Calculate - Hof for chloride ion from the following data- 12H2g-12Cl2g- HClg - H-92-4 KJ HClg-NH2Oexcess - H-aq-Cl-aq - H-74-8 KJ - HofH-aq-0-0 KJ-

Given: 12H_2_( g ) + 12Cl_2_( g )⟶HCl_( g ) ΔH = 92.4 KJ ..... ( 1 ) HCl_( g ) + n H_2O_( l )⟶H^+_( aq. ) + Cl^ _( aq. ) ΔH = 74.8 KJ ..... ( ...

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Heat of Formation

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Question

Calculate $Δ H_{f}^{o}$ for chloride ion from the following data:
$\frac{1}{2} H_{2} (g) + \frac{1}{2} C l_{2} (g) \to H C l (g)$ $Δ H = - 92.4$ KJ
$H C l (g) + N H_{2} O$ (excess) $\to H^{+} (a q) + C l^{-} (a q)$ $Δ H = - 74.8$ KJ
$Δ H_{f}^{o} (H^{+} (a q)) = 0.0$ KJ.

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Δ_{f} H^{o} in kJ

\frac{1}{2} H_{2} (g) + \frac{1}{2} C l_{2} (g) \to H C l (g); Δ_{f} H^{o} = - 92.4 kJ H C l (g) + H_{2} O (l) \to H_{3} O^{+} + (a q) + C l^{-} (a q); Δ H^{o} = - 74.8 kJ

Δ_{f} H^{o} of H^{+} (a q) = 0.0 kJ

△ H_{f}^{\circ}

\frac{1}{2} H_{2} (g) + \frac{1}{2} C l_{2} (g) \to H C l (g); △ H_{f}^{\circ} = - 92.4 k J

H C l (g) + n H_{2} O \to H^{+} (a q) + C l^{-} (a q); △ H^{\circ} = - 74.8 k J

△ H_{f}^{\circ} H^{+} (a q) = 0.0 k J

{Δ H}_{f}^{o}

\frac{1}{2} H_{2} (g) + \frac{1}{2} {C l}_{2} (g ⟶ H C l (g); Δ H = - 92.4 k J {m o l}^{- 1}

H C l (g) + n H_{2} O ⟶ H^{+} (a q .) + {C l}^{-} (a q .); Δ H = - 74.8 k J {m o l}^{- 1}

Δ H_{f (H^{+} (a q .)}^{o} = 0.0 k J {m o l}^{- 1}

N H_{3} (g) + H C l (g) \to N H_{4} C l (s)

N H_{3} (g) + a q \to N H_{3} (a q),

Δ H = - 8.4 K . C a l .

H C l (g) + a q \to H C l (a q),

Δ H = - 17.3 K . C a l .

N H_{3} (a q) + H C l (a q) \to N H_{4} C l (a q), Δ H = - 12.5 K . C a l s .

N H_{4} C l (s) + a q \to N H_{4} C l (a q), Δ H = + 3.9 K . C a l .

;

H_{2} O (l) \to H^{+} (a q) + O H^{-} (a q); Δ H = 57.32

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 286.02

O H^{-}

25^{o}

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