At what temperatures will the following processes be spontaneous? a. \(\Delta H=-18 \mathrm{kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) b. \(\Delta H=+18 \mathrm{kJ}\) and \(\Delta S=+60 . \mathrm{JK}\) c. \(\Delta H=+18 \mathrm{kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) d. \(\Delta H=-18 \mathrm{kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\)

We have the given enthalpy in kJ and the given entropy in J/K. We need to convert them to the same units, so let's convert the enthalpy from kJ to J: \[\Delta H = -18 \mathrm{kJ} \times 1000 = -18000 \mathrm{J}\]

Plug the given values of \(\Delta H\) and \(\Delta S\) into the Gibbs free energy formula and solve for \(T\) when \(\Delta G < 0\): \[\Delta G = \Delta H - T\Delta S\] \[0 > -18000 \mathrm{J} - T(-60 \mathrm{J} / \mathrm{K})\] To find the temperature range for spontaneity, divide both sides by 60 J/K: \[T > \frac{18000}{60} = 300\mathrm{K}\] So, the process will be spontaneous at temperatures greater than 300 K. #b. \(\Delta H=+18 \mathrm{kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\)

We have the given enthalpy in kJ and the given entropy in J/K. We need to convert them to the same units, so let's convert the enthalpy from kJ to J: \[\Delta H = 18 \mathrm{kJ} \times 1000 = 18000 \mathrm{J}\]

Plug the given values of \(\Delta H\) and \(\Delta S\) into the Gibbs free energy formula and solve for \(T\) when \(\Delta G < 0\): \[\Delta G = \Delta H - T\Delta S\] \[0 > 18000 \mathrm{J} - T(60 \mathrm{J} / \mathrm{K})\] To find the temperature range for spontaneity, divide both sides by 60 J/K: \[T > \frac{-18000}{60} = -300\mathrm{K}\] Since the temperature cannot be negative, this process is never spontaneous. #c. \(\Delta H=+18 \mathrm{kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) For this process, both \(\Delta H\) and \(\Delta S\) have positive and negative values, respectively. Since an increase in the temperature would amplify the negative effect of \(\Delta S\) and worsen the positive \(\Delta H\) value, this process will never be spontaneous at any temperature. #d. \(\Delta H=-18 \mathrm{kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\) The given values for \(\Delta H\) and \(\Delta S\) are the same as in part a, but \(\Delta S\) is positive. As \(\Delta H\) is negative and \(\Delta S\) is positive, the process will be spontaneous at all temperatures, as the Gibbs free energy will always be negative.