The equilibrium constant for a certain reaction decreases from 8.84 to $3.25 \times 10^{-2}\( when the temperature increases from \)25^{\circ} \mathrm{C}$ to \(75^{\circ} \mathrm{C}\) . Estimate the temperature where \(K=1.00\) for this reaction. Estimate the value of \(\Delta S^{\circ}\) for this reaction. (Hint: Manipulate the equation in Exercise 85.)

First, we need to find the value of \(\Delta H^{\circ}\) using the given information that K decreases from 8.84 to \(3.25 \times 10^{-2}\) when the temperature increases from 25°C to 75°C. We can rearrange the Van't Hoff equation as \[ \frac{d(\ln K)}{dT} = \frac{1}{RT^2}\Delta H^{\circ} \] Taking the difference of the temperatures as dT and plugging in the values, we can find the value of \(\Delta H^{\circ}\).

Now that we have the value of \(\Delta H^{\circ}\), we can find the temperature where K becomes equal to 1.00 by using the same Van't Hoff equation. We'll first replace "K" with 1.00 and solve the equation for temperature, and then estimate the value of the temperature.

Finally, we will find the value of the standard entropy change (\(\Delta S^{\circ}\)) for the given reaction by using the relationship \[ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \] At equilibrium, \(\Delta G^{\circ}=RT\ln(K)\). With the equilibrium constant equal to 1.00, find the value of \(\Delta S^{\circ}\) using the temperature obtained in step 2 and the calculated value of \(\Delta H^{\circ}\) from step 1. Solution:

Given values: Initial K value: \(K_1 = 8.84\) Final K value: \(K_2 = 3.25 \times 10^{-2}\) Initial temperature T1: 25°C = 298 K Final temperature T2: 75°C = 348 K Implementing the Van't Hoff equation for both temperatures: For T1: \[ \ln(K_1) = -\frac{\Delta H^{\circ}}{R(298)} + \frac{\Delta S^{\circ}}{R} \] For T2: \[ \ln(K_2) = -\frac{\Delta H^{\circ}}{R(348)} + \frac{\Delta S^{\circ}}{R} \] Now, subtract both equations to find the value of \(\Delta H^{\circ}\): \[ -\ln(K_1) + \ln(K_2) = \frac{\Delta H^{\circ}}{R} (\frac{1}{298} - \frac{1}{348}) \] Now we can solve for \(\Delta H^{\circ}\): \[\Delta H^{\circ} = R \left[-\ln(K_1) + \ln(K_2) \right]\left(\frac{1}{298} - \frac{1}{348} \right)^{-1} \] Plugging in the known values, \(R = 8.314 JK^{-1}mol^{-1}\): \[\Delta H^{\circ} = 8.314 \left[-\ln(8.84) + \ln(3.25 \times 10^{-2}) \right]\left(\frac{1}{298} - \frac{1}{348} \right)^{-1} \] \[\Delta H^{\circ} = -44757.6 J/mol\]

Now that we have the value of \(\Delta H^{\circ}\), let's find the temperature for K = 1.00, using the Van't Hoff equation: \[\ln(1) = -\frac{\Delta H^{\circ}}{RT} + \frac{\Delta S^{\circ}}{R} \] Solving for temperature T: \[ T = \frac{\Delta H^{\circ}}{\Delta S^{\circ} - R} \] We cannot find the exact value of the temperature without \(\Delta S^{\circ}\), which will be determined in the next step.

Now let's find the value of \(\Delta S^{\circ}\) for the given reaction using the relationship \[ \Delta S^{\circ} = \frac{\Delta H^{\circ}}{T} + R \] Using the values we calculated in step 1, we have: \[ \Delta S^{\circ} = \frac{-44757.6 J/mol}{T} + 8.314 \] Estimating the temperature from step 2, plug in the temperature T into this equation and solve for \(\Delta S^{\circ}\). Ultimately, the steps listed here provide an approach to estimating the unknown temperature for K = 1.00 and the standard entropy change (\(\Delta S^{\circ}\)) for the given reaction.