The entropy change for the vaporisation of water is 109 `J K^(-1) mol^(-1)`. Calculate the enthalpy change for the vaporisation of water at 373K.

Explanation

To calculate the enthalpy change for the vaporization of water at 373 K, we can use the relationship between enthalpy change (ΔH), temperature (T), and entropy change (ΔS). The formula we will use is derived from the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

At the boiling point of water (373 K), the process is at equilibrium, which means that the change in Gibbs free energy (ΔG) is zero. Therefore, we can set up the equation as follows:

\[ 0 = \Delta H - T \Delta S \]

From this, we can rearrange the equation to solve for the enthalpy change (ΔH):

\[ \Delta H = T \Delta S \]

Now, we can substitute the values we have:

1. **Temperature (T)** = 373 K
2. **Entropy change (ΔS)** = 109 J K⁻¹ mol⁻¹

Now we can calculate the enthalpy change:

\[ \Delta H = 373 \, \text{K} \times 109 \, \text{J K}^{-1} \text{mol}^{-1} \]

Calculating this gives:

\[ \Delta H = 40657 \, \text{J mol}^{-1} \]

To convert this to kilojoules, we divide by 1000:

\[ \Delta H = 40.657 \, \text{kJ mol}^{-1} \]

Thus, the enthalpy change for the vaporization of water at 373 K is approximately **40.66 kJ/mol**.