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

Calculated as follows
`= [DeltaH_("sub"(Na))+(1)/(2)DeltaH_(O_(2)(diss))+(1)/(2)DeltaH_(H_(2)(diss))]-[DeltaH_(NaOH(diss))+DeltaH_(OH(diss))]`
`= (109 +(252)/(2)+(435)/(2)) -(255 +465) =- 268 kJ mol^(-1)`
Now, `-268 =- 427 +x`
`rArr x+159 kJ mol^(-1)`