If H 2 - 1 - 2 O 2 - H 2 O- H-68-09 kcal K- H 2 O-water- KOH aq - 1 - 2 H 2 - H-48-0 kcal KOH-water- KOH aq - H-14-0 kcalWhat is the heat of formation of KOH-

H_2+12O_2 ⇌ H_2O H=68.09 kcal....(i) K+H_2O+water→ KOH(aq)+12H_2 H= 48kcal....(ii) KOH+water → KOH(aq) H= 14 kcal....(iii) We have to fin ...

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Heat of Formation

If H 2 + 1 ...

Question

If $H_{2} + \frac{1}{2} O_{2} ⟶ H_{2} O; △ H = 68.09 k c a l$
$K + H_{2} O + w a t e r ⟶ K O H (a q) + \frac{1}{2} H_{2} : △ H = - 48.0 k c a l$
$K O H + w a t e r ⟶ K O H (a q) : △ H = - 14.0 k c a l$
What is the heat of formation of $K O H$ ?

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H_{2} + \frac{1}{2} O_{2} \to H_{2} O; Δ H = - 68.39 K c a l

K + H_{2} O + w a t e r \to K O H (a q) + \frac{1}{2} H_{2}

Δ H = - 48.0 K c a l

K O H + w a t e r \to K O H (a q) Δ H = - 14.0 K c a l

K O H

H_{2} + \frac{1}{2} O_{2} \to H_{2} O; Δ H = - 68.09 k c a l

K + H_{2} O + w a t e r \to K O H (a q .) + \frac{1}{2} H_{2}; Δ H = - 48.0 k c a l

K O H + w a t e r \to K O H (a q); Δ H = - 14.0 k c a l

K O H

H_{2} + 1 / 2 O_{2} ⟶ H_{2} O; Δ H^{⊖} = - 68 K c a l

K + H_{2} O + a q ⟶ K O H (a q) + 1 / 2 H_{2}; Δ H^{⊖} = - 48 K c a l

K O H + a q ⟶ K O H (a q); Δ H^{⊖} = - 14 K c a l

K O H i n K c a l

H_{2} + \frac{1}{2} O_{2} \to H_{2} O; Δ H = - 68.39

K + H_{2} O \to K O H (a q) + \frac{1}{2} H_{2}; Δ H = - 48.0

K O H + H_{2} O \to K O H (a q); Δ H = - 14.0

K O H

K (s) + H_{2} O (l) + (a q) ⟶ K O H (a q) + \frac{1}{2} H_{2} (g); Δ H = - 48.0 k c a l

H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l); Δ H = - 68.4 k c a l

K O H (s) + (a q) ⟶ K O H (a q); Δ H = - 14.0 k c a l

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