NH4Cls-H2O- NH4Claq- - H-16-3kJ- In the above reaction- - H represents-

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Enthalpy of Solution

NH4Cls+H2O→ N...

Question

$N H_{4} C l (s) + H_{2} O \to N H_{4} C l (a q);$ $Δ H = 16.3 k J$ .
In the above reaction, $Δ H$ represents:

heat of solution

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Solution

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Similar questions

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 .

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

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

N H_{4} C l (s) ⟶ N H_{4} C l (g)

Δ H

Δ U

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