Calculate the heat of formation of KCl from the following data :
`{:((i),KOH(aq)+HCl(aq)rarr KCl(aq)+H_(2)O(l)",",,Delta H=-57.3 kJ mol^(-1)),((ii),H_(2)(g)+(1)/(2)O_(2)(g)rarr H_(2)O(l)",",,Delta H=-286.2 kJ mol^(-1)),((iii),(1)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g)+aq rarr HCl(aq)",",,Delta H=-164.4 kJ mol^(-1)),((iv),K(s)+(1)/(2)O_(2)(g)+(1)/(2)H_(2)(g)+aq rarr KOH(aq)",",,Delta H=-487.4 kJ mol^(-1)),((v),KCl(s)+aq rarr KCl(aq)",",,Delta H=+18.4 kJ mol^(-1)):}`

Explanation

Desired equation is :- `K(s)+(1)/(2)Cl_(2)(g)rarr KCl(s), " " Delta H= ?` …..(vi)
In order to get this thermochemical equation we follow the following two steps :
Step 1. Adding Equation (iii) and (iv) and subtracting Equation (v), we have
`K(s)+(1)/(2)Cl_(2)(g)+H_(2)(g)+(1)/(2)O_(2)(g)rarr KCl(s)+HCl(aq)+KOH(aq)-KCl(aq)`
`Delta H=-487.4+(-164.4)-(18.4)=-670.2 kJ mol^(-1)` ....(vii)
Step 2. To cancel out the terms of this equation which do not appear in the required equation (vi). Adding Equation (i) to Equation (vi) and subtract Equation (ii) from their sum. This gives
`K(s)+(1)/(2)Cl_(2)(g)rarr KCl(s) , Delta H=-670.2-57.3 -(-286.2)=-441.3 kJ mol^(-1)`