H2 - 1-2O2- H2O- - H- - -68 Kcal- K - H2O - aq- KOHaq - 1-2H2- - H- - -48 Kcal- KOH - aq- KOHaq- - H- - -14 KcalFrom the above data- the standard heat of formation of KOH in Kcal is

The correct option is C 68 48 + 14 H_ 2 +1/2O_ 2 ⟶ H_ 2 O nbsp; ...(i) K+H_ 2 O+aq⟶ KOH(aq)+1/2H_ 2 nbsp; ...(ii) KOH+aq⟶ KOH(aq) ...

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Heat of Formation

H2 + 1/2O2⟶ H...

Question

$H_{2} + 1 / 2 O_{2} ⟶ H_{2} O; Δ H^{⊖} = - 68 K c a l$ .
$K + H_{2} O + a q ⟶ K O H (a q) + 1 / 2 H_{2}; Δ H^{⊖} = - 48 K c a l$ .
$K O H + a q ⟶ K O H (a q); Δ H^{⊖} = - 14 K c a l$
From the above data, the standard heat of formation of $K O H i n K c a l$ is

- 68 + 48 - 14

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- 68 - 48 + 14

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68 - 48 + 14

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68 + 48 + 14

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Solution

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Similar questions

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

K O H

K_{(s)} + H_{2} O + a q \to {K O H}_{(a q)} + \frac{1}{2} H_{2}; Δ H = - 48.4 K . C a l

H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O_{(l)}; Δ H = - 68.44 K . C a l

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

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

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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