Question

16.34 g of CuSO4 dissolved in water giving out 55.51 kJ and 25.17 g CuSO4•5H2O absorbs 95.31 kJ. From the following reaction cycle and the experimental data above, calculate the enthalpy of hydration of CuSO4.

Answer 1

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 $CuSO_{4}$ dissolved

${CuSO_{4}}_{(s)} -> {CuSO_{4}}_{(aq)}$ (Eq. 1)

$n_{CuSO_{4}} = \frac{m_{CuSO_{4}}}{M_{CuSO_{4}}}$

$n_{CuSO_{4}} = \frac{16.34}{159.5}$

$n_{CuSO_{4}} = 0.102 mol$

If 0.102 moles of $CuSO_{4}$ releases 55.51kJ of energy, 1 mole will release 541.85kJ/mol

${CuSO_{4}}_{(s)} -> {CuSO_{4}}_{(aq)}$ ΔH = -541.85kJ/mol

Step 2: Determine the energy released per mole of $CuSO_{4}.5H_{2}O$ dissolved

${CuSO_{4}.5H_{2}O}_{(s)} -> {CuSO_{4}}_{(aq)}+{5H_{2}O}_{(l)}$ (Eq. 2)

$n_{CuSO_{4}.5H_{2}O} = \frac{m_{CuSO_{4}.5H_{2}O}}{M_{CuSO_{4}.5H_{2}O}}$

$n_{CuSO_{4}.5H_{2}O} = \frac{25.17}{249.5}$

$n_{CuSO_{4}.5H_{2}O} = 0.101 mol$

If 0.101 moles of $CuSO_{4}.5H_{2}O$ absorbs 95.31kJ of energy, 1 mole will absorb 944.77kJ/mol

${CuSO_{4}.5H_{2}O}_{(s)} -> {CuSO_{4}}_{(aq)}+{5H_{2}O}_{(l)}$ ΔH = 944.77kJ/mol

Step 3: Subtracting Eq. 2 from Eq. 1

${CuSO_{4}}_{(s)} -> {CuSO_{4}}_{(aq)}$ ΔH = -541.85kJ/mol (Eq. 1)

${CuSO_{4}.5H_{2}O}_{(s)} -> {CuSO_{4}}_{(aq)}+{5H_{2}O}_{(l)}$ ΔH = 944.77kJ/mol (Eq. 2)

${CuSO_{4}}_{(s)} -{CuSO_{4}.5H_{2}O}_{(s)} -> {CuSO_{4}}_{(aq)} -{CuSO_{4}}_{(aq)}-{5H_{2}O}_{(l)}$ ΔH = -541.85-944.77

${CuSO_{4}}_{(s)}+{5H_{2}O}_{(l)} -> {CuSO_{4}.5H_{2}O}_{(s)}$ ΔH = -1486.62 kJ/mol