Calculate the magnitude of free energy in `KJ mol^(-1)` when `1` mole of a an ionic salt MX (s) is dissolved in water at `27^(@)C` . Given
Lattice energy of MX `= 780 KJ mol^(-1)`
Hydration energy of MX `=-775.0 KJ mol^(-1)`
Entropy change of dissolution at `27^(@)C = 40Jmol^(-1)K^(-1)`

Explanation

To calculate the magnitude of free energy (ΔG) when 1 mole of an ionic salt MX is dissolved in water at 27°C, we can follow these steps:

### Step 1: Identify the given values
- Lattice energy (ΔH_lattice) of MX = +780 kJ/mol
- Hydration energy (ΔH_hydration) of MX = -775 kJ/mol
- Entropy change (ΔS) of dissolution = 40 J/mol·K
- Temperature (T) = 27°C = 27 + 273.15 = 300.15 K

### Step 2: Calculate the enthalpy change (ΔH) for the dissolution process
The enthalpy change for the dissolution can be calculated using the formula:
\[ \Delta H_{dissolution} = \Delta H_{lattice} + \Delta H_{hydration} \]

Substituting the values:
\[ \Delta H_{dissolution} = 780 \, \text{kJ/mol} + (-775 \, \text{kJ/mol}) = 780 \, \text{kJ/mol} - 775 \, \text{kJ/mol} = 5 \, \text{kJ/mol} \]

### Step 3: Convert the entropy change from J/mol·K to kJ/mol·K
Since we need the units to be consistent, we convert ΔS from J to kJ:
\[ \Delta S = 40 \, \text{J/mol·K} = 0.040 \, \text{kJ/mol·K} \]

### Step 4: Calculate ΔG using the Gibbs free energy equation
The Gibbs free energy change is given by the equation:
\[ \Delta G = \Delta H - T \Delta S \]

Substituting the values:
\[ \Delta G = 5 \, \text{kJ/mol} - (300.15 \, \text{K} \times 0.040 \, \text{kJ/mol·K}) \]
\[ \Delta G = 5 \, \text{kJ/mol} - 12.06 \, \text{kJ/mol} \]
\[ \Delta G = 5 \, \text{kJ/mol} - 12.06 \, \text{kJ/mol} = -7.06 \, \text{kJ/mol} \]

### Step 5: Calculate the magnitude of ΔG
The magnitude of free energy is the absolute value of ΔG:
\[ |\Delta G| = 7.06 \, \text{kJ/mol} \]

### Final Answer
The magnitude of free energy when 1 mole of ionic salt MX is dissolved in water at 27°C is **7.06 kJ/mol**.

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