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Calculate the heat of formation of KCl f...

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)):}`

Text Solution

Verified by Experts

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) `
`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) and subtract eqn. (ii) from their sum. This gives
`K(s) + (1)/(2) Cl_(2)(g) rarrKCl(s) , Delta _(f) H = - 670.2 + 57.3 - (- 286.2) = - 441.3 kJ`
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