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 = - 286.2kJ 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 rarrKCl (aq), Delta H= + 18.4 kJ mol ^(-1)

Calculate the heat of formation of KCl from the following data : (i)KO

1.	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 = - 286.2kJ 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 rarrKCl (aq), Delta H= + 18.4 kJ mol ^(-1)
Answer» Solution :We aim at `: K_(s) + (1)/(2) Cl_(2) (G) rarr KCL (s) , Delta _(f) H = ?` In order to get this thermochemical EQUATION, we follow the following two steps `:` `K(s) + (1)/(2) Cl_(s) (g) + H_(2)(g) + (1)/(2) O_(2)(g) rarr KCl(s) + HCl(aq) + KOH (aq) - KCl(aq) ` `DELTAH = - 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)and subtract eqn. (ii) from theirsum. This gives `K(s) + (1)/(2)Cl_(2)(g) rarrKCl(s) , Delta _(f)H = - 670.2 + 57.3 - (- 286.2) = - 441.3 kJ`