΔS°, we can determine the sign of ΔG° at 25°C using the Gibbs-Helmholtz equation: \[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\] We can predict the sign of ΔG° without calculating it (although we do not have the value of ΔS°). Since ΔH° is positive and ΔS° is positive, at 25°C (298.15 K), the ΔG° will be negative because the TΔS° term will be larger than the ΔH° term. Therefore, the reaction is spontaneous at 25°C, and we can conclude that the signs for ΔG°, ΔH°, and ΔS° at 25°C are as follows: ΔG°: Negative ΔH°: Positive ΔS°: Positive