Table 1 shows enthalpy change data for calcium bromide, CaBr2\text{CaBr}_2CaBr2.
| Process | Enthalpy change / kJ mol−1\text{kJ mol}^{-1}kJ mol−1 |
|---|---|
| Ca2+(g)→Ca2+(aq)\text{Ca}^{2+}(\text{g}) \rightarrow \text{Ca}^{2+}(\text{aq})Ca2+(g)→Ca2+(aq) | -1560 |
| Br−(g)→Br−(aq)\text{Br}^{-}(\text{g}) \rightarrow \text{Br}^{-}(\text{aq})Br−(g)→Br−(aq) | -335 |
| Ca2+(g)+2Br−(g)→CaBr2(s)\text{Ca}^{2+}(\text{g}) + 2\text{Br}^{-}(\text{g}) \rightarrow \text{CaBr}_2(\text{s})Ca2+(g)+2Br−(g)→CaBr2(s) | -2176 |
Use the data in Table 1 to calculate the molar enthalpy change when anhydrous calcium bromide dissolves in water.
Use your answer to part (a) to deduce how the temperature of the water changes when calcium bromide dissolves.
Explain why the enthalpy of hydration of calcium ions, Ca2+\text{Ca}^{2+}Ca2+, is significantly more negative than the enthalpy of hydration of barium ions, Ba2+\text{Ba}^{2+}Ba2+.
