CsOH - HCl- CsCl - H2O- - H - -13-4 Kcal mol -1 -i CsOH - HF- CsF - H2O- - H - -16-4 Kcal mol -1 -iiCalculate - H for the ionisation of HF in H2O

The correct option is C Δ H= 3.0 Kcal Sol. #160; The ionisation of HF is as shown. HF⟶ H^⊕ + F^⊖ .... Δ H = ? #160; #160;The neutralisation of ...

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Hess' Law

CsOH + HCl⟶ C...

Question

$C s O H + H C l ⟶ C s C l + H_{2} O; Δ H = - 13.4 K c a l m o l^{- 1} . . . . (i)$
$C s O H + H F ⟶ C s F + H_{2} O; Δ H = - 16.4 K c a l m o l^{- 1} . . . . (i i)$
Calculate $Δ H$ for the ionisation of $H F$ in $H_{2} O$

Δ H = - 3.0 K c a l

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Δ H = + 3.0 K c a l

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Δ H = - 29.7.0 K c a l

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None of these

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Solution

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C s O H

13.4 K c a l m o l^{- 1}

C s O H

H F

16.4 K c a l m o l^{- 1}

Δ H^{⊖}

H F

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

C H_{3} C O O H + N a O H ⟶ C H_{3} C O O N a + H_{2} O . . . . Δ H = - 13.2 K c a l

H^{\oplus} + ⊖ O H ⟶ H_{2} O; . . . . Δ H = - 13.7 K c a l

N H_{4} O H

H C l + N H_{4} O H ⟶ N H_{4} C l + H_{2} O; Δ H = - 12.27 K c a l

S + O_{2} ⟶ S O_{2}; Δ H = - 298.2 k J

S O_{2} + 1 / 2 O_{2} ⟶ S O_{3}; Δ H = - 98.7 k J

S O_{3} + H_{2} O ⟶ H_{2} S O_{4}; Δ H = - 130.2 k J

H_{2} + 1 / 2 O_{2} ⟶ H_{2} O; Δ H = - 287.3 k J

H_{2} S O_{4} a t 298 K

S + O_{2} \to S O_{2}; Δ H = - 398.2 k J; S O_{2} + \frac{1}{2} O_{2} \to S O_{3}; Δ H = - 98.7 k J

S O_{2} + H_{2} O \to H_{2} S O_{4}; Δ H = - 130.2 k J; H_{2} + \frac{1}{2} O_{2} \to H_{2} O; Δ H = - 227.3 k J

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