Given- NaCls - aq- Na- aq - Cl- - H - 3-9 kJ Na- g - Cl- g- NaCls - H - -788 kJ Cl- g -aq- Cl- aq - H - -394-1 kJThen the enthalpy of hydration of Na- ions is -

The correct option is C 390.0 kJGiven reactions, NaCl(s) + aq⟶ Na^⊕ (aq) + Cl^⊖ #160; #160; Δ H_1 = 3.9 kJ Na^⊕ (g) + Cl^⊖ (g)⟶ NaCl(s) ...

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Byju's Answer

Standard XII

Chemistry

Order of Reaction

Given: NaCls...

Question

Given:
$N a C l (s) + a q ⟶ N a^{\oplus} (a q) + C l^{⊖}$ $Δ H = 3.9 k J$

$N a^{\oplus} (g) + C l^{⊖} (g) ⟶ N a C l (s) Δ H = - 788 k J$

$C l^{⊖} (g) + a q ⟶ C l^{⊖} (a q) Δ H = - 394.1 k J$

Then the enthalpy of hydration of $N a^{\oplus}$ ions is :

-780.0 kJ

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-960.0 kJ

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-390.0 kJ

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-630.0 kJ

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Solution

The correct option is C -390.0 kJ
Given reactions,

$N a C l (s) + a q ⟶ N a^{\oplus} (a q) + C l^{⊖}$ $Δ H_{1} = 3.9 k J$

$N a^{\oplus} (g) + C l^{⊖} (g) ⟶ N a C l (s) Δ H_{2} = - 788 k J$

$C l^{⊖} (g) + a q ⟶ C l^{⊖} (a q) Δ H_{3} = - 394.1 k J$

For the given reaction,
$N a^{\oplus} (g) + a q ⟶ N a^{\oplus} (a q) Δ H = - 390.0 k J$

$Δ H = Δ H_{1} + Δ H_{2} - Δ H_{3}$
$= 3.9 - 788 - (- 394.1)$
$= 3.9 - 788 + 394.1$
$= - 390.0 k J$

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Similar questions

Q. calculate

Δ G_{f}^{⊖}

of the reaction :

A g^{\oplus} (a q) + C l^{⊖} (a q) \to A g C l (s)

Given :

Δ G^{⊖_{A g C l}} = - 109 k J m o l^{- 1}

Δ G_{(C l^{⊖})}^{⊖} = - 129 k J m o l^{- 1}

Δ G_{(A g^{\oplus)}}^{⊖} = - 77 k J m o l^{- 1}

Q. How much heat is liberated when one mole of gaseous

N a^{\oplus}

combines with one mole of

C l^{⊖}

ion to form solid

N a C l

Use the data given below:

N a (s) + \frac{1}{2} C l_{2} (g) ⟶ N a C l (s)

;

Δ H = - 98.23 k c a l

N a (s) ⟶ N a (g)

;

Δ H = + 25.98 k c a l

N a (g) ⟶ N a^{\oplus} + e^{-}

;

Δ H = + 120.0 k c a l

C l_{2} (g) ⟶ 2 C l (g)

;

Δ H = + 58.02 k c a l

C l^{⊖} (g) ⟶ C l (g) + e^{-}

;

Δ H = + 87.3 k c a l

Calculate the enthalpy change for the process

C C l_{4} (g) \to C (g) + 4 C l (g)

and calculate bond enthalpy of C-Cl in

C C l_{4} (g)

Δ_{v a p} H^{⊖} (C C l_{4}) = 30.5

kJ mol

^{- 1}

Δ_{f} H^{⊖} (C C l_{4}) = - 135.5

kJ mol

^{- 1}

Δ_{a} H^{⊖} (C) = 715.0

kJ mol

^{- 1}

where

Δ_{a} H^{⊖}

is enthalpy of atomisation

Δ_{a} H^{⊖} (C l_{2}) = 242

kJ mol

^{- 1}

Q. Use the given bond enthalpy data to estimate the

Δ H (k J)

for the following reaction.

(C - H = 414 k J, H - C l = 431 k J, C l - C l = 243 k J, C - C l = 331 k J)

C H_{4} (g) + 4 {C l}_{2} (g) ⟶ C {C l}_{4} (g) + 4 H C l (g)

Q. The change in enthalpy for hydration of ( i ) the chloride ion ; ( ii ) the iodide ion are :
Given :
* enthalpy change of solution of NaCl(s)

= - 2

kJ / mol.
* enthalpy change of solution of NaI(s)

= + 2

kJ / mol.
* enthalpy change of hydration of

N a^{+} (g) = - 390

kJ / mol.
* lattice enthalpy of NaCl

= - 772

kJ / mol.
* lattice enthalpy of NaI

= - 699

kJ / mol.

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Given:NaCl(s)+aq⟶Na⊕(aq)+Cl⊖ ΔH=3.9kJNa⊕(g)+Cl⊖(g)⟶NaCl(s)ΔH=−788kJCl⊖(g)+aq⟶Cl⊖(aq)ΔH=−394.1kJThen the enthalpy of hydration of Na⊕ ions is :

The correct option is C -390.0 kJGiven reactions,NaCl(s)+aq⟶Na⊕(aq)+Cl⊖ ΔH1=3.9kJNa⊕(g)+Cl⊖(g)⟶NaCl(s)ΔH2=−788kJCl⊖(g)+aq⟶Cl⊖(aq)ΔH3=−394.1kJFor the given reaction,Na⊕(g)+aq⟶Na⊕(aq)ΔH=−390.0kJΔH=ΔH1+ΔH2−ΔH3=3.9−788−(−394.1)=3.9−788+394.1=−390.0kJ

Given:
$N a C l (s) + a q ⟶ N a^{\oplus} (a q) + C l^{⊖}$ $Δ H = 3.9 k J$

$N a^{\oplus} (g) + C l^{⊖} (g) ⟶ N a C l (s) Δ H = - 788 k J$

$C l^{⊖} (g) + a q ⟶ C l^{⊖} (a q) Δ H = - 394.1 k J$

Then the enthalpy of hydration of $N a^{\oplus}$ ions is :