Respuesta :
Answer:
The enthalpy of hydration of copper sulphate is -1486.62 kJ/mol which means 1486.62kJ of energy is absorbed by one mole of copper sulphate during the process of hydration
Explanation:
Step 1: Determine the energy released per mole of [tex]CuSO_{4}[/tex] dissolved
[tex]{CuSO_{4}}_{(s)} -> {CuSO_{4}}_{(aq)}[/tex] (Eq. 1)
[tex]n_{CuSO_{4}} = \frac{m_{CuSO_{4}}}{M_{CuSO_{4}}} [/tex]
[tex]n_{CuSO_{4}} = \frac{16.34}{159.5} [/tex]
[tex]n_{CuSO_{4}} = 0.102 mol [/tex]
If 0.102 moles of [tex]CuSO_{4}[/tex] releases 55.51kJ of energy, 1 mole will release 541.85kJ/mol
[tex]{CuSO_{4}}_{(s)} -> {CuSO_{4}}_{(aq)}[/tex] ΔH = -541.85kJ/mol
Step 2: Determine the energy released per mole of [tex]CuSO_{4}.5H_{2}O[/tex] dissolved
[tex]{CuSO_{4}.5H_{2}O}_{(s)} -> {CuSO_{4}}_{(aq)}+{5H_{2}O}_{(l)}[/tex] (Eq. 2)
[tex]n_{CuSO_{4}.5H_{2}O} = \frac{m_{CuSO_{4}.5H_{2}O}}{M_{CuSO_{4}.5H_{2}O}} [/tex]
[tex]n_{CuSO_{4}.5H_{2}O} = \frac{25.17}{249.5} [/tex]
[tex]n_{CuSO_{4}.5H_{2}O} = 0.101 mol [/tex]
If 0.101 moles of [tex]CuSO_{4}.5H_{2}O[/tex] absorbs 95.31kJ of energy, 1 mole will absorb 944.77kJ/mol
[tex]{CuSO_{4}.5H_{2}O}_{(s)} -> {CuSO_{4}}_{(aq)}+{5H_{2}O}_{(l)}[/tex] ΔH = 944.77kJ/mol
Step 3: Subtracting Eq. 2 from Eq. 1
[tex]{CuSO_{4}}_{(s)} -> {CuSO_{4}}_{(aq)}[/tex] ΔH = -541.85kJ/mol (Eq. 1)
[tex]{CuSO_{4}.5H_{2}O}_{(s)} -> {CuSO_{4}}_{(aq)}+{5H_{2}O}_{(l)}[/tex] ΔH = 944.77kJ/mol (Eq. 2)
[tex]{CuSO_{4}}_{(s)} -{CuSO_{4}.5H_{2}O}_{(s)} -> {CuSO_{4}}_{(aq)} -{CuSO_{4}}_{(aq)}-{5H_{2}O}_{(l)}[/tex] ΔH = -541.85-944.77
[tex]{CuSO_{4}}_{(s)}+{5H_{2}O}_{(l)} -> {CuSO_{4}.5H_{2}O}_{(s)} [/tex] ΔH = -1486.62 kJ/mol
