[" If "H_(2)+1/2O_(2)longrightarrow H_(2)O,,Delta H=-68kcal],[K+H_(2)O+" water "longrightarrow KOH(aq)+1/2H_(2),,Delta H=-48kcal],[KOH+" water "longrightarrow KOH(aq),,Delta H=-14kcal],[" Find the heat of formation of KOH."]

If H_(2)+1//2O_(2)rarr H_(2)O , Delta =-68.39 Kcal K+H_(2)O+ "water" rarr KOH(aq)+1//2H_(2) , Delta H=-48.0 Kcal KOH+"water" rarr KOH(aq)Delta H=-14.0 Kcal the heat of formation of KOH is -

If H_(2)+1//2O_(2)rarr H_(2)O , Delta =-68.39 Kcal K+H_(2)O+ "water" rarr KOH(aq)+1//2H_(2) , Delta H=-48.0 Kcal KOH+"water" rarr KOH(aq)Delta H=-14.0 Kcal the heat of formation of KOH is -

If {:(H_(2)+1//2O_(2)rarrH_(2)O",",,,,DeltaH= -68 kcal),(K+H_(2)OrarrKOH(aq)+1//2H_(2)",",,,,DeltaH= -48 kcal),(KOH+"water"rarrKOH(aq)"," ,,,,DeltaH= -14 kcal):} Find the heat of formation of KOH . Find the heat of formation of KOH .

H_(2) + (1)/(2) O_(2) rarr H_(2)O, Delta = -68.39 kcal K + H_(2)O rarr KOH + (1)/(2) H_(2), Delta H = -48 kcal KOH + Water rarr KOH (aq), Delta H = -14 kcal The heat of formation of KOH in kcal is :

H _(2 ) + 1/2 O _(2) to H _(2) O, Delta H =- 68 kcal K + H _(2) O + aq to KOH _((aq)) + (1)/(2) H _(2) , Delta H =- 48 kcal KOH + aq to KOH _((aq)) , Delta H =- 14 kcal From the above data Delta H _((f)) of KOH in kcal is

H_2+1/2O_2rarrH2O , triangle H =-68.39 kcal (i) K+aqrarrKOH(aq)+1/2H_2 , triangleH=-48 kcal (ii) KOH+aqrarrKOH(aq): triangleH=-14 kcal (iii)T he heat of formation(in kcal) of KOH is: