d) For the process of potassium chlorate decomposition: 2KClO_(3)arrow 2KCl+3O_(2) Delta _(f)H^0_(298)=-89.38kJ/mol and Delta _(r)S_(200)^0=349.51J/moltimes K Calculate Delta G_(298)^0 at 298K and make the conclusion about spontaneity of this reaction.

To calculate the change in Gibbs free energy ($\Delta G_{298}^{0}$) for the decomposition of potassium chlorate at 298K, we can use the following equation:<br /><br />$\Delta G_{298}^{0} = \Delta H_{298}^{0} - T\Delta S_{298}^{0}$<br /><br />where:<br />$\Delta H_{298}^{0}$ is the change in enthalpy at 298K,<br />$T$ is the temperature in Kelvin, and<br />$\Delta S_{298}^{0}$ is the change in entropy at 298K.<br /><br />Given that $\Delta H_{f}^{0}_{298} = -89.38kJ/mol$ and $\Delta S_{298}^{0} = 349.51J/mol\times K$, we can substitute these values into the equation:<br /><br />$\Delta G_{298}^{0} = -89.38kJ/mol - (298K \times 349.51J/mol\times K) / 1000$<br /><br />$\Delta G_{298}^{0} = -89.38kJ/mol - 98.23kJ/mol$<br /><br />$\Delta G_{298}^{0} = -187.61kJ/mol$<br /><br />Since $\Delta G_{298}^{0}$ is negative, the reaction is spontaneous at 298K.