The equilibrium constant \(K\) for the reaction $$2 \mathrm{Cl}(g) \rightleftharpoons \mathrm{Cl}_{2}(g)$$ was measured as a function of temperature (Kelvin). A graph of \(\ln (K)\) versus \(1 / T\) for this reaction gives a straight line with a slope of \(1.352 \times 10^{4} \mathrm{K}\) and a \(y\) -intercept of \(-14.51 .\) Determine the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. See Exercise 79.

The Van't Hoff equation is given by the following formula: \[\ln K = -\frac{\Delta H^{\circ}}{RT} + \frac{\Delta S^{\circ}}{R}\] Here, \(K\) is the equilibrium constant, \(R\) is the ideal gas constant, which is equal to \(8.314 \textrm{ J K}^{-1}\textrm{mol}^{-1}\), \(T\) is the temperature in Kelvin, ΔH⁰ is the change in enthalpy, and ΔS⁰ is the change in entropy.

We can rewrite the Van't Hoff equation as: \[\ln K = \left(-\frac{\Delta H^{\circ}}{R}\right) \times \frac{1}{T} +\frac{\Delta S^{\circ}}{R}\] Now, comparing it with the linear equation form, \(y = mx + b\), we can identify the following: \(y = \ln K\) \(m = -\frac{\Delta H^{\circ}}{R}\) \(x = \frac{1}{T}\) \(b = \frac{\Delta S^{\circ}}{R}\)

We are given the slope, m, as \(1.352 \times 10^4 \textrm{K}\), and the y-intercept, b, as \(-14.51\). Therefore, we can find ΔH⁰ and ΔS⁰ as follows: First, find ΔH⁰ using the slope: \[m = -\frac{\Delta H^{\circ}}{R} \implies \Delta H^{\circ} = -m \times R\] Using the given slope, \(m = 1.352 \times 10^4 \textrm{K}\) and the gas constant, \(R = 8.314 \textrm{ J K}^{-1}\textrm{mol}^{-1}\), we get: \[\Delta H^{\circ} = -(1.352 \times 10^4 \textrm{K}) \times (8.314 \textrm{ J K}^{-1}\textrm{mol}^{-1}) = -1.124 \times 10^5 \textrm{ J mol}^{-1}\] Next, find ΔS⁰ using the y-intercept: \[b = \frac{\Delta S^{\circ}}{R} \implies \Delta S^{\circ} = b \times R\] Using the given y-intercept, \(b = -14.51\) and the gas constant, \(R = 8.314 \textrm{ J K}^{-1}\textrm{mol}^{-1}\), we get: \[\Delta S^{\circ} = (-14.51) \times (8.314 \textrm{ J K}^{-1}\textrm{mol}^{-1}) = -120.6 \textrm{ J K}^{-1}\textrm{mol}^{-1}\] Thus, the values of ΔH⁰ and ΔS⁰ for this reaction are \(-1.124 \times 10^5 \textrm{ J mol}^{-1}\) and \(-120.6 \textrm{ J K}^{-1}\textrm{mol}^{-1}\), respectively.