Finish the table by putting one tick (✓) in each row to show the sign of each kind of energy change under standard conditions.
Define, in words, the meaning of enthalpy change of solution.
The following enthalpy changes are provided: - standard enthalpy change of formation, $\Delta H_f^\circ$, for $\text{K}_3\text{PO}_4(\text{s})$: $-2035\ \text{kJ mol}^{-1}$ - standard enthalpy change, $\Delta H^\circ$, for $\text{P}(\text{s}) + 2\text{O}_2(\text{g}) + 3\text{e}^- \rightarrow \text{PO}_4^{3-}(\text{aq})$: $-1284\ \text{kJ mol}^{-1}$ - standard enthalpy change, $\Delta H^\circ$, for $\text{K}(\text{s}) \rightarrow \text{K}^+(\text{aq}) + \text{e}^-$: $-251\ \text{kJ mol}^{-1}$ Determine the standard enthalpy change of solution for potassium phosphate, $\text{K}_3\text{PO}_4(\text{s})$. A labelled energy cycle may help.
Some lattice energy values are listed in the table. $\text{CaBr}_2(s)$ has a lattice energy value of $-2176\ \text{kJ mol}^{-1}$. $\text{KBr}(s)$ has a lattice energy value of $-679\ \text{kJ mol}^{-1}$. Suggest an explanation for why $\Delta H_{lat}^{\circ}\ \text{CaBr}_2$ is more exothermic than $\Delta H_{lat}^{\circ}\ \text{KBr}$.
Write the equation linking $\Delta G^{\circ}$ with $\Delta H^{\circ}$ and $\Delta S^{\circ}$.
Use this equation to explain why $\Delta G^{\circ}$ becomes less positive as temperature rises in this reaction.