Find `DeltaH` of the process
`NaOH(s) rarr NaOH(g)`
Given: `Delta_(diss)H^(Theta)of O_(2) = 151 kJ mol^(-1)`
`Delta_(diss)H^(Theta) of H_(2) = 435 kJ mol^(-1)`
`Delta_(diss)H^(Theta) of O-H = 465 kJ mol^(-1)`
`Delta_(diss)H^(Theta) of Na -O = 255 kJ mol^(-1)`
`Delta_(soln)H^(Theta)of NaOH = - 46 kJ mol^(-1)`
`Delta_(f)H^(Theta) of NaOH(s) =- 427 kJ mol^(-1)`
`Delta_("sub")H^(Theta) of Na(s) = 109 kJ mol^(-1)`

Explanation

To find the change in enthalpy (ΔH) for the process of converting solid sodium hydroxide (NaOH) to gaseous sodium hydroxide (NaOH), we will follow these steps:

### Step 1: Write the Reaction
The reaction we are considering is:
\[ \text{NaOH(s)} \rightarrow \text{NaOH(g)} \]

### Step 2: Use Hess's Law
We will apply Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps.
...